Sort a list of pairs representing an acyclic, partial automorphism
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
I have this sample data:
let trips = [
{
from: "DEN",
to: "JFK"
},
{
from: "SEA",
to: "DEN"
},
{
from: 'JFK',
to: 'SEA'
},
];
and my origin is 'JFK', I want to sort the list by how I have traveled. So for instance, this should be the end result:
let trips = [
{
from: 'JFK',
to: 'SEA'
},
{
from: "SEA",
to: "DEN"
},
{
from: "DEN",
to: "JFK"
},
];
My solution works but it's not very well written, but I tried!
function sortByLinked(trips, origin = 'JFK') {
let sortedArray = ;
let first = trips.filter(trip => trip.from === origin)[0];
sortedArray.push(first);
for(var i = 0; i < trips.length; i++) {
if(sortedArray[0].to === trips[i].from) {
sortedArray.push(trips[i]);
}
}
for(var i = 0; i < trips.length; i++) {
if(sortedArray[1].to === trips[i].from) {
sortedArray.push(trips[i]);
}
}
return sortedArray;
}
sortByLinked(trips)
javascript algorithm sorting graph
$endgroup$
migrated from stackoverflow.com yesterday
This question came from our site for professional and enthusiast programmers.
|
show 1 more comment
$begingroup$
I have this sample data:
let trips = [
{
from: "DEN",
to: "JFK"
},
{
from: "SEA",
to: "DEN"
},
{
from: 'JFK',
to: 'SEA'
},
];
and my origin is 'JFK', I want to sort the list by how I have traveled. So for instance, this should be the end result:
let trips = [
{
from: 'JFK',
to: 'SEA'
},
{
from: "SEA",
to: "DEN"
},
{
from: "DEN",
to: "JFK"
},
];
My solution works but it's not very well written, but I tried!
function sortByLinked(trips, origin = 'JFK') {
let sortedArray = ;
let first = trips.filter(trip => trip.from === origin)[0];
sortedArray.push(first);
for(var i = 0; i < trips.length; i++) {
if(sortedArray[0].to === trips[i].from) {
sortedArray.push(trips[i]);
}
}
for(var i = 0; i < trips.length; i++) {
if(sortedArray[1].to === trips[i].from) {
sortedArray.push(trips[i]);
}
}
return sortedArray;
}
sortByLinked(trips)
javascript algorithm sorting graph
$endgroup$
migrated from stackoverflow.com yesterday
This question came from our site for professional and enthusiast programmers.
$begingroup$
is it like the origin and final destination always the same? and how many intermediate trips are going to be present?
$endgroup$
– karthick
yesterday
$begingroup$
Isfrom
unique for all elements?
$endgroup$
– Taplar
yesterday
$begingroup$
@karthick yes the origin and final should be the same.
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
@Taplar yes from is unique
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
The task you want to perform is a simple kind of topological sorting.
$endgroup$
– 200_success
yesterday
|
show 1 more comment
$begingroup$
I have this sample data:
let trips = [
{
from: "DEN",
to: "JFK"
},
{
from: "SEA",
to: "DEN"
},
{
from: 'JFK',
to: 'SEA'
},
];
and my origin is 'JFK', I want to sort the list by how I have traveled. So for instance, this should be the end result:
let trips = [
{
from: 'JFK',
to: 'SEA'
},
{
from: "SEA",
to: "DEN"
},
{
from: "DEN",
to: "JFK"
},
];
My solution works but it's not very well written, but I tried!
function sortByLinked(trips, origin = 'JFK') {
let sortedArray = ;
let first = trips.filter(trip => trip.from === origin)[0];
sortedArray.push(first);
for(var i = 0; i < trips.length; i++) {
if(sortedArray[0].to === trips[i].from) {
sortedArray.push(trips[i]);
}
}
for(var i = 0; i < trips.length; i++) {
if(sortedArray[1].to === trips[i].from) {
sortedArray.push(trips[i]);
}
}
return sortedArray;
}
sortByLinked(trips)
javascript algorithm sorting graph
$endgroup$
I have this sample data:
let trips = [
{
from: "DEN",
to: "JFK"
},
{
from: "SEA",
to: "DEN"
},
{
from: 'JFK',
to: 'SEA'
},
];
and my origin is 'JFK', I want to sort the list by how I have traveled. So for instance, this should be the end result:
let trips = [
{
from: 'JFK',
to: 'SEA'
},
{
from: "SEA",
to: "DEN"
},
{
from: "DEN",
to: "JFK"
},
];
My solution works but it's not very well written, but I tried!
function sortByLinked(trips, origin = 'JFK') {
let sortedArray = ;
let first = trips.filter(trip => trip.from === origin)[0];
sortedArray.push(first);
for(var i = 0; i < trips.length; i++) {
if(sortedArray[0].to === trips[i].from) {
sortedArray.push(trips[i]);
}
}
for(var i = 0; i < trips.length; i++) {
if(sortedArray[1].to === trips[i].from) {
sortedArray.push(trips[i]);
}
}
return sortedArray;
}
sortByLinked(trips)
javascript algorithm sorting graph
javascript algorithm sorting graph
edited 2 hours ago
Solomon Ucko
1,1891415
1,1891415
asked yesterday
Shivam BhallaShivam Bhalla
21715
21715
migrated from stackoverflow.com yesterday
This question came from our site for professional and enthusiast programmers.
migrated from stackoverflow.com yesterday
This question came from our site for professional and enthusiast programmers.
$begingroup$
is it like the origin and final destination always the same? and how many intermediate trips are going to be present?
$endgroup$
– karthick
yesterday
$begingroup$
Isfrom
unique for all elements?
$endgroup$
– Taplar
yesterday
$begingroup$
@karthick yes the origin and final should be the same.
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
@Taplar yes from is unique
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
The task you want to perform is a simple kind of topological sorting.
$endgroup$
– 200_success
yesterday
|
show 1 more comment
$begingroup$
is it like the origin and final destination always the same? and how many intermediate trips are going to be present?
$endgroup$
– karthick
yesterday
$begingroup$
Isfrom
unique for all elements?
$endgroup$
– Taplar
yesterday
$begingroup$
@karthick yes the origin and final should be the same.
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
@Taplar yes from is unique
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
The task you want to perform is a simple kind of topological sorting.
$endgroup$
– 200_success
yesterday
$begingroup$
is it like the origin and final destination always the same? and how many intermediate trips are going to be present?
$endgroup$
– karthick
yesterday
$begingroup$
is it like the origin and final destination always the same? and how many intermediate trips are going to be present?
$endgroup$
– karthick
yesterday
$begingroup$
Is
from
unique for all elements?$endgroup$
– Taplar
yesterday
$begingroup$
Is
from
unique for all elements?$endgroup$
– Taplar
yesterday
$begingroup$
@karthick yes the origin and final should be the same.
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
@karthick yes the origin and final should be the same.
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
@Taplar yes from is unique
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
@Taplar yes from is unique
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
The task you want to perform is a simple kind of topological sorting.
$endgroup$
– 200_success
yesterday
$begingroup$
The task you want to perform is a simple kind of topological sorting.
$endgroup$
– 200_success
yesterday
|
show 1 more comment
5 Answers
5
active
oldest
votes
$begingroup$
Code style
Use constants for variables that do not change. Eg
const sortedArray = ;
Don't include the type in the name, Eg
const sortedArray = ;
can be Egconst sorted = ;
The default parameter in this case seams inappropriate as it is likely that
"JFK"
is not an origin in all calls to this function. If no origin is given one could assume that the first item in trips contains the origin.function sortByLinked(trips, origin = trips[0].from) {
which will throw an error if trips is empty so the function should not be called with an empty trips array if you don't pass thefrom
parameter.
However in this example best to leave the default as
undefined
if not passed as that will return an empty array which is more fitting the input parameters.
The name
sort
is inappropriate as in JS it implies that the array be sorted in place, that all items be sorted (may not be possible).You have declared
i
two times. As avar
you should put the declaration at the top of the function and not in thefor
loop.Rather than use
Array.filter
you can useArray.find
. It will find the first instance.Using
for...of
rather thanfor(;;)
reduces the code complexity.No point continuing the search inside the for loops when you have found a match. Use the
break
token to stop a loop earlyPut a space between
if
and(
Don't forget to add the
;
where appropriate. It is missing from the callsortByLinked(trips)
You call the origin
origin
andfrom
this can get confusing. Keep the naming unambiguous. As the trip items usefrom
then that would be the better name for the second input argument.
Using the above points to modify your code we get
function tripFrom(trips, from) {
const sorted = ;
const first = trips.find(trip => trip.from === from);
sorted.push(first);
for (const trip of trips) {
if (first.to === trip.from) {
sorted.push(trip);
break;
}
}
for (const trip of trips) {
if (sorted[sorted.length - 1].to === trip.from) {
sorted.push(trips);
break;
}
}
return sorted;
}
sortByLinked(trips, "JFK");
This is still not a good solution. Its not at all DRY (don't repeat yourself) and is hard coded to a single use case.
Improving the function.
It can all be done within a single loop and work for any length array.
To create the function we must add some constraints on the array trips
and what to do when we encounter any problems.
- That the array
trips
contains objects that each have the propertyfrom
andto
that are correctly formatted static strings. The resulting array is erroneous or indeterminate if not so. - That the array does not contain circular trips shorter than the array length.
- That a complete trip length is no longer than the array, or when a matching
trip.to
can not be found. The returned array can be 0 totrips.length
in size. - Locations are case sensitive.
- If there is more than one matching
trip.from
it is assumed that the first match in trips is the correct one. (It would be interesting to extract the longest possible trip from? or the shortest trip that returns to the origin?)
Example
function tripFrom(trips, from) {
const result = ;
while (result.length < trips.length) {
const trip = trips.find(trip => trip.from === from);
if (!trip) { break }
from = trip.to;
result.push(trip);
}
return result;
}
tripFrom(trips, "JFK");
Or if it is known that the trip is the same length as the input array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
return res;
}
tripFrom(trips, "JFK");
It is unclear if you want the array sorted in place. If that is a requirement then the above version can be modified to do that by simply copying the results array res
to the trips
array. You can empty an array by setting its length to zero. The spread ...
operator in this case spreads the array items over the functions arguments trips.push(...res)
thus pushing all the items to the array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
trips.length = 0;
trips.push(...res);
return trips;
}
tripFrom(trips, "JFK");
$endgroup$
$begingroup$
It's common practice to avoid using braces in inline if-statements (your first example). Also, you make a comment about a missing semicolon and I'd like to point out, just for the sake of having options, that completely getting rid of semicolons is a growing practice since it generally looks nicer.
$endgroup$
– Adam
10 hours ago
$begingroup$
@Adam It may be a common practice to not delimit single line statement blocks, but that does not make it a good practice.Consistency is the most important part of good style. If you use semicolon then use them, not just sometimes. Generally I advice that if you are not going to use them then you should know every case where automatic insertion will not happen but is required to mark the end of line. Eg "(()=>{});n()" is not the same as "(()=>{})n()" (as string to show new line location)
$endgroup$
– Blindman67
8 hours ago
$begingroup$
Right. I'm not saying he shouldn't use a semicolon in that specific spot. I'm just making it known that completely getting rid of semicolon delimiters is a very common practice nowadays.
$endgroup$
– Adam
8 hours ago
add a comment |
$begingroup$
It would be good if your function could work for more than 3 trips.
And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map
first, so that you can access a trip by its from
property in constant time:
function sortByLinked(trips, origin = "JFK") {
const map = new Map(trips.map(trip => [trip.from, trip]));
const result = ;
for (let trip; trip = map.get(origin); origin = trip.to) {
result.push(trip);
map.delete(origin);
}
return result;
}
const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];
const sorted = sortByLinked(trips, "JFK");
console.log(sorted);
$endgroup$
$begingroup$
Interesting. It doesn't work with cycles, though, does it? e.g.[{from: "JFK",to: "XXX"},{from: "XXX",to: "YYY"},{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}]
$endgroup$
– Eric Duminil
12 hours ago
$begingroup$
Indeed, it is not expecting multiple trip objects with the samefrom
property. That would need a more comprehensive tree traversal.
$endgroup$
– trincot
12 hours ago
add a comment |
$begingroup$
Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for
loop like you're doing there are other ways to find the next trip like using Array.map()
or Array.filter()
.
Here's one way to sort it in place and can handle any number of trips greater than 1.
function sortByLinked(trips, origin = 'JFK') {
// this will be useful
function swap(array, index1, index2){
let temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
// find first one
let first = trips.filter(trip => trip.from === origin)[0];
// put him in the front of the list
swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);
// sort it in place
for(let i=1; i<trips.length; i++){
swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));
}
}
New contributor
$endgroup$
$begingroup$
Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
$endgroup$
– Sᴀᴍ Onᴇᴌᴀ
yesterday
add a comment |
$begingroup$
You could also solve the problem by introducing an object mo
(as a makeshift associative array) like shown in the following example
const trav=[{from:"bos",to:"sfo"},{from:"jer",to:"jfk"},{from:"zur",to:"brm"},{from:"haj",to:"cdg"},{from:"pma",to:"mlg"},{from:"sfo",to:"zur"},
{from:"cdg",to:"hav"},{from:"jfk",to:"haj"},{from:"man",to:"cdg"},
{from:"mlg",to:"jer"},{from:"brm",to:"pma"},{from:"hav",to:"bos"}],
mo={},srt=;
var fr='jfk',v,n=20;
trav.forEach(v=>mo[v.from]=v);
while (n--&&(v=mo[fr])){srt.push(v);fr=v.to;}
console.log(JSON.stringify(srt));
The n
is a "safety switch` that gets you out when the sequence turns out to be circular (as is the case in my example).
New contributor
$endgroup$
2
$begingroup$
Welcome to Code Reviews, cars10m. Providing an alternative implementation is interesting and handy on some level, but it doesn't constitute a code review. Was there something so wrong with the original that you needed to rewrite it instead of tweaking it in smaller ways?
$endgroup$
– chicks
13 hours ago
add a comment |
$begingroup$
This problem is related to graph theory, and you're looking for an Eulerian cycle:
In graph theory, an Eulerian trail (or Eulerian path) is a trail in a
finite graph which visits every edge exactly once. Similarly, an
Eulerian circuit or Eulerian cycle is an Eulerian trail which starts
and ends on the same vertex.
You can find a Eulerian cycle if and only if every airport is connected to an even number of airports. To find the cycle in linear time, you can use Hierholzer's algorithm.
You could use a graph library for Javascript, e.g. Cytoscape.
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Code style
Use constants for variables that do not change. Eg
const sortedArray = ;
Don't include the type in the name, Eg
const sortedArray = ;
can be Egconst sorted = ;
The default parameter in this case seams inappropriate as it is likely that
"JFK"
is not an origin in all calls to this function. If no origin is given one could assume that the first item in trips contains the origin.function sortByLinked(trips, origin = trips[0].from) {
which will throw an error if trips is empty so the function should not be called with an empty trips array if you don't pass thefrom
parameter.
However in this example best to leave the default as
undefined
if not passed as that will return an empty array which is more fitting the input parameters.
The name
sort
is inappropriate as in JS it implies that the array be sorted in place, that all items be sorted (may not be possible).You have declared
i
two times. As avar
you should put the declaration at the top of the function and not in thefor
loop.Rather than use
Array.filter
you can useArray.find
. It will find the first instance.Using
for...of
rather thanfor(;;)
reduces the code complexity.No point continuing the search inside the for loops when you have found a match. Use the
break
token to stop a loop earlyPut a space between
if
and(
Don't forget to add the
;
where appropriate. It is missing from the callsortByLinked(trips)
You call the origin
origin
andfrom
this can get confusing. Keep the naming unambiguous. As the trip items usefrom
then that would be the better name for the second input argument.
Using the above points to modify your code we get
function tripFrom(trips, from) {
const sorted = ;
const first = trips.find(trip => trip.from === from);
sorted.push(first);
for (const trip of trips) {
if (first.to === trip.from) {
sorted.push(trip);
break;
}
}
for (const trip of trips) {
if (sorted[sorted.length - 1].to === trip.from) {
sorted.push(trips);
break;
}
}
return sorted;
}
sortByLinked(trips, "JFK");
This is still not a good solution. Its not at all DRY (don't repeat yourself) and is hard coded to a single use case.
Improving the function.
It can all be done within a single loop and work for any length array.
To create the function we must add some constraints on the array trips
and what to do when we encounter any problems.
- That the array
trips
contains objects that each have the propertyfrom
andto
that are correctly formatted static strings. The resulting array is erroneous or indeterminate if not so. - That the array does not contain circular trips shorter than the array length.
- That a complete trip length is no longer than the array, or when a matching
trip.to
can not be found. The returned array can be 0 totrips.length
in size. - Locations are case sensitive.
- If there is more than one matching
trip.from
it is assumed that the first match in trips is the correct one. (It would be interesting to extract the longest possible trip from? or the shortest trip that returns to the origin?)
Example
function tripFrom(trips, from) {
const result = ;
while (result.length < trips.length) {
const trip = trips.find(trip => trip.from === from);
if (!trip) { break }
from = trip.to;
result.push(trip);
}
return result;
}
tripFrom(trips, "JFK");
Or if it is known that the trip is the same length as the input array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
return res;
}
tripFrom(trips, "JFK");
It is unclear if you want the array sorted in place. If that is a requirement then the above version can be modified to do that by simply copying the results array res
to the trips
array. You can empty an array by setting its length to zero. The spread ...
operator in this case spreads the array items over the functions arguments trips.push(...res)
thus pushing all the items to the array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
trips.length = 0;
trips.push(...res);
return trips;
}
tripFrom(trips, "JFK");
$endgroup$
$begingroup$
It's common practice to avoid using braces in inline if-statements (your first example). Also, you make a comment about a missing semicolon and I'd like to point out, just for the sake of having options, that completely getting rid of semicolons is a growing practice since it generally looks nicer.
$endgroup$
– Adam
10 hours ago
$begingroup$
@Adam It may be a common practice to not delimit single line statement blocks, but that does not make it a good practice.Consistency is the most important part of good style. If you use semicolon then use them, not just sometimes. Generally I advice that if you are not going to use them then you should know every case where automatic insertion will not happen but is required to mark the end of line. Eg "(()=>{});n()" is not the same as "(()=>{})n()" (as string to show new line location)
$endgroup$
– Blindman67
8 hours ago
$begingroup$
Right. I'm not saying he shouldn't use a semicolon in that specific spot. I'm just making it known that completely getting rid of semicolon delimiters is a very common practice nowadays.
$endgroup$
– Adam
8 hours ago
add a comment |
$begingroup$
Code style
Use constants for variables that do not change. Eg
const sortedArray = ;
Don't include the type in the name, Eg
const sortedArray = ;
can be Egconst sorted = ;
The default parameter in this case seams inappropriate as it is likely that
"JFK"
is not an origin in all calls to this function. If no origin is given one could assume that the first item in trips contains the origin.function sortByLinked(trips, origin = trips[0].from) {
which will throw an error if trips is empty so the function should not be called with an empty trips array if you don't pass thefrom
parameter.
However in this example best to leave the default as
undefined
if not passed as that will return an empty array which is more fitting the input parameters.
The name
sort
is inappropriate as in JS it implies that the array be sorted in place, that all items be sorted (may not be possible).You have declared
i
two times. As avar
you should put the declaration at the top of the function and not in thefor
loop.Rather than use
Array.filter
you can useArray.find
. It will find the first instance.Using
for...of
rather thanfor(;;)
reduces the code complexity.No point continuing the search inside the for loops when you have found a match. Use the
break
token to stop a loop earlyPut a space between
if
and(
Don't forget to add the
;
where appropriate. It is missing from the callsortByLinked(trips)
You call the origin
origin
andfrom
this can get confusing. Keep the naming unambiguous. As the trip items usefrom
then that would be the better name for the second input argument.
Using the above points to modify your code we get
function tripFrom(trips, from) {
const sorted = ;
const first = trips.find(trip => trip.from === from);
sorted.push(first);
for (const trip of trips) {
if (first.to === trip.from) {
sorted.push(trip);
break;
}
}
for (const trip of trips) {
if (sorted[sorted.length - 1].to === trip.from) {
sorted.push(trips);
break;
}
}
return sorted;
}
sortByLinked(trips, "JFK");
This is still not a good solution. Its not at all DRY (don't repeat yourself) and is hard coded to a single use case.
Improving the function.
It can all be done within a single loop and work for any length array.
To create the function we must add some constraints on the array trips
and what to do when we encounter any problems.
- That the array
trips
contains objects that each have the propertyfrom
andto
that are correctly formatted static strings. The resulting array is erroneous or indeterminate if not so. - That the array does not contain circular trips shorter than the array length.
- That a complete trip length is no longer than the array, or when a matching
trip.to
can not be found. The returned array can be 0 totrips.length
in size. - Locations are case sensitive.
- If there is more than one matching
trip.from
it is assumed that the first match in trips is the correct one. (It would be interesting to extract the longest possible trip from? or the shortest trip that returns to the origin?)
Example
function tripFrom(trips, from) {
const result = ;
while (result.length < trips.length) {
const trip = trips.find(trip => trip.from === from);
if (!trip) { break }
from = trip.to;
result.push(trip);
}
return result;
}
tripFrom(trips, "JFK");
Or if it is known that the trip is the same length as the input array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
return res;
}
tripFrom(trips, "JFK");
It is unclear if you want the array sorted in place. If that is a requirement then the above version can be modified to do that by simply copying the results array res
to the trips
array. You can empty an array by setting its length to zero. The spread ...
operator in this case spreads the array items over the functions arguments trips.push(...res)
thus pushing all the items to the array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
trips.length = 0;
trips.push(...res);
return trips;
}
tripFrom(trips, "JFK");
$endgroup$
$begingroup$
It's common practice to avoid using braces in inline if-statements (your first example). Also, you make a comment about a missing semicolon and I'd like to point out, just for the sake of having options, that completely getting rid of semicolons is a growing practice since it generally looks nicer.
$endgroup$
– Adam
10 hours ago
$begingroup$
@Adam It may be a common practice to not delimit single line statement blocks, but that does not make it a good practice.Consistency is the most important part of good style. If you use semicolon then use them, not just sometimes. Generally I advice that if you are not going to use them then you should know every case where automatic insertion will not happen but is required to mark the end of line. Eg "(()=>{});n()" is not the same as "(()=>{})n()" (as string to show new line location)
$endgroup$
– Blindman67
8 hours ago
$begingroup$
Right. I'm not saying he shouldn't use a semicolon in that specific spot. I'm just making it known that completely getting rid of semicolon delimiters is a very common practice nowadays.
$endgroup$
– Adam
8 hours ago
add a comment |
$begingroup$
Code style
Use constants for variables that do not change. Eg
const sortedArray = ;
Don't include the type in the name, Eg
const sortedArray = ;
can be Egconst sorted = ;
The default parameter in this case seams inappropriate as it is likely that
"JFK"
is not an origin in all calls to this function. If no origin is given one could assume that the first item in trips contains the origin.function sortByLinked(trips, origin = trips[0].from) {
which will throw an error if trips is empty so the function should not be called with an empty trips array if you don't pass thefrom
parameter.
However in this example best to leave the default as
undefined
if not passed as that will return an empty array which is more fitting the input parameters.
The name
sort
is inappropriate as in JS it implies that the array be sorted in place, that all items be sorted (may not be possible).You have declared
i
two times. As avar
you should put the declaration at the top of the function and not in thefor
loop.Rather than use
Array.filter
you can useArray.find
. It will find the first instance.Using
for...of
rather thanfor(;;)
reduces the code complexity.No point continuing the search inside the for loops when you have found a match. Use the
break
token to stop a loop earlyPut a space between
if
and(
Don't forget to add the
;
where appropriate. It is missing from the callsortByLinked(trips)
You call the origin
origin
andfrom
this can get confusing. Keep the naming unambiguous. As the trip items usefrom
then that would be the better name for the second input argument.
Using the above points to modify your code we get
function tripFrom(trips, from) {
const sorted = ;
const first = trips.find(trip => trip.from === from);
sorted.push(first);
for (const trip of trips) {
if (first.to === trip.from) {
sorted.push(trip);
break;
}
}
for (const trip of trips) {
if (sorted[sorted.length - 1].to === trip.from) {
sorted.push(trips);
break;
}
}
return sorted;
}
sortByLinked(trips, "JFK");
This is still not a good solution. Its not at all DRY (don't repeat yourself) and is hard coded to a single use case.
Improving the function.
It can all be done within a single loop and work for any length array.
To create the function we must add some constraints on the array trips
and what to do when we encounter any problems.
- That the array
trips
contains objects that each have the propertyfrom
andto
that are correctly formatted static strings. The resulting array is erroneous or indeterminate if not so. - That the array does not contain circular trips shorter than the array length.
- That a complete trip length is no longer than the array, or when a matching
trip.to
can not be found. The returned array can be 0 totrips.length
in size. - Locations are case sensitive.
- If there is more than one matching
trip.from
it is assumed that the first match in trips is the correct one. (It would be interesting to extract the longest possible trip from? or the shortest trip that returns to the origin?)
Example
function tripFrom(trips, from) {
const result = ;
while (result.length < trips.length) {
const trip = trips.find(trip => trip.from === from);
if (!trip) { break }
from = trip.to;
result.push(trip);
}
return result;
}
tripFrom(trips, "JFK");
Or if it is known that the trip is the same length as the input array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
return res;
}
tripFrom(trips, "JFK");
It is unclear if you want the array sorted in place. If that is a requirement then the above version can be modified to do that by simply copying the results array res
to the trips
array. You can empty an array by setting its length to zero. The spread ...
operator in this case spreads the array items over the functions arguments trips.push(...res)
thus pushing all the items to the array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
trips.length = 0;
trips.push(...res);
return trips;
}
tripFrom(trips, "JFK");
$endgroup$
Code style
Use constants for variables that do not change. Eg
const sortedArray = ;
Don't include the type in the name, Eg
const sortedArray = ;
can be Egconst sorted = ;
The default parameter in this case seams inappropriate as it is likely that
"JFK"
is not an origin in all calls to this function. If no origin is given one could assume that the first item in trips contains the origin.function sortByLinked(trips, origin = trips[0].from) {
which will throw an error if trips is empty so the function should not be called with an empty trips array if you don't pass thefrom
parameter.
However in this example best to leave the default as
undefined
if not passed as that will return an empty array which is more fitting the input parameters.
The name
sort
is inappropriate as in JS it implies that the array be sorted in place, that all items be sorted (may not be possible).You have declared
i
two times. As avar
you should put the declaration at the top of the function and not in thefor
loop.Rather than use
Array.filter
you can useArray.find
. It will find the first instance.Using
for...of
rather thanfor(;;)
reduces the code complexity.No point continuing the search inside the for loops when you have found a match. Use the
break
token to stop a loop earlyPut a space between
if
and(
Don't forget to add the
;
where appropriate. It is missing from the callsortByLinked(trips)
You call the origin
origin
andfrom
this can get confusing. Keep the naming unambiguous. As the trip items usefrom
then that would be the better name for the second input argument.
Using the above points to modify your code we get
function tripFrom(trips, from) {
const sorted = ;
const first = trips.find(trip => trip.from === from);
sorted.push(first);
for (const trip of trips) {
if (first.to === trip.from) {
sorted.push(trip);
break;
}
}
for (const trip of trips) {
if (sorted[sorted.length - 1].to === trip.from) {
sorted.push(trips);
break;
}
}
return sorted;
}
sortByLinked(trips, "JFK");
This is still not a good solution. Its not at all DRY (don't repeat yourself) and is hard coded to a single use case.
Improving the function.
It can all be done within a single loop and work for any length array.
To create the function we must add some constraints on the array trips
and what to do when we encounter any problems.
- That the array
trips
contains objects that each have the propertyfrom
andto
that are correctly formatted static strings. The resulting array is erroneous or indeterminate if not so. - That the array does not contain circular trips shorter than the array length.
- That a complete trip length is no longer than the array, or when a matching
trip.to
can not be found. The returned array can be 0 totrips.length
in size. - Locations are case sensitive.
- If there is more than one matching
trip.from
it is assumed that the first match in trips is the correct one. (It would be interesting to extract the longest possible trip from? or the shortest trip that returns to the origin?)
Example
function tripFrom(trips, from) {
const result = ;
while (result.length < trips.length) {
const trip = trips.find(trip => trip.from === from);
if (!trip) { break }
from = trip.to;
result.push(trip);
}
return result;
}
tripFrom(trips, "JFK");
Or if it is known that the trip is the same length as the input array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
return res;
}
tripFrom(trips, "JFK");
It is unclear if you want the array sorted in place. If that is a requirement then the above version can be modified to do that by simply copying the results array res
to the trips
array. You can empty an array by setting its length to zero. The spread ...
operator in this case spreads the array items over the functions arguments trips.push(...res)
thus pushing all the items to the array.
function tripFrom(trips, from) {
const res = ;
while (res.length < trips.length) {
from = (res[res.length] = trips.find(trip => trip.from === from)).to;
}
trips.length = 0;
trips.push(...res);
return trips;
}
tripFrom(trips, "JFK");
answered yesterday
Blindman67Blindman67
9,4651622
9,4651622
$begingroup$
It's common practice to avoid using braces in inline if-statements (your first example). Also, you make a comment about a missing semicolon and I'd like to point out, just for the sake of having options, that completely getting rid of semicolons is a growing practice since it generally looks nicer.
$endgroup$
– Adam
10 hours ago
$begingroup$
@Adam It may be a common practice to not delimit single line statement blocks, but that does not make it a good practice.Consistency is the most important part of good style. If you use semicolon then use them, not just sometimes. Generally I advice that if you are not going to use them then you should know every case where automatic insertion will not happen but is required to mark the end of line. Eg "(()=>{});n()" is not the same as "(()=>{})n()" (as string to show new line location)
$endgroup$
– Blindman67
8 hours ago
$begingroup$
Right. I'm not saying he shouldn't use a semicolon in that specific spot. I'm just making it known that completely getting rid of semicolon delimiters is a very common practice nowadays.
$endgroup$
– Adam
8 hours ago
add a comment |
$begingroup$
It's common practice to avoid using braces in inline if-statements (your first example). Also, you make a comment about a missing semicolon and I'd like to point out, just for the sake of having options, that completely getting rid of semicolons is a growing practice since it generally looks nicer.
$endgroup$
– Adam
10 hours ago
$begingroup$
@Adam It may be a common practice to not delimit single line statement blocks, but that does not make it a good practice.Consistency is the most important part of good style. If you use semicolon then use them, not just sometimes. Generally I advice that if you are not going to use them then you should know every case where automatic insertion will not happen but is required to mark the end of line. Eg "(()=>{});n()" is not the same as "(()=>{})n()" (as string to show new line location)
$endgroup$
– Blindman67
8 hours ago
$begingroup$
Right. I'm not saying he shouldn't use a semicolon in that specific spot. I'm just making it known that completely getting rid of semicolon delimiters is a very common practice nowadays.
$endgroup$
– Adam
8 hours ago
$begingroup$
It's common practice to avoid using braces in inline if-statements (your first example). Also, you make a comment about a missing semicolon and I'd like to point out, just for the sake of having options, that completely getting rid of semicolons is a growing practice since it generally looks nicer.
$endgroup$
– Adam
10 hours ago
$begingroup$
It's common practice to avoid using braces in inline if-statements (your first example). Also, you make a comment about a missing semicolon and I'd like to point out, just for the sake of having options, that completely getting rid of semicolons is a growing practice since it generally looks nicer.
$endgroup$
– Adam
10 hours ago
$begingroup$
@Adam It may be a common practice to not delimit single line statement blocks, but that does not make it a good practice.Consistency is the most important part of good style. If you use semicolon then use them, not just sometimes. Generally I advice that if you are not going to use them then you should know every case where automatic insertion will not happen but is required to mark the end of line. Eg "(()=>{});n()" is not the same as "(()=>{})n()" (as string to show new line location)
$endgroup$
– Blindman67
8 hours ago
$begingroup$
@Adam It may be a common practice to not delimit single line statement blocks, but that does not make it a good practice.Consistency is the most important part of good style. If you use semicolon then use them, not just sometimes. Generally I advice that if you are not going to use them then you should know every case where automatic insertion will not happen but is required to mark the end of line. Eg "(()=>{});n()" is not the same as "(()=>{})n()" (as string to show new line location)
$endgroup$
– Blindman67
8 hours ago
$begingroup$
Right. I'm not saying he shouldn't use a semicolon in that specific spot. I'm just making it known that completely getting rid of semicolon delimiters is a very common practice nowadays.
$endgroup$
– Adam
8 hours ago
$begingroup$
Right. I'm not saying he shouldn't use a semicolon in that specific spot. I'm just making it known that completely getting rid of semicolon delimiters is a very common practice nowadays.
$endgroup$
– Adam
8 hours ago
add a comment |
$begingroup$
It would be good if your function could work for more than 3 trips.
And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map
first, so that you can access a trip by its from
property in constant time:
function sortByLinked(trips, origin = "JFK") {
const map = new Map(trips.map(trip => [trip.from, trip]));
const result = ;
for (let trip; trip = map.get(origin); origin = trip.to) {
result.push(trip);
map.delete(origin);
}
return result;
}
const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];
const sorted = sortByLinked(trips, "JFK");
console.log(sorted);
$endgroup$
$begingroup$
Interesting. It doesn't work with cycles, though, does it? e.g.[{from: "JFK",to: "XXX"},{from: "XXX",to: "YYY"},{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}]
$endgroup$
– Eric Duminil
12 hours ago
$begingroup$
Indeed, it is not expecting multiple trip objects with the samefrom
property. That would need a more comprehensive tree traversal.
$endgroup$
– trincot
12 hours ago
add a comment |
$begingroup$
It would be good if your function could work for more than 3 trips.
And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map
first, so that you can access a trip by its from
property in constant time:
function sortByLinked(trips, origin = "JFK") {
const map = new Map(trips.map(trip => [trip.from, trip]));
const result = ;
for (let trip; trip = map.get(origin); origin = trip.to) {
result.push(trip);
map.delete(origin);
}
return result;
}
const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];
const sorted = sortByLinked(trips, "JFK");
console.log(sorted);
$endgroup$
$begingroup$
Interesting. It doesn't work with cycles, though, does it? e.g.[{from: "JFK",to: "XXX"},{from: "XXX",to: "YYY"},{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}]
$endgroup$
– Eric Duminil
12 hours ago
$begingroup$
Indeed, it is not expecting multiple trip objects with the samefrom
property. That would need a more comprehensive tree traversal.
$endgroup$
– trincot
12 hours ago
add a comment |
$begingroup$
It would be good if your function could work for more than 3 trips.
And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map
first, so that you can access a trip by its from
property in constant time:
function sortByLinked(trips, origin = "JFK") {
const map = new Map(trips.map(trip => [trip.from, trip]));
const result = ;
for (let trip; trip = map.get(origin); origin = trip.to) {
result.push(trip);
map.delete(origin);
}
return result;
}
const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];
const sorted = sortByLinked(trips, "JFK");
console.log(sorted);
$endgroup$
It would be good if your function could work for more than 3 trips.
And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map
first, so that you can access a trip by its from
property in constant time:
function sortByLinked(trips, origin = "JFK") {
const map = new Map(trips.map(trip => [trip.from, trip]));
const result = ;
for (let trip; trip = map.get(origin); origin = trip.to) {
result.push(trip);
map.delete(origin);
}
return result;
}
const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];
const sorted = sortByLinked(trips, "JFK");
console.log(sorted);
function sortByLinked(trips, origin = "JFK") {
const map = new Map(trips.map(trip => [trip.from, trip]));
const result = ;
for (let trip; trip = map.get(origin); origin = trip.to) {
result.push(trip);
map.delete(origin);
}
return result;
}
const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];
const sorted = sortByLinked(trips, "JFK");
console.log(sorted);
function sortByLinked(trips, origin = "JFK") {
const map = new Map(trips.map(trip => [trip.from, trip]));
const result = ;
for (let trip; trip = map.get(origin); origin = trip.to) {
result.push(trip);
map.delete(origin);
}
return result;
}
const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];
const sorted = sortByLinked(trips, "JFK");
console.log(sorted);
answered yesterday
trincottrincot
48737
48737
$begingroup$
Interesting. It doesn't work with cycles, though, does it? e.g.[{from: "JFK",to: "XXX"},{from: "XXX",to: "YYY"},{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}]
$endgroup$
– Eric Duminil
12 hours ago
$begingroup$
Indeed, it is not expecting multiple trip objects with the samefrom
property. That would need a more comprehensive tree traversal.
$endgroup$
– trincot
12 hours ago
add a comment |
$begingroup$
Interesting. It doesn't work with cycles, though, does it? e.g.[{from: "JFK",to: "XXX"},{from: "XXX",to: "YYY"},{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}]
$endgroup$
– Eric Duminil
12 hours ago
$begingroup$
Indeed, it is not expecting multiple trip objects with the samefrom
property. That would need a more comprehensive tree traversal.
$endgroup$
– trincot
12 hours ago
$begingroup$
Interesting. It doesn't work with cycles, though, does it? e.g.
[{from: "JFK",to: "XXX"},{from: "XXX",to: "YYY"},{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}]
$endgroup$
– Eric Duminil
12 hours ago
$begingroup$
Interesting. It doesn't work with cycles, though, does it? e.g.
[{from: "JFK",to: "XXX"},{from: "XXX",to: "YYY"},{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}]
$endgroup$
– Eric Duminil
12 hours ago
$begingroup$
Indeed, it is not expecting multiple trip objects with the same
from
property. That would need a more comprehensive tree traversal.$endgroup$
– trincot
12 hours ago
$begingroup$
Indeed, it is not expecting multiple trip objects with the same
from
property. That would need a more comprehensive tree traversal.$endgroup$
– trincot
12 hours ago
add a comment |
$begingroup$
Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for
loop like you're doing there are other ways to find the next trip like using Array.map()
or Array.filter()
.
Here's one way to sort it in place and can handle any number of trips greater than 1.
function sortByLinked(trips, origin = 'JFK') {
// this will be useful
function swap(array, index1, index2){
let temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
// find first one
let first = trips.filter(trip => trip.from === origin)[0];
// put him in the front of the list
swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);
// sort it in place
for(let i=1; i<trips.length; i++){
swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));
}
}
New contributor
$endgroup$
$begingroup$
Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
$endgroup$
– Sᴀᴍ Onᴇᴌᴀ
yesterday
add a comment |
$begingroup$
Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for
loop like you're doing there are other ways to find the next trip like using Array.map()
or Array.filter()
.
Here's one way to sort it in place and can handle any number of trips greater than 1.
function sortByLinked(trips, origin = 'JFK') {
// this will be useful
function swap(array, index1, index2){
let temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
// find first one
let first = trips.filter(trip => trip.from === origin)[0];
// put him in the front of the list
swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);
// sort it in place
for(let i=1; i<trips.length; i++){
swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));
}
}
New contributor
$endgroup$
$begingroup$
Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
$endgroup$
– Sᴀᴍ Onᴇᴌᴀ
yesterday
add a comment |
$begingroup$
Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for
loop like you're doing there are other ways to find the next trip like using Array.map()
or Array.filter()
.
Here's one way to sort it in place and can handle any number of trips greater than 1.
function sortByLinked(trips, origin = 'JFK') {
// this will be useful
function swap(array, index1, index2){
let temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
// find first one
let first = trips.filter(trip => trip.from === origin)[0];
// put him in the front of the list
swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);
// sort it in place
for(let i=1; i<trips.length; i++){
swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));
}
}
New contributor
$endgroup$
Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for
loop like you're doing there are other ways to find the next trip like using Array.map()
or Array.filter()
.
Here's one way to sort it in place and can handle any number of trips greater than 1.
function sortByLinked(trips, origin = 'JFK') {
// this will be useful
function swap(array, index1, index2){
let temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
// find first one
let first = trips.filter(trip => trip.from === origin)[0];
// put him in the front of the list
swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);
// sort it in place
for(let i=1; i<trips.length; i++){
swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));
}
}
function sortByLinked(trips, origin = 'JFK') {
// this will be useful
function swap(array, index1, index2){
let temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
// find first one
let first = trips.filter(trip => trip.from === origin)[0];
// put him in the front of the list
swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);
// sort it in place
for(let i=1; i<trips.length; i++){
swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));
}
}
function sortByLinked(trips, origin = 'JFK') {
// this will be useful
function swap(array, index1, index2){
let temp = array[index1];
array[index1] = array[index2];
array[index2] = temp;
}
// find first one
let first = trips.filter(trip => trip.from === origin)[0];
// put him in the front of the list
swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);
// sort it in place
for(let i=1; i<trips.length; i++){
swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));
}
}
New contributor
edited yesterday
Sᴀᴍ Onᴇᴌᴀ
10.3k62168
10.3k62168
New contributor
answered yesterday
matthewlam.jsmatthewlam.js
411
411
New contributor
New contributor
$begingroup$
Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
$endgroup$
– Sᴀᴍ Onᴇᴌᴀ
yesterday
add a comment |
$begingroup$
Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
$endgroup$
– Sᴀᴍ Onᴇᴌᴀ
yesterday
$begingroup$
Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
$endgroup$
– Sᴀᴍ Onᴇᴌᴀ
yesterday
$begingroup$
Because ecmascript-6 is used here: "Two variables values can be swapped in one destructuring expression."
$endgroup$
– Sᴀᴍ Onᴇᴌᴀ
yesterday
add a comment |
$begingroup$
You could also solve the problem by introducing an object mo
(as a makeshift associative array) like shown in the following example
const trav=[{from:"bos",to:"sfo"},{from:"jer",to:"jfk"},{from:"zur",to:"brm"},{from:"haj",to:"cdg"},{from:"pma",to:"mlg"},{from:"sfo",to:"zur"},
{from:"cdg",to:"hav"},{from:"jfk",to:"haj"},{from:"man",to:"cdg"},
{from:"mlg",to:"jer"},{from:"brm",to:"pma"},{from:"hav",to:"bos"}],
mo={},srt=;
var fr='jfk',v,n=20;
trav.forEach(v=>mo[v.from]=v);
while (n--&&(v=mo[fr])){srt.push(v);fr=v.to;}
console.log(JSON.stringify(srt));
The n
is a "safety switch` that gets you out when the sequence turns out to be circular (as is the case in my example).
New contributor
$endgroup$
2
$begingroup$
Welcome to Code Reviews, cars10m. Providing an alternative implementation is interesting and handy on some level, but it doesn't constitute a code review. Was there something so wrong with the original that you needed to rewrite it instead of tweaking it in smaller ways?
$endgroup$
– chicks
13 hours ago
add a comment |
$begingroup$
You could also solve the problem by introducing an object mo
(as a makeshift associative array) like shown in the following example
const trav=[{from:"bos",to:"sfo"},{from:"jer",to:"jfk"},{from:"zur",to:"brm"},{from:"haj",to:"cdg"},{from:"pma",to:"mlg"},{from:"sfo",to:"zur"},
{from:"cdg",to:"hav"},{from:"jfk",to:"haj"},{from:"man",to:"cdg"},
{from:"mlg",to:"jer"},{from:"brm",to:"pma"},{from:"hav",to:"bos"}],
mo={},srt=;
var fr='jfk',v,n=20;
trav.forEach(v=>mo[v.from]=v);
while (n--&&(v=mo[fr])){srt.push(v);fr=v.to;}
console.log(JSON.stringify(srt));
The n
is a "safety switch` that gets you out when the sequence turns out to be circular (as is the case in my example).
New contributor
$endgroup$
2
$begingroup$
Welcome to Code Reviews, cars10m. Providing an alternative implementation is interesting and handy on some level, but it doesn't constitute a code review. Was there something so wrong with the original that you needed to rewrite it instead of tweaking it in smaller ways?
$endgroup$
– chicks
13 hours ago
add a comment |
$begingroup$
You could also solve the problem by introducing an object mo
(as a makeshift associative array) like shown in the following example
const trav=[{from:"bos",to:"sfo"},{from:"jer",to:"jfk"},{from:"zur",to:"brm"},{from:"haj",to:"cdg"},{from:"pma",to:"mlg"},{from:"sfo",to:"zur"},
{from:"cdg",to:"hav"},{from:"jfk",to:"haj"},{from:"man",to:"cdg"},
{from:"mlg",to:"jer"},{from:"brm",to:"pma"},{from:"hav",to:"bos"}],
mo={},srt=;
var fr='jfk',v,n=20;
trav.forEach(v=>mo[v.from]=v);
while (n--&&(v=mo[fr])){srt.push(v);fr=v.to;}
console.log(JSON.stringify(srt));
The n
is a "safety switch` that gets you out when the sequence turns out to be circular (as is the case in my example).
New contributor
$endgroup$
You could also solve the problem by introducing an object mo
(as a makeshift associative array) like shown in the following example
const trav=[{from:"bos",to:"sfo"},{from:"jer",to:"jfk"},{from:"zur",to:"brm"},{from:"haj",to:"cdg"},{from:"pma",to:"mlg"},{from:"sfo",to:"zur"},
{from:"cdg",to:"hav"},{from:"jfk",to:"haj"},{from:"man",to:"cdg"},
{from:"mlg",to:"jer"},{from:"brm",to:"pma"},{from:"hav",to:"bos"}],
mo={},srt=;
var fr='jfk',v,n=20;
trav.forEach(v=>mo[v.from]=v);
while (n--&&(v=mo[fr])){srt.push(v);fr=v.to;}
console.log(JSON.stringify(srt));
The n
is a "safety switch` that gets you out when the sequence turns out to be circular (as is the case in my example).
New contributor
edited 16 hours ago
New contributor
answered 16 hours ago
cars10mcars10m
1012
1012
New contributor
New contributor
2
$begingroup$
Welcome to Code Reviews, cars10m. Providing an alternative implementation is interesting and handy on some level, but it doesn't constitute a code review. Was there something so wrong with the original that you needed to rewrite it instead of tweaking it in smaller ways?
$endgroup$
– chicks
13 hours ago
add a comment |
2
$begingroup$
Welcome to Code Reviews, cars10m. Providing an alternative implementation is interesting and handy on some level, but it doesn't constitute a code review. Was there something so wrong with the original that you needed to rewrite it instead of tweaking it in smaller ways?
$endgroup$
– chicks
13 hours ago
2
2
$begingroup$
Welcome to Code Reviews, cars10m. Providing an alternative implementation is interesting and handy on some level, but it doesn't constitute a code review. Was there something so wrong with the original that you needed to rewrite it instead of tweaking it in smaller ways?
$endgroup$
– chicks
13 hours ago
$begingroup$
Welcome to Code Reviews, cars10m. Providing an alternative implementation is interesting and handy on some level, but it doesn't constitute a code review. Was there something so wrong with the original that you needed to rewrite it instead of tweaking it in smaller ways?
$endgroup$
– chicks
13 hours ago
add a comment |
$begingroup$
This problem is related to graph theory, and you're looking for an Eulerian cycle:
In graph theory, an Eulerian trail (or Eulerian path) is a trail in a
finite graph which visits every edge exactly once. Similarly, an
Eulerian circuit or Eulerian cycle is an Eulerian trail which starts
and ends on the same vertex.
You can find a Eulerian cycle if and only if every airport is connected to an even number of airports. To find the cycle in linear time, you can use Hierholzer's algorithm.
You could use a graph library for Javascript, e.g. Cytoscape.
$endgroup$
add a comment |
$begingroup$
This problem is related to graph theory, and you're looking for an Eulerian cycle:
In graph theory, an Eulerian trail (or Eulerian path) is a trail in a
finite graph which visits every edge exactly once. Similarly, an
Eulerian circuit or Eulerian cycle is an Eulerian trail which starts
and ends on the same vertex.
You can find a Eulerian cycle if and only if every airport is connected to an even number of airports. To find the cycle in linear time, you can use Hierholzer's algorithm.
You could use a graph library for Javascript, e.g. Cytoscape.
$endgroup$
add a comment |
$begingroup$
This problem is related to graph theory, and you're looking for an Eulerian cycle:
In graph theory, an Eulerian trail (or Eulerian path) is a trail in a
finite graph which visits every edge exactly once. Similarly, an
Eulerian circuit or Eulerian cycle is an Eulerian trail which starts
and ends on the same vertex.
You can find a Eulerian cycle if and only if every airport is connected to an even number of airports. To find the cycle in linear time, you can use Hierholzer's algorithm.
You could use a graph library for Javascript, e.g. Cytoscape.
$endgroup$
This problem is related to graph theory, and you're looking for an Eulerian cycle:
In graph theory, an Eulerian trail (or Eulerian path) is a trail in a
finite graph which visits every edge exactly once. Similarly, an
Eulerian circuit or Eulerian cycle is an Eulerian trail which starts
and ends on the same vertex.
You can find a Eulerian cycle if and only if every airport is connected to an even number of airports. To find the cycle in linear time, you can use Hierholzer's algorithm.
You could use a graph library for Javascript, e.g. Cytoscape.
answered 12 hours ago
Eric DuminilEric Duminil
2,1111613
2,1111613
add a comment |
add a comment |
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$begingroup$
is it like the origin and final destination always the same? and how many intermediate trips are going to be present?
$endgroup$
– karthick
yesterday
$begingroup$
Is
from
unique for all elements?$endgroup$
– Taplar
yesterday
$begingroup$
@karthick yes the origin and final should be the same.
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
@Taplar yes from is unique
$endgroup$
– Shivam Bhalla
yesterday
$begingroup$
The task you want to perform is a simple kind of topological sorting.
$endgroup$
– 200_success
yesterday