Reverse int within the 32-bit signed integer range: [−2^31, 2^31 − 1]
$begingroup$
Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.
Feedback
Looking for any ways I can optimize this with modern c++ features overall. I hope my use of const
correctness, exception handling, and assertions is implemented well here, please let me know. Is there any way I can use byte operations to reverse the int and keep track of the sign possibly?
Based on the submission feedback from LeetCode, is it safe to say that the time complexity is O(n)
and space complexity is O(n)
? If I can reduce the complexity in anyway would love to know! Thanks for the feedback in advance.
#include <cassert>
#include <climits>
#include <stdexcept>
#include <string>
class Solution
{
public:
int reverse(int i) {
bool is_signed = false;
if(i < 0) { is_signed = true; }
auto i_string = std::to_string(i);
std::string reversed = "";
while(!i_string.empty()) {
reversed.push_back(i_string.back());
i_string.pop_back();
}
try {
i = std::stoi(reversed);
} catch (const std::out_of_range& e) {
return 0;
}
if(is_signed) { i *= -1; }
return i;
}
};
int main()
{
Solution s;
assert(s.reverse(1) == 1);
assert(s.reverse(0) == 0);
assert(s.reverse(123) == 321);
assert(s.reverse(120) == 21);
assert(s.reverse(-123) == -321);
assert(s.reverse(1207) == 7021);
assert(s.reverse(INT_MAX) == 0);
assert(s.reverse(INT_MIN) == 0);
}
c++ c++11 interview-questions integer
$endgroup$
add a comment |
$begingroup$
Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.
Feedback
Looking for any ways I can optimize this with modern c++ features overall. I hope my use of const
correctness, exception handling, and assertions is implemented well here, please let me know. Is there any way I can use byte operations to reverse the int and keep track of the sign possibly?
Based on the submission feedback from LeetCode, is it safe to say that the time complexity is O(n)
and space complexity is O(n)
? If I can reduce the complexity in anyway would love to know! Thanks for the feedback in advance.
#include <cassert>
#include <climits>
#include <stdexcept>
#include <string>
class Solution
{
public:
int reverse(int i) {
bool is_signed = false;
if(i < 0) { is_signed = true; }
auto i_string = std::to_string(i);
std::string reversed = "";
while(!i_string.empty()) {
reversed.push_back(i_string.back());
i_string.pop_back();
}
try {
i = std::stoi(reversed);
} catch (const std::out_of_range& e) {
return 0;
}
if(is_signed) { i *= -1; }
return i;
}
};
int main()
{
Solution s;
assert(s.reverse(1) == 1);
assert(s.reverse(0) == 0);
assert(s.reverse(123) == 321);
assert(s.reverse(120) == 21);
assert(s.reverse(-123) == -321);
assert(s.reverse(1207) == 7021);
assert(s.reverse(INT_MAX) == 0);
assert(s.reverse(INT_MIN) == 0);
}
c++ c++11 interview-questions integer
$endgroup$
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
3 hours ago
$begingroup$
Yes that's the one
$endgroup$
– greg
3 hours ago
add a comment |
$begingroup$
Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.
Feedback
Looking for any ways I can optimize this with modern c++ features overall. I hope my use of const
correctness, exception handling, and assertions is implemented well here, please let me know. Is there any way I can use byte operations to reverse the int and keep track of the sign possibly?
Based on the submission feedback from LeetCode, is it safe to say that the time complexity is O(n)
and space complexity is O(n)
? If I can reduce the complexity in anyway would love to know! Thanks for the feedback in advance.
#include <cassert>
#include <climits>
#include <stdexcept>
#include <string>
class Solution
{
public:
int reverse(int i) {
bool is_signed = false;
if(i < 0) { is_signed = true; }
auto i_string = std::to_string(i);
std::string reversed = "";
while(!i_string.empty()) {
reversed.push_back(i_string.back());
i_string.pop_back();
}
try {
i = std::stoi(reversed);
} catch (const std::out_of_range& e) {
return 0;
}
if(is_signed) { i *= -1; }
return i;
}
};
int main()
{
Solution s;
assert(s.reverse(1) == 1);
assert(s.reverse(0) == 0);
assert(s.reverse(123) == 321);
assert(s.reverse(120) == 21);
assert(s.reverse(-123) == -321);
assert(s.reverse(1207) == 7021);
assert(s.reverse(INT_MAX) == 0);
assert(s.reverse(INT_MIN) == 0);
}
c++ c++11 interview-questions integer
$endgroup$
Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0.
Feedback
Looking for any ways I can optimize this with modern c++ features overall. I hope my use of const
correctness, exception handling, and assertions is implemented well here, please let me know. Is there any way I can use byte operations to reverse the int and keep track of the sign possibly?
Based on the submission feedback from LeetCode, is it safe to say that the time complexity is O(n)
and space complexity is O(n)
? If I can reduce the complexity in anyway would love to know! Thanks for the feedback in advance.
#include <cassert>
#include <climits>
#include <stdexcept>
#include <string>
class Solution
{
public:
int reverse(int i) {
bool is_signed = false;
if(i < 0) { is_signed = true; }
auto i_string = std::to_string(i);
std::string reversed = "";
while(!i_string.empty()) {
reversed.push_back(i_string.back());
i_string.pop_back();
}
try {
i = std::stoi(reversed);
} catch (const std::out_of_range& e) {
return 0;
}
if(is_signed) { i *= -1; }
return i;
}
};
int main()
{
Solution s;
assert(s.reverse(1) == 1);
assert(s.reverse(0) == 0);
assert(s.reverse(123) == 321);
assert(s.reverse(120) == 21);
assert(s.reverse(-123) == -321);
assert(s.reverse(1207) == 7021);
assert(s.reverse(INT_MAX) == 0);
assert(s.reverse(INT_MIN) == 0);
}
c++ c++11 interview-questions integer
c++ c++11 interview-questions integer
edited 4 hours ago
Martin R
16.2k12366
16.2k12366
asked 4 hours ago
greggreg
37018
37018
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
3 hours ago
$begingroup$
Yes that's the one
$endgroup$
– greg
3 hours ago
add a comment |
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
3 hours ago
$begingroup$
Yes that's the one
$endgroup$
– greg
3 hours ago
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
3 hours ago
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
3 hours ago
$begingroup$
Yes that's the one
$endgroup$
– greg
3 hours ago
$begingroup$
Yes that's the one
$endgroup$
– greg
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
General comments
There is no reason to use a class. Instead, the functionality should be made into a free function.
Your code is overly complicated. There is no reason to make new string from which you erase characters one-by-one. Instead, you can convert the input integer to a string and use a standard function to reverse that.
Also, pay attention to const correctness. This protects from unintended mistakes and helps the compiler optimize more.
I would simplify your function to just:
int reverse(int i)
{
try
{
auto reversed{ std::to_string(i) };
std::reverse(reversed.begin(), reversed.end());
const auto result{ std::stoi(reversed) };
return i < 0 ? -1 * result : result;
}
catch (const std::out_of_range& e)
{
return 0;
}
}
Further comments
If you want to have a fast solution, you should avoid
std::string
altogether. This you can do by "iterating" through the digits using arithmetic operations (division and modulus), as in (usingstd::string
to only show you what is happening):
int x = 1234;
std::string s;
while (x > 0)
{
s.push_back('0' + (x % 10));
x /= 10;
}
std::cout << s << "n"; // Prints 4321
I will let you take it from here to use these ideas to make your program even faster.
Regarding your theoretical question concerning complexity, if we assume that the input is treated as a string of n characters, there is Omega(n) lower bound by a trivial adversary argument. Basically, if you don't spend at least n time, you can't read the whole of the input, and then you cannot guarantee correct output on every instance.
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
General comments
There is no reason to use a class. Instead, the functionality should be made into a free function.
Your code is overly complicated. There is no reason to make new string from which you erase characters one-by-one. Instead, you can convert the input integer to a string and use a standard function to reverse that.
Also, pay attention to const correctness. This protects from unintended mistakes and helps the compiler optimize more.
I would simplify your function to just:
int reverse(int i)
{
try
{
auto reversed{ std::to_string(i) };
std::reverse(reversed.begin(), reversed.end());
const auto result{ std::stoi(reversed) };
return i < 0 ? -1 * result : result;
}
catch (const std::out_of_range& e)
{
return 0;
}
}
Further comments
If you want to have a fast solution, you should avoid
std::string
altogether. This you can do by "iterating" through the digits using arithmetic operations (division and modulus), as in (usingstd::string
to only show you what is happening):
int x = 1234;
std::string s;
while (x > 0)
{
s.push_back('0' + (x % 10));
x /= 10;
}
std::cout << s << "n"; // Prints 4321
I will let you take it from here to use these ideas to make your program even faster.
Regarding your theoretical question concerning complexity, if we assume that the input is treated as a string of n characters, there is Omega(n) lower bound by a trivial adversary argument. Basically, if you don't spend at least n time, you can't read the whole of the input, and then you cannot guarantee correct output on every instance.
$endgroup$
add a comment |
$begingroup$
General comments
There is no reason to use a class. Instead, the functionality should be made into a free function.
Your code is overly complicated. There is no reason to make new string from which you erase characters one-by-one. Instead, you can convert the input integer to a string and use a standard function to reverse that.
Also, pay attention to const correctness. This protects from unintended mistakes and helps the compiler optimize more.
I would simplify your function to just:
int reverse(int i)
{
try
{
auto reversed{ std::to_string(i) };
std::reverse(reversed.begin(), reversed.end());
const auto result{ std::stoi(reversed) };
return i < 0 ? -1 * result : result;
}
catch (const std::out_of_range& e)
{
return 0;
}
}
Further comments
If you want to have a fast solution, you should avoid
std::string
altogether. This you can do by "iterating" through the digits using arithmetic operations (division and modulus), as in (usingstd::string
to only show you what is happening):
int x = 1234;
std::string s;
while (x > 0)
{
s.push_back('0' + (x % 10));
x /= 10;
}
std::cout << s << "n"; // Prints 4321
I will let you take it from here to use these ideas to make your program even faster.
Regarding your theoretical question concerning complexity, if we assume that the input is treated as a string of n characters, there is Omega(n) lower bound by a trivial adversary argument. Basically, if you don't spend at least n time, you can't read the whole of the input, and then you cannot guarantee correct output on every instance.
$endgroup$
add a comment |
$begingroup$
General comments
There is no reason to use a class. Instead, the functionality should be made into a free function.
Your code is overly complicated. There is no reason to make new string from which you erase characters one-by-one. Instead, you can convert the input integer to a string and use a standard function to reverse that.
Also, pay attention to const correctness. This protects from unintended mistakes and helps the compiler optimize more.
I would simplify your function to just:
int reverse(int i)
{
try
{
auto reversed{ std::to_string(i) };
std::reverse(reversed.begin(), reversed.end());
const auto result{ std::stoi(reversed) };
return i < 0 ? -1 * result : result;
}
catch (const std::out_of_range& e)
{
return 0;
}
}
Further comments
If you want to have a fast solution, you should avoid
std::string
altogether. This you can do by "iterating" through the digits using arithmetic operations (division and modulus), as in (usingstd::string
to only show you what is happening):
int x = 1234;
std::string s;
while (x > 0)
{
s.push_back('0' + (x % 10));
x /= 10;
}
std::cout << s << "n"; // Prints 4321
I will let you take it from here to use these ideas to make your program even faster.
Regarding your theoretical question concerning complexity, if we assume that the input is treated as a string of n characters, there is Omega(n) lower bound by a trivial adversary argument. Basically, if you don't spend at least n time, you can't read the whole of the input, and then you cannot guarantee correct output on every instance.
$endgroup$
General comments
There is no reason to use a class. Instead, the functionality should be made into a free function.
Your code is overly complicated. There is no reason to make new string from which you erase characters one-by-one. Instead, you can convert the input integer to a string and use a standard function to reverse that.
Also, pay attention to const correctness. This protects from unintended mistakes and helps the compiler optimize more.
I would simplify your function to just:
int reverse(int i)
{
try
{
auto reversed{ std::to_string(i) };
std::reverse(reversed.begin(), reversed.end());
const auto result{ std::stoi(reversed) };
return i < 0 ? -1 * result : result;
}
catch (const std::out_of_range& e)
{
return 0;
}
}
Further comments
If you want to have a fast solution, you should avoid
std::string
altogether. This you can do by "iterating" through the digits using arithmetic operations (division and modulus), as in (usingstd::string
to only show you what is happening):
int x = 1234;
std::string s;
while (x > 0)
{
s.push_back('0' + (x % 10));
x /= 10;
}
std::cout << s << "n"; // Prints 4321
I will let you take it from here to use these ideas to make your program even faster.
Regarding your theoretical question concerning complexity, if we assume that the input is treated as a string of n characters, there is Omega(n) lower bound by a trivial adversary argument. Basically, if you don't spend at least n time, you can't read the whole of the input, and then you cannot guarantee correct output on every instance.
edited 2 hours ago
answered 3 hours ago
JuhoJuho
1,461612
1,461612
add a comment |
add a comment |
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$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
3 hours ago
$begingroup$
Yes that's the one
$endgroup$
– greg
3 hours ago