Reverse int within the 32-bit signed integer rangeReverse int within the 32-bit signed integer range: $[−2^31, 2^31 − 1]$ OptimizedCombining two 32-bit integers into one 64-bit integerDetermine if an int is within rangeLossy packing 32 bit integer to 16 bitComputing the square root of a 64-bit integerKeeping integer addition within boundsSafe multiplication of two 64-bit signed integersLeetcode 10: Regular Expression MatchingReverse the digits of an Integer“Add two numbers given in reverse order from a linked list”Reverse int within the 32-bit signed integer range: $[−2^31, 2^31 − 1]$ Optimized
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Reverse int within the 32-bit signed integer range
Reverse int within the 32-bit signed integer range: $[−2^31, 2^31 − 1]$ OptimizedCombining two 32-bit integers into one 64-bit integerDetermine if an int is within rangeLossy packing 32 bit integer to 16 bitComputing the square root of a 64-bit integerKeeping integer addition within boundsSafe multiplication of two 64-bit signed integersLeetcode 10: Regular Expression MatchingReverse the digits of an Integer“Add two numbers given in reverse order from a linked list”Reverse int within the 32-bit signed integer range: $[−2^31, 2^31 − 1]$ Optimized
$begingroup$
Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0. Optimized code here.
Feedback
I'm 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!
#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. Optimized code here.
Feedback
I'm 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!
#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
Mar 23 at 18:39
$begingroup$
Yes that's the one
$endgroup$
– greg
Mar 23 at 19:03
$begingroup$
The problem is ill-posed. If 'the' reverse of120
is21
, how can you know whether 'the' reverse of21
should be120
or12
? What makes 'the' reverse of1
number1
and not10000
? How can you tell your code solves the problem if the correct solution is not defined?
$endgroup$
– CiaPan
yesterday
add a comment |
$begingroup$
Problem
Reverse digits of a 32-bit signed integer. When the reversed integer overflows return 0. Optimized code here.
Feedback
I'm 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!
#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. Optimized code here.
Feedback
I'm 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!
#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 15 mins ago
Jamal♦
30.4k11121227
30.4k11121227
asked Mar 23 at 18:31
greggreg
39928
39928
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
Mar 23 at 18:39
$begingroup$
Yes that's the one
$endgroup$
– greg
Mar 23 at 19:03
$begingroup$
The problem is ill-posed. If 'the' reverse of120
is21
, how can you know whether 'the' reverse of21
should be120
or12
? What makes 'the' reverse of1
number1
and not10000
? How can you tell your code solves the problem if the correct solution is not defined?
$endgroup$
– CiaPan
yesterday
add a comment |
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
Mar 23 at 18:39
$begingroup$
Yes that's the one
$endgroup$
– greg
Mar 23 at 19:03
$begingroup$
The problem is ill-posed. If 'the' reverse of120
is21
, how can you know whether 'the' reverse of21
should be120
or12
? What makes 'the' reverse of1
number1
and not10000
? How can you tell your code solves the problem if the correct solution is not defined?
$endgroup$
– CiaPan
yesterday
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
Mar 23 at 18:39
$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
Mar 23 at 18:39
$begingroup$
Yes that's the one
$endgroup$
– greg
Mar 23 at 19:03
$begingroup$
Yes that's the one
$endgroup$
– greg
Mar 23 at 19:03
$begingroup$
The problem is ill-posed. If 'the' reverse of
120
is 21
, how can you know whether 'the' reverse of 21
should be 120
or 12
? What makes 'the' reverse of 1
number 1
and not 10000
? How can you tell your code solves the problem if the correct solution is not defined?$endgroup$
– CiaPan
yesterday
$begingroup$
The problem is ill-posed. If 'the' reverse of
120
is 21
, how can you know whether 'the' reverse of 21
should be 120
or 12
? What makes 'the' reverse of 1
number 1
and not 10000
? How can you tell your code solves the problem if the correct solution is not defined?$endgroup$
– CiaPan
yesterday
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 4321I 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.
$endgroup$
$begingroup$
Trying the digit iteration now, great concept that I'm adding to my toolbox. Fantastic feedback.
$endgroup$
– greg
Mar 25 at 1:59
$begingroup$
if you'd like to see the optimized version I've edited my post
$endgroup$
– greg
2 days ago
1
$begingroup$
@greg I'd be happy to comment on that as well, but you shouldn't change your code after you've received an answer. Instead, you should post it as a separate question.
$endgroup$
– Juho
2 days ago
1
$begingroup$
@esote Great, thanks for the edits! I know how to use Latex now on this site too :-)
$endgroup$
– Juho
2 days ago
$begingroup$
Updated the code here codereview.stackexchange.com/questions/216300/… learning math operations to pop and push digits has been very valuable in optimizing this
$endgroup$
– greg
yesterday
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
votes
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oldest
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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 4321I 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.
$endgroup$
$begingroup$
Trying the digit iteration now, great concept that I'm adding to my toolbox. Fantastic feedback.
$endgroup$
– greg
Mar 25 at 1:59
$begingroup$
if you'd like to see the optimized version I've edited my post
$endgroup$
– greg
2 days ago
1
$begingroup$
@greg I'd be happy to comment on that as well, but you shouldn't change your code after you've received an answer. Instead, you should post it as a separate question.
$endgroup$
– Juho
2 days ago
1
$begingroup$
@esote Great, thanks for the edits! I know how to use Latex now on this site too :-)
$endgroup$
– Juho
2 days ago
$begingroup$
Updated the code here codereview.stackexchange.com/questions/216300/… learning math operations to pop and push digits has been very valuable in optimizing this
$endgroup$
– greg
yesterday
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 4321I 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.
$endgroup$
$begingroup$
Trying the digit iteration now, great concept that I'm adding to my toolbox. Fantastic feedback.
$endgroup$
– greg
Mar 25 at 1:59
$begingroup$
if you'd like to see the optimized version I've edited my post
$endgroup$
– greg
2 days ago
1
$begingroup$
@greg I'd be happy to comment on that as well, but you shouldn't change your code after you've received an answer. Instead, you should post it as a separate question.
$endgroup$
– Juho
2 days ago
1
$begingroup$
@esote Great, thanks for the edits! I know how to use Latex now on this site too :-)
$endgroup$
– Juho
2 days ago
$begingroup$
Updated the code here codereview.stackexchange.com/questions/216300/… learning math operations to pop and push digits has been very valuable in optimizing this
$endgroup$
– greg
yesterday
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 4321I 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 4321I 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 days ago
esote
2,89611038
2,89611038
answered Mar 23 at 18:50
JuhoJuho
1,536612
1,536612
$begingroup$
Trying the digit iteration now, great concept that I'm adding to my toolbox. Fantastic feedback.
$endgroup$
– greg
Mar 25 at 1:59
$begingroup$
if you'd like to see the optimized version I've edited my post
$endgroup$
– greg
2 days ago
1
$begingroup$
@greg I'd be happy to comment on that as well, but you shouldn't change your code after you've received an answer. Instead, you should post it as a separate question.
$endgroup$
– Juho
2 days ago
1
$begingroup$
@esote Great, thanks for the edits! I know how to use Latex now on this site too :-)
$endgroup$
– Juho
2 days ago
$begingroup$
Updated the code here codereview.stackexchange.com/questions/216300/… learning math operations to pop and push digits has been very valuable in optimizing this
$endgroup$
– greg
yesterday
add a comment |
$begingroup$
Trying the digit iteration now, great concept that I'm adding to my toolbox. Fantastic feedback.
$endgroup$
– greg
Mar 25 at 1:59
$begingroup$
if you'd like to see the optimized version I've edited my post
$endgroup$
– greg
2 days ago
1
$begingroup$
@greg I'd be happy to comment on that as well, but you shouldn't change your code after you've received an answer. Instead, you should post it as a separate question.
$endgroup$
– Juho
2 days ago
1
$begingroup$
@esote Great, thanks for the edits! I know how to use Latex now on this site too :-)
$endgroup$
– Juho
2 days ago
$begingroup$
Updated the code here codereview.stackexchange.com/questions/216300/… learning math operations to pop and push digits has been very valuable in optimizing this
$endgroup$
– greg
yesterday
$begingroup$
Trying the digit iteration now, great concept that I'm adding to my toolbox. Fantastic feedback.
$endgroup$
– greg
Mar 25 at 1:59
$begingroup$
Trying the digit iteration now, great concept that I'm adding to my toolbox. Fantastic feedback.
$endgroup$
– greg
Mar 25 at 1:59
$begingroup$
if you'd like to see the optimized version I've edited my post
$endgroup$
– greg
2 days ago
$begingroup$
if you'd like to see the optimized version I've edited my post
$endgroup$
– greg
2 days ago
1
1
$begingroup$
@greg I'd be happy to comment on that as well, but you shouldn't change your code after you've received an answer. Instead, you should post it as a separate question.
$endgroup$
– Juho
2 days ago
$begingroup$
@greg I'd be happy to comment on that as well, but you shouldn't change your code after you've received an answer. Instead, you should post it as a separate question.
$endgroup$
– Juho
2 days ago
1
1
$begingroup$
@esote Great, thanks for the edits! I know how to use Latex now on this site too :-)
$endgroup$
– Juho
2 days ago
$begingroup$
@esote Great, thanks for the edits! I know how to use Latex now on this site too :-)
$endgroup$
– Juho
2 days ago
$begingroup$
Updated the code here codereview.stackexchange.com/questions/216300/… learning math operations to pop and push digits has been very valuable in optimizing this
$endgroup$
– greg
yesterday
$begingroup$
Updated the code here codereview.stackexchange.com/questions/216300/… learning math operations to pop and push digits has been very valuable in optimizing this
$endgroup$
– greg
yesterday
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$begingroup$
This leetcode.com/problems/reverse-integer ?
$endgroup$
– Martin R
Mar 23 at 18:39
$begingroup$
Yes that's the one
$endgroup$
– greg
Mar 23 at 19:03
$begingroup$
The problem is ill-posed. If 'the' reverse of
120
is21
, how can you know whether 'the' reverse of21
should be120
or12
? What makes 'the' reverse of1
number1
and not10000
? How can you tell your code solves the problem if the correct solution is not defined?$endgroup$
– CiaPan
yesterday