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In this article, we will explore an efficient algorithm to reverse a 32 bit Integer. This involve edge cases where the reverse integer is out of bounds.
Table of contents:
 Problem Statement
 Algorithm to reverse a Number
 Pseudocode
 Code
 Explanation
 Time and Space Complexity
Let us get started with reversing a 32 bit Integer.
Problem Statement
Given a 32 bit number we have to reverse digit of that 32 bit integer and print reversed 32 bit number.
Example:
Input: 34924
Output: 42943
Input: 93432
Output: 23439
Algorithm to reverse a Number
Implementation of this problem is easy.
We will run a while loop till the number is greater than 0
inside while loop we will run following code

reverseNo = reverseNo*10 + last digit of n

reduce number by dividing with 10

If number is negative then we can reverse number by making that positive and at last we can multiply that number with 1 to convert it into negative number

But, we have edge cases here
as in question it is mentioned that reverse number is also 32 bit
Therefore we need to check that the reverse number can be stored in 32 bit int or not. 
If we can store reverse number 32 bit then, we will print that number
else we will print "can't store reverse number in 32 bit" 
Also, of N is INT_MIN (2147483648) then as number is negative so we can convert any negative number to positive by multiplying it with 1
But here positive of N is 2147483648 which can be stored in int as max range of int is 2147483647
So, this is also an exceptional case.
Pseudocode

Take number N as input

Check if N is INT32_MIN
 If Yes,
 print "we can't store reverse of that number in 32 bit"
 return
 If Yes,

define boolean isNegative = false

check whether N is negative or not

If yes,
 then isNegative = true

else,
 isNegative = false

while(N is greater than 0)
 check if reverse>INT32_MAX/10
 if yes,
 print "we can't store reverse of that number in 32 bit"
 return
 reverse = reverse*10 + N%10
 N = N/10

print reverse
Code
Following is the implementation in C++:
#include<bits/stdc++.h>
using namespace std;
void solve() {
int n, reverse=0; cin>>n;
if(n==INT_MIN) {
cout<<"reverse number can't be stored in 32 bits"<<endl;
return;
}
bool isNegative = false;
if(n<0) isNegative = true, n=1*n;
while(n>0) {
if(reverse>INT32_MAX/10) {
cout<<"reverse number can't be stored in 32 bits"<<endl;
return;
}
reverse = reverse*10 + n%10;
n /= 10;
}
if(isNegative) cout<<1*reverse<<endl;
else cout<<reverse<<endl;
}
int main() {
solve();
return 0;
}
Input 1
746384741
Output 1
147483647
Input 2
2147483647
Output 2
reverse number can't be stored in 32 bits
Explanation
lets N = INT32_MAX = 2147483647
 first we will check if the number is
INT_MIN
or not as it is not then move forward  as number is positive so
isNegative = false
 then we will run while loop
 Inside while loop first we will check if number is greater than
INT_MAX/10
Reason for doing this and writing this before performing reverse*10 + n%10
operation is
if number will be greater than INT_MAX/10
and if we multiply that number with 10
then it can be stored in int so its will get overflowed
Example:
if any int K = 2147483647
which is greater then INT_MAX/10=2147483647/10=214748364
if we multiply this by 10 : K = K * 10;
then if we print K then it will show K = 2147483626 in C++
Hence it get overflowed
so we need to first check if N > INT_MAX/10
if it is greater than INT_MAX/10
then we will just break the loop
if N < = INT_MAX/10
then we can run further iterations
for N => INT32_MAX => 2147483647
the following will be the 10 iterations
Initially,n= 2147483647 , reverse: 0
 n= 214748364 , reverse: 7
 n= 21474836 , reverse: 74
 n= 2147483 , reverse: 746
 n= 214748 , reverse: 7463
 n= 21474 , reverse: 74638
 n= 2147 , reverse: 746384
 n= 214 , reverse: 7463847
 n= 21 , reverse: 74638474
 n= 2 , reverse: 746384741
and in 10th
iteration when n= 2 , reverse: 746384741
we will check if (reverse > INT_MAX/10)
=> (746384741 > 2147483647/10)
so answer is true
Hence, we will break the loop and print â€śwe can't reverse the number because we can't store the reverse of N in intâ€ť.
as we all know if we perform operation reverse*10 + n%10 without checking
then,
reverse = 1126087180
Hence if we perform operation without checking reverse will get overflowed
Time and Space Complexity
Time Complexity
 O(No of digits in number)
 As we are traversing loop till all the digits are visited
Space Complexity
 O(1)
 As one variable is created for storing reverse number