Reverse Integer

Easy
0/40
Average time to solve is 20m
profile
Contributed by
21 upvotes
Asked in companies
AppleAdobeFreshworks

Problem statement

You are given a 32-bit signed integer ‘N’. So, the integer will lie in the range [-2^31, 2^31 - 1]. Your task is to return the reverse of the given integer. If reversing causes overflow, then return -1.

Note:

(1) Do not use data types with the capacity of more than 32-bit like ‘long int’ or ‘long long int’ in C++. The problem is meant to reverse the integer using a 32-bit data type only.

(2) You should assume that the environment does not allow storing signed or unsigned 64-bit integers.
Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line of input contains an integer 'T' representing the number of test cases.

The first and the only line of each test case contains a 32-bit signed integer ‘N’ representing the integer to be reversed.

Output Format:

For each test case, return the reverse of integer, ’N’. 

The output of each test case will be printed in a separate line.
Note: 
You do not need to print anything, it has already been taken care of. Just implement the given function.
Constraints: 
1 <= T <= 5
-2^31 <= N <= 2^31 - 1   

Time limit: 1 sec
Sample Input 1:
2
8596396
1253489646
Sample Output 1:
6936958
-1
Explanation of Sample Output 1:
In test case 1, Given ‘N’ = 8596396

On reversing the digits of ‘N’, we will get 6936958. So the output is 6936958.

In test case 2, Given ‘N’ = 1253489646

On reversing the digits of ‘N’, we will get 6469843521 which is greater than 2^31 - 1. Thus, it will cause an overflow. So the output is -1.
Sample Input 2:
2
-803
-1
Sample Output 2:
-308
-1
Explanation of Sample Output 1:
In test case 1, Given ‘N’ = -803

On reversing the digits of ‘N’, we will get -308. So the output is -308.

In test case 2, Given ‘N’ = -1

On reversing the digits of ‘N’, we will get -1. So the output is -1.
Hint

Can you exploit the fact that the given number is a 32 bit signed integer?

Approaches (1)
Implementation

The idea is to use the fact that the input ‘N’ is a 32 bit signed integer.  So, as mentioned, -2147483648 <= ‘N’ <= 2147483647. In this range, for ‘N’ = 10  *  x + y, the ‘N’ can never attain a value when |x| = INT_MAX / 10, y = 8 or 9 for x > 0 and y = -9, for x < 0.

 

Therefore, it is sufficient to check |x| > INT_MAX / 10 (i.e. x > INT_MAX / 10 or x < INT_MIN / 10) for overflow.

 

The steps are as follows:

 

  • Declare an integer variable, ‘RESULT’, and initialized it to zero, to store the reversed value of ‘N’.
  • Run a loop while ‘N’ is not zero.
    • Extract the last digit of ‘N’ and store it in an integer variable, ‘LASTDIGIT’.
    • Check for overflow. If ‘RESULT’ is greater than INT_MAX / 10’ or ‘RESULT’ is less than INT_MIN / 10’ that shows that appending the ‘LASTDIGIT’ to the ‘RESULT’ will cause an overflow. So, return -1.
    • Else, append the ‘lastDigit’ to the ‘RESULT’.
    • Divide ‘N’ by 10 to remove the last digit.
  • Return the ‘RESULT’.
Time Complexity

O(log | N |), Where ‘N’ is the 32-bit signed integer.

 

Since number ‘N’ has approximately log|N| digits. Since we are dividing the integer ‘N’ in each iteration of the while loop, the while loop will execute log|N| times. Thus the time complexity will be O(log | N |).

Space Complexity

O(1).

 

Since no auxiliary space has been used to perform the reversal. Thus the space complexity will be O(1).

Code Solution
(100% EXP penalty)
Reverse Integer
Full screen
Console