Write a program to reverse an integer assuming that the input is a 32-bit integer. If the reversed integer overflows, print -1 as the output.
Let us see a simple approach to reverse digits of an integer.
C++
// A simple C program to reverse digits of // an integer. #include <stdio.h> int reversDigits( int num) { int rev_num = 0; while (num > 0) { rev_num = rev_num*10 + num%10; num = num/10; } return rev_num; } /* Driver program to test reversDigits */ int main() { int num = 5896; printf ( "Reverse of no. is %d" , reversDigits(num)); return 0; } |
Java
// Java program to reverse a number class GFG { /* Iterative function to reverse digits of num*/ static int reversDigits( int num) { int rev_num = 0 ; while (num > 0 ) { rev_num = rev_num * 10 + num % 10 ; num = num / 10 ; } return rev_num; } // Driver code public static void main (String[] args) { int num = 4562 ; System.out.println( "Reverse of no. is " + reversDigits(num)); } } // This code is contributed by Anant Agarwal. |
Python
# Python program to reverse a number n = 4562 ; rev = 0 while (n > 0 ): a = n % 10 rev = rev * 10 + a n = n / 10 print (rev) # This code is contributed by Shariq Raza |
C#
// C# program to reverse a number using System; class GFG { // Iterative function to // reverse digits of num static int reversDigits( int num) { int rev_num = 0; while (num > 0) { rev_num = rev_num * 10 + num % 10; num = num / 10; } return rev_num; } // Driver code public static void Main() { int num = 4562; Console.Write( "Reverse of no. is " + reversDigits(num)); } } // This code is contributed by Sam007 |
PHP
<?php // Iterative function to // reverse digits of num function reversDigits( $num ) { $rev_num = 0; while ( $num > 1) { $rev_num = $rev_num * 10 + $num % 10; $num = (int) $num / 10; } return $rev_num ; } // Driver Code $num = 4562; echo "Reverse of no. is " , reversDigits( $num ); // This code is contributed by aj_36 ?> |
Output:
6985
However, if the number is large such that the reverse overflows, the output is some garbage value. If we run the code above with input as any large number say 1000000045, then the output is some garbage value like 1105032705 or any other garbage value. See this for the output.
How to handle overflow?
The idea is to store previous value of the sum can be stored in a variable which can be checked every time to see if the reverse overflowed or not.
Below is the implementation to deal with such a situation.
C++
// C++ program to reverse digits // of a number #include <bits/stdc++.h> using namespace std; /* Iterative function to reverse digits of num*/ int reversDigits( int num) { // Handling negative numbers bool negativeFlag = false ; if (num < 0) { negativeFlag = true ; num = -num ; } int prev_rev_num = 0, rev_num = 0; while (num != 0) { int curr_digit = num % 10; rev_num = (rev_num * 10) + curr_digit; // checking if the reverse overflowed or not. // The values of (rev_num - curr_digit)/10 and // prev_rev_num must be same if there was no // problem. if ((rev_num - curr_digit) / 10 != prev_rev_num) { cout << "WARNING OVERFLOWED!!!" << endl; return 0; } prev_rev_num = rev_num; num = num / 10; } return (negativeFlag == true ) ? -rev_num : rev_num; } // Driver Code int main() { int num = 12345; cout << "Reverse of no. is " << reversDigits(num) << endl; num = 1000000045; cout << "Reverse of no. is " << reversDigits(num) << endl; return 0; } // This code is contributed // by Akanksha Rai(Abby_akku) |
C
// C program to reverse digits of a number #include <stdio.h> /* Iterative function to reverse digits of num*/ int reversDigits( int num) { // Handling negative numbers bool negativeFlag = false ; if (num < 0) { negativeFlag = true ; num = -num ; } int prev_rev_num = 0, rev_num = 0; while (num != 0) { int curr_digit = num%10; rev_num = (rev_num*10) + curr_digit; // checking if the reverse overflowed or not. // The values of (rev_num - curr_digit)/10 and // prev_rev_num must be same if there was no // problem. if ((rev_num - curr_digit)/10 != prev_rev_num) { printf ( "WARNING OVERFLOWED!!!\n" ); return 0; } prev_rev_num = rev_num; num = num/10; } return (negativeFlag == true )? -rev_num : rev_num; } /* Driver program to test reverse Digits */ int main() { int num = 12345; printf ( "Reverse of no. is %d\n" , reversDigits(num)); num = 1000000045; printf ( "Reverse of no. is %d\n" , reversDigits(num)); return 0; } |
Java
// Java program to reverse digits of a number class ReverseDigits { /* Iterative function to reverse digits of num*/ static int reversDigits( int num) { // Handling negative numbers boolean negativeFlag = false ; if (num < 0 ) { negativeFlag = true ; num = -num ; } int prev_rev_num = 0 , rev_num = 0 ; while (num != 0 ) { int curr_digit = num% 10 ; rev_num = (rev_num* 10 ) + curr_digit; // checking if the reverse overflowed or not. // The values of (rev_num - curr_digit)/10 and // prev_rev_num must be same if there was no // problem. if ((rev_num - curr_digit)/ 10 != prev_rev_num) { System.out.println( "WARNING OVERFLOWED!!!" ); return 0 ; } prev_rev_num = rev_num; num = num/ 10 ; } return (negativeFlag == true )? -rev_num : rev_num; } public static void main (String[] args) { int num = 12345 ; System.out.println( "Reverse of no. is " + reversDigits(num)); num = 1000000045 ; System.out.println( "Reverse of no. is " + reversDigits(num)); } } |
Python3
# Python program to reverse digits # of a number """ Iterative function to reverse digits of num""" def reversDigits(num): # Handling negative numbers negativeFlag = False if (num < 0 ): negativeFlag = True num = - num prev_rev_num = 0 rev_num = 0 while (num ! = 0 ): curr_digit = num % 10 rev_num = (rev_num * 10 ) + curr_digit # checking if the reverse overflowed or not. # The values of (rev_num - curr_digit)/10 and # prev_rev_num must be same if there was no # problem. if (rev_num > = 2147483647 or rev_num < = - 2147483648 ): rev_num = 0 if ((rev_num - curr_digit) / / 10 ! = prev_rev_num): print ( "WARNING OVERFLOWED!!!" ) return 0 prev_rev_num = rev_num num = num / / 10 return - rev_num if (negativeFlag = = True ) else rev_num # Driver code if __name__ = = "__main__" : num = 12345 print ( "Reverse of no. is " ,reversDigits(num)) num = 1000000045 print ( "Reverse of no. is " ,reversDigits(num)) # This code is contributed # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# program to reverse digits // of a number using System; class GFG { /* Iterative function to reverse digits of num*/ static int reversDigits( int num) { // Handling negative numbers bool negativeFlag = false ; if (num < 0) { negativeFlag = true ; num = -num ; } int prev_rev_num = 0, rev_num = 0; while (num != 0) { int curr_digit = num % 10; rev_num = (rev_num * 10) + curr_digit; // checking if the reverse overflowed // or not. The values of (rev_num - // curr_digit)/10 and prev_rev_num must // be same if there was no problem. if ((rev_num - curr_digit) / 10 != prev_rev_num) { Console.WriteLine( "WARNING OVERFLOWED!!!" ); return 0; } prev_rev_num = rev_num; num = num / 10; } return (negativeFlag == true ) ? -rev_num : rev_num; } // Driver Code static public void Main () { int num = 12345; Console.WriteLine( "Reverse of no. is " + reversDigits(num)); num = 1000000045; Console.WriteLine( "Reverse of no. is " + reversDigits(num)); } } // This code is contributed by ajit |
Output:
Reverse of no. is 54321 WARNING OVERFLOWED!!! Reverse of no. is 0
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