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 <bits/stdc++.h>using namespace std;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; cout << "Reverse of no. is " << reversDigits(num); return 0;}// This code is contributed by Sania Kumari Gupta (kriSania804)// This code is improved by Md. Owais Ashraf (professor_011) |
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;}// This code is contributed by Sania Kumari Gupta (kriSania804)// This code is improved by Md. Owais Ashraf (professor_011) |
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 = 5896; System.out.println("Reverse of no. is " + reversDigits(num)); }}// This code is contributed by Anant Agarwal.// This code is improved by Md. Owais Ashraf (professor_011)// This code is improved by Jeet Purohit (jeetpurohit989) |
Python3
# Python program to reverse a number n = 5896;rev = 0while(n != 0): a = n % 10 rev = rev * 10 + a n = n // 10 print(rev)# This code is contributed by Shariq Raza # This code is improved by Jeet Purohit (jeetpurohit989) |
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 = 5896; Console.Write("Reverse of no. is " + reversDigits(num)); }}// This code is contributed by Sam007// This code is improved by Md. Owais Ashraf (professor_011) |
PHP
<?php// Iterative function to // reverse digits of numfunction reversDigits($num){ $rev_num = 0; while($num != 0) { $rev_num = $rev_num * 10 + $num % 10; $num = (int)$num / 10; } return $rev_num/10;}// Driver Code$num = 5896;echo "Reverse of no. is ", reversDigits($num);// This code is contributed by aj_36// This code is improved by Md. Owais Ashraf (professor_011)// This code is improved by Jeet Purohit (jeetpurohit989)?> |
Javascript
<script>// A simple Javascript program to reverse digits of // an integer. function reversDigits(num) { let rev_num = 0; while (num != 0) { rev_num = rev_num*10 + num%10; num = Math.floor(num/10); } return rev_num; } /* Driver program to test reversDigits */ let num = 5896; document.write("Reverse of no. is "+ reversDigits(num)); // This code is contributed by Mayank Tyagi// This code is improved by Md. Owais Ashraf (professor_011)// This code is improved by Jeet Purohit (jeetpurohit989)</script> |
Output
Reverse of no. is 6985
Time Complexity: O(log(num))
Auxiliary Space: O(1)
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 the previous value of the sum can be stored in a variable that 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 Codeint 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 numberclass 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 numberusing 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 Codestatic 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 |
Javascript
<script>// JavaScript program to reverse digits// of a number/* Iterative function to reversedigits of num*/function reversDigits(num){ // Handling negative numbers let negativeFlag = false; if (num < 0) { negativeFlag = true; num = -num ; } let prev_rev_num = 0, rev_num = 0; while (num != 0) { let 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 || rev_num <= -2147483648) rev_num = 0; if (Math.floor((rev_num - curr_digit) / 10) != prev_rev_num) { document.write("WARNING OVERFLOWED!!!" + "<br>"); return 0; } prev_rev_num = rev_num; num = Math.floor(num / 10); } return (negativeFlag == true) ? -rev_num : rev_num;}// Driver Code let num = 12345; document.write("Reverse of no. is " + reversDigits(num) + "<br>"); num = 1000000045; document.write("Reverse of no. is " + reversDigits(num) + "<br>");// This code is contributed by Surbhi Tyagi.</script> |
Output
Reverse of no. is 54321 Reverse of no. is 1105032705
Time Complexity: O(log(num))
Auxiliary Space: O(1)
Efficient Approach :
The above approach won’t work if we are given a signed 32-bit integer x, and return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 – 1], then return 0. So we cannot multiply the number*10 and then check if the number overflows or not.
We must check the overflow condition before multiplying by 10 by using the following logic :
You are checking the boundary case before you do the operation. (reversed >INT_MAX ) wouldn’t work because reversed will overflow and become negative if it goes past MAX_VALUE. Dividing MAX_VALUE by 10 lets you check the condition without overflowing INT_MAX is equal 2147483647. INT_MIN is equal -2147483648. The last digits are 7 and 8. This is the reason why we also check rem > 7 and rem < -8 conditions
C++
// C++ program to reverse digits // of a number#include <bits/stdc++.h>using namespace std;int reversDigits(int num) { int rev = 0 ; while(num != 0){ int rem = num % 10 ; num /= 10 ; if(rev > INT_MAX/10 || rev == INT_MAX/10 && rem > 7){ return 0 ; } if(rev < INT_MIN/10 || rev == INT_MIN/10 && rem < -8){ return 0 ; } rev = rev*10 + rem ; } return rev ; }// Driver Codeint main(){ int num = 12345; cout << "Reverse of no. is " << reversDigits(num) << endl; num = 1000000045; cout << "Reverse of no. is " << reversDigits(num) << endl; return 0;} |
Java
// Java program to reverse digits// of a numberpublic class GFG{ static int reversDigits(int num) { int rev = 0 ; while(num != 0){ int rem = num % 10 ; num /= 10 ; if(rev > Integer.MAX_VALUE/10 || rev == Integer.MAX_VALUE/10 && rem > 7){ return 0 ; } if(rev < Integer.MIN_VALUE/10 || rev == Integer.MIN_VALUE/10 && rem < -8){ return 0 ; } rev = rev*10 + rem ; } return rev ; } // Driver code 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) ); }}// This code is contributed by jana_sayantan. |
Python3
def reverse_digits(num): # Initialize reverse number as 0 rev = 0 # Check sign of the number sign = -1 if num < 0 else 1 # Take absolute value of the number num = abs(num) # Reverse the digits while num != 0: # Get the last digit rem = num % 10 # Remove the last digit from the number num = num // 10 # Check if the reverse number will overflow 32-bit integer if rev > 2**31//10 or (rev == 2**31//10 and rem > 7): return 0 # Check if the reverse number will underflow 32-bit integer if rev < -2**31//10 or (rev == -2**31//10 and rem < -8): return 0 # Update the reverse number rev = rev * 10 + rem # Return the reverse number with the original sign return sign * revif __name__ == '__main__': num = 12345 print("Reverse of no. is", reverse_digits(num)) num = 1000000045 print("Reverse of no. is", reverse_digits(num)) |
C#
// C# program to reverse digits// of a numberusing System;public class GFG{ static int reversDigits(int num) { int rev = 0 ; while(num != 0){ int rem = num % 10 ; num /= 10 ; if(rev > Int32.MaxValue/10 || rev == Int32.MaxValue/10 && rem > 7){ return 0 ; } if(rev < Int32.MinValue/10 || rev == Int32.MinValue/10 && rem < -8){ return 0 ; } rev = rev*10 + rem ; } return rev ; } // Driver code public static void Main (string[] args) { 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 phasing17 |
Javascript
<script>// JavaScript program to reverse digits // of a numberconst INT_MAX = 2147483647;const INT_MIN = -2147483648;function reversDigits(num) { let rev = 0; while(num > 0){ let rem = num % 10 ; num = Math.floor(num/10) ; if(rev > INT_MAX/10 || rev == INT_MAX/10 && rem > 7){ return 0 ; } if(rev < INT_MIN/10 || rev == INT_MIN/10 && rem < -8){ return 0 ; } rev = rev*10 + rem ; } return rev ; }// Driver Codelet num = 12345;document.write(`Reverse of no. is ${reversDigits(num)}`,"</br>");num = 1000000045;document.write(`Reverse of no. is ${reversDigits(num)}`,"</br>");// This code is contributed by shinjanpatra</script> |
Output
Reverse of no. is 54321 Reverse of no. is 0
Time Complexity: O(log(num))
Auxiliary Space: O(1)
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METHOD 4:Using the slice approach
APPROACH:
We convert the integer to a string, reverse the string using the slice operator, and then convert the reversed string back to an integer.
ALGORITHM:
1. Convert the integer to a string using the str() function
2.Use the slice operator [::-1] to reverse the string
3. Convert the reversed string back to an integer using the int() function
Python3
def reverse_num_string_slice(num): return int(str(num)[::-1])num = 12345rev_num_string_slice = reverse_num_string_slice(num)print(f"Reverse of {num} is {rev_num_string_slice}") |
Output
Reverse of 12345 is 54321
Time complexity: O(n), where n is the number of digits in the integer.
Space complexity: O(n), where n is the number of digits in the integer, since we need to create a string of length n to store the reversed integer.
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