Replace all ‘0’ with ‘5’ in an input Integer
Last Updated :
23 Jul, 2025
Given an integer as input and replace all the ‘0’ with ‘5’ in the integer.
Examples:
Input: 102
Output: 152
Explanation: All the digits which are '0' is replaced by '5'
Input: 1020
Output: 1525
Explanation: All the digits which are '0' is replaced by '5'
The use of an array to store all digits is not allowed.
Iterative Approach-1: By observing the test cases it is evident that all the 0 digits are replaced by 5. For Example, for input = 1020, output = 1525. The idea is simple, we assign a variable 'temp' to 0, we get the last digit using mod operator '%'. If the digit is 0, we replace it with 5, otherwise, keep it as it is. Then we multiply the 'temp' with 10 and add the digit got by mod operation. After that, we divide the original number by 10 to get the other digits. In this way, we will have a number in which all the '0's are assigned with '5's. If we reverse this number, we will get the desired answer.
Algorithm:
- if the number is 0, directly return 5.
- else do the steps below.
- Create a variable temp= 0 to store the reversed number having all '0's assigned to '5's.
- Find the last digit using the mod operator '%'. If the digit is '0', then make the last digit '5'.
- Multiply temp with 10 and add the last digit.
- Divide the number by 10 to get more digits by mod operation.
- Then reverse this number.
- return the reversed number.
Implementation:
C++
// C++ program to replace all ‘0’
// with ‘5’ in an input Integer
#include <iostream>
using namespace std;
// A iterative function to reverse a number
int reverseTheNumber(int temp)
{
int ans = 0;
while (temp > 0) {
int rem = temp % 10;
ans = ans * 10 + rem;
temp = temp / 10;
}
return ans;
}
int convert0To5(int num)
{
// if num is 0 return 5
if (num == 0)
return 5;
// Extract the last digit and
// change it if needed
else {
int temp = 0;
while (num > 0) {
int digit = num % 10;
//if digit is 0, make it 5
if (digit == 0)
digit = 5;
temp = temp * 10 + digit;
num = num / 10;
}
// call the function reverseTheNumber by passing
// temp
return reverseTheNumber(temp);
}
}
// Driver code
int main()
{
int num = 10120;
cout << convert0To5(num);
return 0;
}
// This code is contributed by Vrashank Rao M.
// Java program for the above approach
import java.io.*;
class GFG {
// A iterative function to reverse a number
static int reverseTheNumber(int temp)
{
int ans = 0;
while (temp > 0) {
int rem = temp % 10;
ans = ans * 10 + rem;
temp = temp / 10;
}
return ans;
}
static int convert0To5(int num)
{
// if num is 0 return 5
if (num == 0)
return 5;
// Extract the last digit and
// change it if needed
else {
int temp = 0;
while (num > 0) {
int digit = num % 10;
//if digit is 0, make it 5
if (digit == 0)
digit = 5;
temp = temp * 10 + digit;
num = num / 10;
}
// call the function reverseTheNumber by passing
// temp
return reverseTheNumber(temp);
}
}
// Driver program
public static void main(String args[])
{
int num = 10120;
System.out.println(convert0To5(num));
}
}
// This code is contributed by sanjoy_62.
# Python program for the above approach
# A iterative function to reverse a number
def reverseTheNumber(temp):
ans = 0;
while (temp > 0):
rem = temp % 10;
ans = ans * 10 + rem;
temp = temp // 10;
return ans;
def convert0To5(num):
# if num is 0 return 5
if (num == 0):
return 5;
# Extract the last digit and
# change it if needed
else:
temp = 0;
while (num > 0):
digit = num % 10;
# if digit is 0, make it 5
if (digit == 0):
digit = 5;
temp = temp * 10 + digit;
num = num // 10;
# call the function reverseTheNumber by passing
# temp
return reverseTheNumber(temp);
# Driver program
if __name__ == '__main__':
num = 10120;
print(convert0To5(num));
# This code is contributed by umadevi9616
// C# program for the above approach
using System;
public class GFG {
// A iterative function to reverse a number
static int reverseTheNumber(int temp)
{
int ans = 0;
while (temp > 0) {
int rem = temp % 10;
ans = ans * 10 + rem;
temp = temp / 10;
}
return ans;
}
static int convert0To5(int num)
{
// if num is 0 return 5
if (num == 0)
return 5;
// Extract the last digit and
// change it if needed
else {
int temp = 0;
while (num > 0) {
int digit = num % 10;
//if digit is 0, make it 5
if (digit == 0)
digit = 5;
temp = temp * 10 + digit;
num = num / 10;
}
// call the function reverseTheNumber by passing
// temp
return reverseTheNumber(temp);
}
}
// Driver Code
public static void Main (string[] args) {
int num = 10120;
Console.Write(convert0To5(num));
}
}
// This code is contributed by splevel62.
<script>
// javascript program for the above approach
// A iterative function to reverse a number
function reverseTheNumber(temp) {
var ans = 0;
while (temp > 0) {
var rem = temp % 10;
ans = ans * 10 + rem;
temp = parseInt(temp / 10);
}
return ans;
}
function convert0To5(num)
{
// if num is 0 return 5
if (num == 0)
return 5;
// Extract the last digit and
// change it if needed
else {
var temp = 0;
while (num > 0) {
var digit = num % 10;
// if digit is 0, make it 5
if (digit == 0)
digit = 5;
temp = temp * 10 + digit;
num = parseInt(num / 10);
}
// call the function reverseTheNumber by passing
// temp
return reverseTheNumber(temp);
}
}
// Driver program
var num = 10120;
document.write(convert0To5(num));
// This code is contributed by umadevi9616
</script>
Complexity Analysis:
- Time Complexity: O(n), where n is number of digits in the number.
- Auxiliary Space: O(1), no extra space is required.
Iterative Approach-2: By observing the test cases it is evident that all the 0 digits are replaced by 5. For Example, for input = 1020, output = 1525, which can be written as 1020 + 505, which can be further written as 1020 + 5*(10^2) + 5*(10^0). So the solution can be formed in an iterative way where if a '0' digit is encountered find the place value of that digit and multiply it with 5 and find the sum for all 0's in the number. Add that sum to the input number to find the output number.
Algorithm:
- Create a variable sum = 0 to store the sum, place = 1 to store the place value of the current digit, and create a copy of the input variable
- If the number is zero return 5
- Iterate the next step while the input variable is greater than 0
- Extract the last digit (n%10) and if the digit is zero, then update sum = sum + place*5, remove the last digit from the number n = n/10 and update place = place * 10
- Return the sum.
Implementation:
C++
#include<bits/stdc++.h>
using namespace std;
// Returns the number to be added to the
// input to replace all zeroes with five
int calculateAddedValue(int number)
{
// Amount to be added
int result = 0;
// Unit decimal place
int decimalPlace = 1;
if (number == 0)
{
result += (5 * decimalPlace);
}
while (number > 0)
{
if (number % 10 == 0)
{
// A number divisible by 10, then
// this is a zero occurrence in
// the input
result += (5 * decimalPlace);
}
// Move one decimal place
number /= 10;
decimalPlace *= 10;
}
return result;
}
int replace0with5(int number)
{
return number += calculateAddedValue(number);
}
// Driver code
int main()
{
cout << replace0with5(1020);
}
// This code is contributed by avanitrachhadiya2155
import java.io.*;
public class ReplaceDigits {
static int replace0with5(int number)
{
return number += calculateAddedValue(number);
}
// returns the number to be added to the
// input to replace all zeroes with five
private static int calculateAddedValue(int number)
{
// amount to be added
int result = 0;
// unit decimal place
int decimalPlace = 1;
if (number == 0) {
result += (5 * decimalPlace);
}
while (number > 0) {
if (number % 10 == 0)
// a number divisible by 10, then
// this is a zero occurrence in the input
result += (5 * decimalPlace);
// move one decimal place
number /= 10;
decimalPlace *= 10;
}
return result;
}
public static void main(String[] args)
{
System.out.print(replace0with5(1020));
}
}
def replace0with5(number):
number += calculateAddedValue(number)
return number
# returns the number to be added to the
# input to replace all zeroes with five
def calculateAddedValue(number):
# amount to be added
result = 0
# unit decimal place
decimalPlace = 1
if (number == 0):
result += (5 * decimalPlace)
while (number > 0):
if (number % 10 == 0):
# a number divisible by 10, then
# this is a zero occurrence in the input
result += (5 * decimalPlace)
# move one decimal place
number //= 10
decimalPlace *= 10
return result
# Driver code
print(replace0with5(1020))
# This code is contributed by shubhmasingh10
using System;
class GFG{
static int replace0with5(int number)
{
return number += calculateAddedValue(number);
}
// Returns the number to be added to the
// input to replace all zeroes with five
static int calculateAddedValue(int number)
{
// Amount to be added
int result = 0;
// Unit decimal place
int decimalPlace = 1;
if (number == 0)
{
result += (5 * decimalPlace);
}
while (number > 0)
{
if (number % 10 == 0)
// A number divisible by 10, then
// this is a zero occurrence in the input
result += (5 * decimalPlace);
// Move one decimal place
number /= 10;
decimalPlace *= 10;
}
return result;
}
// Driver Code
static public void Main()
{
Console.WriteLine(replace0with5(1020));
}
}
// This code is contributed by rag2127
<script>
// Returns the number to be added to the
// input to replace all zeroes with five
function calculateAddedValue(number){
// Amount to be added
let result = 0;
// Unit decimal place
let decimalPlace = 1;
if (number == 0) {
result += (5 * decimalPlace);
}
while (number > 0){
if (number % 10 == 0){
// A number divisible by 10, then
// this is a zero occurrence in
// the input
result += (5 * decimalPlace);
}
// Move one decimal place
number = Math.floor(number/10);
decimalPlace *= 10;
}
return result;
}
function replace0with5(number){
return number += calculateAddedValue(number);
}
// Driver code
document.write(replace0with5(1020));
</script>
Complexity Analysis:
- Time Complexity: O(k), the loops run only k times, where k is the number of digits of the number.
- Auxiliary Space: O(1), no extra space is required.
Recursive Approach: The idea is simple, we get the last digit using the mod operator '%'. If the digit is 0, we replace it with 5, otherwise, keep it as it is. Then we recur for the remaining digits. The approach remains the same, the basic difference is the loop is replaced by a recursive function.
Algorithm:
- Check a base case when the number is 0 return 5, and for all other cases, form a recursive function.
- The function (solve(int n))can be defined as follows, if the number passed is 0 then return 0, else extract the last digit i.e. n = n/10 and remove the last digit. If the last digit is zero assigns 5 to it.
- Now return the value by calling the recursive function for n, i.e return solve(n)*10 + digit.
- Print the answer.
Implementation:
C++
// C++ program to replace all ‘0’
// with ‘5’ in an input Integer
#include <bits/stdc++.h>
using namespace std;
// A recursive function to replace all 0s
// with 5s in an input number It doesn't
// work if input number itself is 0.
int convert0To5Rec(int num)
{
// Base case for recursion termination
if (num == 0)
return 0;
// Extract the last digit and
// change it if needed
int digit = num % 10;
if (digit == 0)
digit = 5;
// Convert remaining digits and
// append the last digit
return convert0To5Rec(num / 10) * 10 + digit;
}
// It handles 0 and calls convert0To5Rec()
// for other numbers
int convert0To5(int num)
{
if (num == 0)
return 5;
else
return convert0To5Rec(num);
}
// Driver code
int main()
{
int num = 10120;
cout << convert0To5(num);
return 0;
}
// This code is contributed by Code_Mech.
// C program to replace all ‘0’
// with ‘5’ in an input Integer
#include <stdio.h>
// A recursive function to replace
// all 0s with 5s in an input number
// It doesn't work if input number itself is 0.
int convert0To5Rec(int num)
{
// Base case for recursion termination
if (num == 0)
return 0;
// Extract the last digit and change it if needed
int digit = num % 10;
if (digit == 0)
digit = 5;
// Convert remaining digits
// and append the last digit
return convert0To5Rec(num / 10) * 10 + digit;
}
// It handles 0 and calls
// convert0To5Rec() for other numbers
int convert0To5(int num)
{
if (num == 0)
return 5;
else
return convert0To5Rec(num);
}
// Driver program to test above function
int main()
{
int num = 10120;
printf("%d", convert0To5(num));
return 0;
}
import java.io.*;
// Java code for Replace all 0 with
// 5 in an input Integer
public class GFG {
// A recursive function to replace all 0s with 5s in
// an input number. It doesn't work if input number
// itself is 0.
static int convert0To5Rec(int num)
{
// Base case
if (num == 0)
return 0;
// Extract the last digit and change it if needed
int digit = num % 10;
if (digit == 0)
digit = 5;
// Convert remaining digits and append the
// last digit
return convert0To5Rec(num / 10) * 10 + digit;
}
// It handles 0 and calls convert0To5Rec() for
// other numbers
static int convert0To5(int num)
{
if (num == 0)
return 5;
else
return convert0To5Rec(num);
}
// Driver function
public static void main(String[] args)
{
System.out.println(convert0To5(10120));
}
}
// This code is contributed by Kamal Rawal
# Python program to replace all
# 0 with 5 in given integer
# A recursive function to replace all 0s
# with 5s in an integer
# Does'nt work if the given number is 0 itself
def convert0to5rec(num):
# Base case for recursion termination
if num == 0:
return 0
# Extract the last digit and change it if needed
digit = num % 10
if digit == 0:
digit = 5
# Convert remaining digits and append the last digit
return convert0to5rec(num // 10) * 10 + digit
# It handles 0 to 5 calls convert0to5rec()
# for other numbers
def convert0to5(num):
if num == 0:
return 5
else:
return convert0to5rec(num)
# Driver Program
num = 10120
print (convert0to5(num))
# Contributed by Harshit Agrawal
// C# code for Replace all 0
// with 5 in an input Integer
using System;
class GFG {
// A recursive function to replace
// all 0s with 5s in an input number.
// It doesn't work if input number
// itself is 0.
static int convert0To5Rec(int num)
{
// Base case
if (num == 0)
return 0;
// Extract the last digit and
// change it if needed
int digit = num % 10;
if (digit == 0)
digit = 5;
// Convert remaining digits
// and append the last digit
return convert0To5Rec(num / 10) * 10 + digit;
}
// It handles 0 and calls
// convert0To5Rec() for other numbers
static int convert0To5(int num)
{
if (num == 0)
return 5;
else
return convert0To5Rec(num);
}
// Driver Code
static public void Main()
{
Console.Write(convert0To5(10120));
}
}
// This code is contributed by Raj
<?php
// PHP program to replace all 0 with 5
// in given integer
// A recursive function to replace all 0s
// with 5s in an integer. Does'nt work if
// the given number is 0 itself
function convert0to5rec($num)
{
// Base case for recursion termination
if ($num == 0)
return 0;
// Extract the last digit and
// change it if needed
$digit = ($num % 10);
if ($digit == 0)
$digit = 5;
// Convert remaining digits and append
// the last digit
return convert0to5rec((int)($num / 10)) *
10 + $digit;
}
// It handles 0 to 5 calls convert0to5rec()
// for other numbers
function convert0to5($num)
{
if ($num == 0)
return 5;
else
return convert0to5rec($num);
}
// Driver Code
$num = 10120;
print(convert0to5($num));
// This code is contributed by mits
?>
<script>
// javascript code for Replace all 0 with
// 5 in an input Integer
// A recursive function to replace all 0s with 5s in
// an input number. It doesn't work if input number
// itself is 0.
function convert0To5Rec(num) {
// Base case
if (num == 0)
return 0;
// Extract the last digit and change it if needed
var digit = num % 10;
if (digit == 0)
digit = 5;
// Convert remaining digits and append the
// last digit
return convert0To5Rec(parseInt(num / 10)) * 10 + digit;
}
// It handles 0 and calls convert0To5Rec() for
// other numbers
function convert0To5(num) {
if (num == 0)
return 5;
else
return convert0To5Rec(num);
}
// Driver function
document.write(convert0To5(10120));
// This code contributed by gauravrajput1
</script>
Complexity Analysis:
- Time Complexity: O(k), the recursive function is called only k times, where k is the number of digits of the number
- Auxiliary Space: O(1), no extra space is required.
Approach (Using builtin function replace())
C++
#include <bits/stdc++.h>
using namespace std;
string change(int n)
{
string temp = to_string(n) + "";
replace(temp.begin(), temp.end(), '0', '5');
return temp;
}
int main()
{
int num = 10120;
cout << (change(num));
return 0;
}
// This code is contributed by akashish__
/*package whatever //do not write package name here */
import java.io.*;
class GFG {
static String change(int n){
String temp = n + "";
return temp.replace('0','5');
}
public static void main (String[] args) {
int num = 10120;
System.out.println(change(num));
}
}
// This code is contributed by aadityaburujwale.
# Python program for the above approach
# Function to replace all 0s with 5s
def change(num):
s = str(num)
s = s.replace('0', '5')
return s
# Driver code
if __name__ == '__main__':
num = 10120
print(change(num))
using System;
public class GFG {
static String change(int n)
{
String temp = n + "";
return temp.Replace('0', '5');
}
static public void Main()
{
// Code
int num = 10120;
Console.WriteLine(change(num));
}
}
// This code is contributed by karandeep1234
function change(n)
{
let temp = n.toString();
temp = temp.replaceAll('0', '5');
return temp;
}
let num = 10120;
console.log((change(num)));
// This code is contributed by akashish__
Time Complexity: O(log(num)), where num is the number of digits in num variable.
Auxiliary Space: O(num)
Replace all 0's with 5 | DSA Practice Problem
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