Replace all ‘0’ with ‘5’ in an input Integer

Given an integer as input and replace all the ‘0’ with ‘5’ in the integer.

Examples:

Input: 102 
Output: 152
Explantion: All the digits which are '0' is replaced by '5' 

Input: 1020 
Output: 1525
Explantion: All the digits which are '0' is replaced by '5'

Use of array to store all digits is not allowed.
Source: Amazon interview Experience | Set 136 (For SDE-T)

Iterative Approach: 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 current digit and create a copy of 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 numbern = n/10 and update place = place * 10
Return the sum.

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

Java

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));
    }
}

Python3

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

Output:

Complexity Analysis:

Time Complexity: O(k), the loops runs only k times, where k is the number of digits of the number.
Space Complexity: O(1), no extra space is required.

Recursive Approach: The idea is simple, 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 recur for 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, 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 the assign 5 to it.
Now return the value by calling the recursive function for n, i.e return solve(n)*10 + digit.
print the answer.

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;
 
    // Extraxt 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

// 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;
}

Java

// Java code for Replace all 0 with
// 5 in an input Integer
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;
 
        // Extraxt 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

# 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 recurssion 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#

// 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;
 
        // Extraxt 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
// 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 recurssion 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
?>

Output:

Complexity Analysis:

Time Complexity: O(k), the recusrsive function is called only k times, where k is the number of digits of the number
Space Complexity: O(1), no extra space is required.

https://www.youtube.com/watch?v=bFBqTgC4MlM

This article is contributed by Sai Kiran. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

C++

Java

Python3

C++

C

Java

Python

C#

PHP

Recommended Posts: