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

Given an integer n. Convert all zeroes of n to 5.

Examples:

Input: n = 1004
Output: 1554
Explanation: There are two zeroes in 1004. On replacing all zeroes with 5, the new number will be 1554.

Input: n = 121
Output: 121
Explanation: Since there are no zeroes in 121, the number remains as 121.

[Iterative Approach] Digit Extraction with Place Value - O(log n) Time and O(1) Space

Digits of n can be extracted one at a time from the rightmost end using modulo and division by 10. Each extracted 0 is replaced with 5, and then placed into the result at its correct position by multiplying it with the current place value (starting at 1 and growing by a factor of 10 with each step).

Illustration:

• Take n = 1004.
• placeValue = 1, result = 0. Extract 1004 % 10 = 4, keep as 4. result = 0 + 4×1 = 4. placeValue becomes 10. n becomes 100.
• Extract 100 % 10 = 0, replace with 5. result = 4 + 5×10 = 54. placeValue becomes 100. n becomes 10.
• Extract 10 % 10 = 0, replace with 5. result = 54 + 5×100 = 554. placeValue becomes 1000. n becomes 1.
• Extract 1 % 10 = 1, keep as 1. result = 554 + 1×1000 = 1554. placeValue becomes 10000. n becomes 0, loop ends.
• Final result is 1554.

#include <bits/stdc++.h>
using namespace std;

int convert0To5Iter(int num) {
    int result = 0;
    int placeValue = 1;

    while (num > 0) {
        int digit = num % 10;

        // replace the digit with 5 if it is 0, otherwise keep it unchanged
        if (digit == 0)
            digit = 5;

        // place the digit at its correct position using the growing place value
        result = result + digit * placeValue;
        placeValue *= 10;
        num /= 10;
    }

    return result;
}

int convertFive(int n) {
    // n = 0 is itself a single zero digit, handled directly
    if (n == 0)
        return 5;
    else
        return convert0To5Iter(n);
}

int main() {
    int n = 1004;

    cout << convertFive(n) << endl;

    return 0;
}

class GfG {
    static int convert0To5Iter(int num) {
        int result = 0;
        int placeValue = 1;

        while (num > 0) {
            int digit = num % 10;

            // replace the digit with 5 if it is 0, otherwise keep it unchanged
            if (digit == 0)
                digit = 5;

            // place the digit at its correct position using the growing place value
            result = result + digit * placeValue;
            placeValue *= 10;
            num /= 10;
        }

        return result;
    }

    static int convertFive(int n) {
        // n = 0 is itself a single zero digit, handled directly
        if (n == 0)
            return 5;
        else
            return convert0To5Iter(n);
    }

    public static void main(String[] args) {
        int n = 1004;

        System.out.println(convertFive(n));
    }
}

def convert0to5Iter(num):
    result = 0
    placeValue = 1

    while num > 0:
        digit = num % 10

        # replace the digit with 5 if it is 0, otherwise keep it unchanged
        if digit == 0:
            digit = 5

        # place the digit at its correct position using the growing place value
        result = result + digit * placeValue
        placeValue *= 10
        num //= 10

    return result

def convertFive(n):
    # n = 0 is itself a single zero digit, handled directly
    if n == 0:
        return 5
    else:
        return convert0to5Iter(n)

n = 1004
print(convertFive(n))

using System;

class GfG {
    static int convert0To5Iter(int num) {
        int result = 0;
        int placeValue = 1;

        while (num > 0) {
            int digit = num % 10;

            // replace the digit with 5 if it is 0, otherwise keep it unchanged
            if (digit == 0)
                digit = 5;

            // place the digit at its correct position using the growing place value
            result = result + digit * placeValue;
            placeValue *= 10;
            num /= 10;
        }

        return result;
    }

    static int convertFive(int n) {
        // n = 0 is itself a single zero digit, handled directly
        if (n == 0)
            return 5;
        else
            return convert0To5Iter(n);
    }

    static void Main() {
        int n = 1004;

        Console.WriteLine(convertFive(n));
    }
}

function convert0To5Iter(num) {
    let result = 0;
    let placeValue = 1;

    while (num > 0) {
        let digit = num % 10;

        // replace the digit with 5 if it is 0, otherwise keep it unchanged
        if (digit === 0)
            digit = 5;

        // place the digit at its correct position using the growing place value
        result = result + digit * placeValue;
        placeValue *= 10;
        num = Math.floor(num / 10);
    }

    return result;
}

function convertFive(n) {
    // n = 0 is itself a single zero digit, handled directly
    if (n === 0)
        return 5;
    else
        return convert0To5Iter(n);
}

// Driver Code
const n = 1004;
console.log(convertFive(n));

1554

[Recursive Approach] Digit Extraction with Recursive Rebuild - O(log n) Time and O(log n) Space

Instead of tracking a place value explicitly, the digits can be rebuilt using recursion: the function recurses down to the last digit first (the base case, when the number becomes 0), and then as each recursive call returns, it multiplies the already-built result from deeper calls by 10 and appends its own digit.

Illustration:

• Take n = 1004.
• convert0To5Rec(1004) extracts the last digit (4, kept as is) and recurses on 100.
• convert0To5Rec(100) extracts the last digit (0, replaced with 5) and recurses on 10.
• convert0To5Rec(10) extracts the last digit (0, replaced with 5) and recurses on 1.
• convert0To5Rec(1) extracts the last digit (1, kept as is) and recurses on 0.
• convert0To5Rec(0) hits the base case and returns 0.
• Unwinding back up: convert0To5Rec(1) returns 0 × 10 + 1 = 1. convert0To5Rec(10) returns 1 × 10 + 5 = 15. convert0To5Rec(100) returns 15 × 10 + 5 = 155. convert0To5Rec(1004) returns 155 × 10 + 4 = 1554.

#include <bits/stdc++.h>
using namespace std;

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;
}

int convertFive(int n) {
    // n = 0 is itself a single zero digit, handled directly
    if (n == 0)
        return 5;
    else
        return convert0To5Rec(n);
}

int main() {
    int n = 1004;

    cout << convertFive(n) << endl;

    return 0;
}

class GfG {
    static 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;
    }

    static int convertFive(int n) {
        // n = 0 is itself a single zero digit, handled directly
        if (n == 0)
            return 5;
        else
            return convert0To5Rec(n);
    }

    public static void main(String[] args) {
        int n = 1004;

        System.out.println(convertFive(n));
    }
}

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

def convertFive(n):
    # n = 0 is itself a single zero digit, handled directly
    if n == 0:
        return 5
    else:
        return convert0to5Rec(n)

n = 1004
print(convertFive(n))

using System;

class GfG {
    static 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;
    }

    static int convertFive(int n) {
        // n = 0 is itself a single zero digit, handled directly
        if (n == 0)
            return 5;
        else
            return convert0To5Rec(n);
    }

    static void Main() {
        int n = 1004;

        Console.WriteLine(convertFive(n));
    }
}

function convert0To5Rec(num) {
    // base case for recursion termination
    if (num === 0)
        return 0;

    // extract the last digit and change it if needed
    let digit = num % 10;
    if (digit === 0)
        digit = 5;

    // convert remaining digits and append the last digit
    return convert0To5Rec(Math.floor(num / 10)) * 10 + digit;
}

function convertFive(n) {
    // n = 0 is itself a single zero digit, handled directly
    if (n === 0)
        return 5;
    else
        return convert0To5Rec(n);
}

// Driver Code
const n = 1004;
console.log(convertFive(n));

1554

[Alternate Approach] Using Built In Methods

#include <bits/stdc++.h>
using namespace std;

int convertFive(int n)
{
  string temp = to_string(n) + "";
  replace(temp.begin(), temp.end(), '0', '5');
  return stoi(temp);
}

int main()
{
  int n = 10120;
  cout << (convertFive(n));
  return 0;
}

import java.util.stream.Collectors;

public class Main {
  public static int convertFive(int n) {
    String temp = Integer.toString(n);
    temp = temp.replace('0', '5');
    return Integer.parseInt(temp);
  }

  public static void main(String[] args) {
    int num = 10120;
    System.out.println(convertFive(num));
  }
}

def convertFive(n):
  temp = str(n)
  temp = temp.replace('0', '5')
  return int(temp)

if __name__ == '__main__':
  num = 10120
  print(convertFive(num))

using System;

public class Program
{
  public static int convertFive(int n)
  {
    string temp = n.ToString();
    temp = temp.Replace('0', '5');
    return int.Parse(temp);
  }

  public static void Main()
  {
    int num = 10120;
    Console.WriteLine(convertFive(num));
  }
}

function convertFive(n) {
  let temp = n.toString();
  temp = temp.replace(/0/g, '5');
  return parseInt(temp);
}

// Main function
let num = 10120;
console.log(convertFive(num));