Replace All 0's with 5

Problem

You are given an integer N. You need to convert all zeroes of N to 5.

Examples:

Example 1:

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

Example 2:

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

Your task:

Your task is to complete the function convertFive() which takes an integer N as an argument and replaces all zeros in the number N with 5. Your function should return the converted number.

• Expected Time Complexity: $O(K)$, where K is the number of digits in N
• Expected Auxiliary Space: $O(1)$

Constraints:

• $1 <= n <= 10000$

Solution

Python

def convertFive(n):
    n_str = str(n)
    modified_digits = []
    for char in n_str:
        if char == '0':
            modified_digits.append('5')
        else:
            modified_digits.append(char)
    modified_str = ''.join(modified_digits)
    modified_n = int(modified_str)
    return modified_n

Java

int convertfive(int num) {
    String numStr = String.valueOf(num);
    StringBuilder modifiedDigits = new StringBuilder();
    for (int i = 0; i < numStr.length(); i++) {
        char ch = numStr.charAt(i);
        if (ch == '0') {
            modifiedDigits.append('5');
        } else {
            modifiedDigits.append(ch); 
        }
    }
    int modifiednum = Integer.parseInt(modifiedDigits.toString());
    return modifiednum;
}

C++

int convertFive(int n) {
    string nStr = to_string(n);
    string modifiedStr = "";
    for (char ch : nStr) {
        if (ch == '0') {
            modifiedStr += '5';
        } else {
            modifiedStr += ch; 
        }
    }
    int modifiedN = stoi(modifiedStr);
    return modifiedN;
}

C

int convertFive(int num) {
    if (num == 0) {
        return 5;
    }
    int result = 0;
    int decimalPlace = 1;
    while (num > 0) {
        int digit = num % 10; 
        if (digit == 0) {
            digit = 5; 
        }
        result = digit * decimalPlace + result;
        decimalPlace *= 10; 
        num /= 10;
    }
    return result;
}

• Time Complexity: $O(K)$, where K is the number of digits in N
• Auxiliary Space: $O(1)$