Solution Explanation for Maximum 69 Number

The problem asks to find the maximum number possible by changing at most one digit in a given number (consisting only of digits 6 and 9) where a 6 can become a 9 and vice-versa. The most efficient approach is a greedy one.

Approach:

The optimal strategy is to find the first occurrence of the digit 6 from the left (most significant digit) and replace it with 9. Replacing a 6 with a 9 results in the largest possible increase in the number's value. If no 6 exists, the original number is already maximal, and no change is needed.

Algorithm:

1. Convert to String: Convert the input integer num into a string. This allows easy access to individual digits.

2. Iterate and Replace: Iterate through the string representation of the number. If a '6' is found, replace it with a '9' and immediately break the loop (since we're allowed to change only one digit).

3. Convert Back to Integer: Convert the modified string back into an integer and return it.

Time Complexity Analysis:

• Conversion to String: O(log₁₀(num)), as the number of digits is proportional to the logarithm base 10 of the number.
• Iteration and Replacement: O(log₁₀(num)), as we iterate through the digits of the string once.
• Conversion back to Integer: O(log₁₀(num)), similar to the string conversion.

Therefore, the overall time complexity is O(log₁₀(num)), which is linear in the number of digits. This is very efficient.

Space Complexity Analysis:

The space complexity is O(log₁₀(num)) because the space used is proportional to the length of the string representation of the number. This is also linear in the number of digits.

Code Examples (with detailed comments):

Python:

class Solution:
    def maximum69Number(self, num: int) -> int:
        s_num = str(num)  # Convert to string for easy digit access
        
        modified_num = ""  # Initialize the modified string
 
        found_six = False   # Flag to track if a '6' has been replaced
        for digit in s_num:
            if digit == '6' and not found_six:
                modified_num += '9' #Replace the first '6' with '9'
                found_six = True
            else:
                modified_num += digit #Append rest of digits as they are
 
        return int(modified_num) # Convert back to an integer
 

Java:

class Solution {
    public int maximum69Number(int num) {
        String sNum = Integer.toString(num); // Convert to String
        StringBuilder sb = new StringBuilder(); //More efficient than String concatenation
 
        boolean sixFound = false;
        for (char c : sNum.toCharArray()) {
            if (c == '6' && !sixFound) {
                sb.append('9');
                sixFound = true;
            } else {
                sb.append(c);
            }
        }
        return Integer.parseInt(sb.toString()); // Convert back to integer
    }
}

The other languages (C++, Go, TypeScript, Rust, PHP, C) follow similar logic, primarily focusing on converting the number to a string, modifying the string by replacing the first '6' with '9', and then converting it back to an integer. The choice of using string manipulation is the most readable and concise way to solve this problem. The C example demonstrates a slightly different approach that avoids string manipulation using mathematical operations. It's less readable but equally efficient.