1323. Maximum 69 Number

Hello, reader 👋🏽 ! Welcome to day 37 of the series on Problem Solving. Through this series, I aim to pick up at least one question everyday and share my approach for solving it.

Today, I will be picking up LeetCode's daily challenge problem: 1323. Maximum 69 Number.

🤔 Problem Statement

• Given a positive number containing only digits 6 and 9, return the maximum number that can be obtained after modifying at most one digit.

• A number can be modified from 6 to 9 or 9 to 6.

• E.g.:

  • num = 9669 => 9969
  • num = 9999 => 9999
  • num = 669 => 969

💬 Thought Process - Convert to String

• From intuition, we can get the maximum number by converting the first 6 to 9.
• For e.g., we can obtain the maximum number from this 9669 by modifying the 6 at 100's place to 9.
• The easiest way to change this would be to convert it to string and then to a character array.
• We will scan from left to right, update the first 6 we see to 9, and then convert the array back to string and then to integer and return.

👩🏽‍💻 Solution - Convert to String

class Solution {
    public int maximum69Number (int num) {
        char[] nums = Integer.toString(num).toCharArray();

        for(int i = 0; i<nums.length; i++) {
            if(nums[i] == '6') {
                nums[i] = '9';
                break;
            }
        }

        return Integer.valueOf(new String(nums));
    }
}

Time Complexity: O(N)
  N = number of digits in num
Space Complexity: O(N)
  N = number of digits in num
      - We will have to use additional space to convert the number to an Interger and character array.

💬 Thought Process - Get Index of First 6

• This is the optimisation you might have to come up if you get this in an interview.
• Instead of using extra space, we will keep getting the number at 1s, 10s, 100s, 1000s, place and so on.
• Whenever we encounter a 6, we will note down the digit number. Then we will add 3 * 10^digit to the original number.
• For e.g.: num = 996.
  • the first digit index for 6 would be at 0.
  • we add 3 * 10^0 = 3 to 996. Hence 996 + 3 = 999.
  • 999 is the greatest number that can be formed by modifying at most 1 number in 996.

• For e.g.: num = 9669.

  • the first 6 is found at index 2.
  • we add 3 * 10^2 = 300 to 9669. Hence 9669 + 300 = 9969.
  • 9969 is the greatest number that can be formed by modifying at most 1 number in 9669.

• One edge case is if we have no 6s in the number.

  • To address this, we will initialise the index with -1.
  • Then once we have looped through the number, if the index is still -1, we return the original number itself.

👩🏽‍💻 Solution - Get Index of First 6

class Solution {
    public int maximum69Number (int num) {        
        int n = num;
        int index = -1;
        int currentDigit = 0;

        while(n > 0) {
            int dig = n % 10;

            if(dig == 6) {
                index = currentDigit;
            }

            currentDigit++;
            n /= 10;
        }

        if(index == -1) return num;

        int tens = (int)(Math.pow(10, index));
        return num + 3 * tens;
    }
}

Time Complexity: O(N)
  N = number of digits in num
Space Complexity: O(1)

• You can find the link to the GitHub repo for this question here: 1323. Maximum 69 Number.

Conclusion

That's a wrap for today's problem. If you liked my explanation then please do drop a like/ comment. Also, please correct me if I've made any mistakes or if you want me to improve something!

Thank you for reading!