LeetCode 2224: Minimum Number of Operations to Convert Time Solution

Master LeetCode problem 2224 (Minimum Number of Operations to Convert Time), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2224. Minimum Number of Operations to Convert Time

Easy

String Greedy

Problem Explanation

Explanation

To solve this problem, we can iterate through each digit of the current and correct times and calculate the difference between them. Based on this difference, we can determine the minimum number of operations needed to convert the current time to the correct time. The possible operations are adding 1, 5, 15, or 60 minutes.

Parse the current and correct times into hours and minutes.
Calculate the total minutes for both times.
Calculate the difference in minutes between the current and correct times.
Determine the minimum number of operations needed based on the difference.
Return the minimum number of operations.

Time Complexity: O(1)
Space Complexity: O(1)

Solution Code

class Solution {
    public int minOperations(String current, String correct) {
        int currentHours = Integer.parseInt(current.substring(0, 2));
        int currentMinutes = Integer.parseInt(current.substring(3));
        int correctHours = Integer.parseInt(correct.substring(0, 2));
        int correctMinutes = Integer.parseInt(correct.substring(3));

        int currentTotalMinutes = currentHours * 60 + currentMinutes;
        int correctTotalMinutes = correctHours * 60 + correctMinutes;

        int diff = Math.abs(currentTotalMinutes - correctTotalMinutes);

        int operations = 0;
        while (diff > 0) {
            if (diff >= 60) {
                operations += diff / 60;
                diff %= 60;
            } else if (diff >= 15) {
                operations += diff / 15;
                diff %= 15;
            } else if (diff >= 5) {
                operations += diff / 5;
                diff %= 5;
            } else {
                operations += diff;
                diff = 0;
            }
        }

        return operations;
    }
}

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