LeetCode 2437: Number of Valid Clock Times Solution

Master LeetCode problem 2437 (Number of Valid Clock Times), 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.

2437. Number of Valid Clock Times

Easy

String Enumeration

Problem Explanation

Explanation

To solve this problem, we need to iterate over all possible combinations of digits for the unknown characters in the input time string. We can do this by replacing each '?' character with digits from 0 to 9 and checking if the resulting time is a valid time. For each valid time, we increment a counter. Finally, we return the count of valid times.

Algorithm:

Initialize a counter to keep track of the number of valid times.
Iterate over all possible digits from 0 to 9 for each '?' character in the input time string.
Check if the resulting time is a valid time.
If the time is valid, increment the counter.
Return the count of valid times.

Time Complexity

The time complexity of this algorithm is O(1) because we are iterating over a fixed number of positions and checking a fixed number of possibilities at each position.

Space Complexity

The space complexity of this algorithm is O(1) as we are not using any extra space that grows with the input size.

Solution Code

class Solution {
    public int numberOfValidTimes(String time) {
        int count = 0;
        for (char h1 = '0'; h1 <= '2'; h1++) {
            for (char h2 = '0'; h2 <= '9'; h2++) {
                for (char m1 = '0'; m1 <= '5'; m1++) {
                    for (char m2 = '0'; m2 <= '9'; m2++) {
                        if ((h1 == '2' && h2 > '3') || (m1 > '5')) continue;
                        String currTime = "" + h1 + h2 + ":" + m1 + m2;
                        if (isValidTime(currTime)) {
                            count++;
                        }
                    }
                }
            }
        }
        return count;
    }

    private boolean isValidTime(String time) {
        return time.compareTo("24:00") < 0;
    }
}

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