g2401_2500.s2437_number_of_valid_clock_times.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2401_2500.s2437_number_of_valid_clock_times;

// #Easy #String #Enumeration #2022_12_07_Time_0_ms_(100.00%)_Space_42.2_MB_(29.58%)

/**
 * 2437 - Number of Valid Clock Times\.
 *
 * Easy
 *
 * You are given a string of length `5` called `time`, representing the current time on a digital clock in the format `"hh:mm"`. The **earliest** possible time is `"00:00"` and the **latest** possible time is `"23:59"`.
 *
 * In the string `time`, the digits represented by the `?` symbol are **unknown** , and must be **replaced** with a digit from `0` to `9`.
 *
 * Return _an integer_ `answer`_, the number of valid clock times that can be created by replacing every_ `?`_with a digit from_ `0` _to_ `9`.
 *
 * **Example 1:**
 *
 * **Input:** time = "?5:00"
 *
 * **Output:** 2
 *
 * **Explanation:** We can replace the ? with either a 0 or 1, producing "05:00" or "15:00". Note that we cannot replace it with a 2, since the time "25:00" is invalid. In total, we have two choices.
 *
 * **Example 2:**
 *
 * **Input:** time = "0?:0?"
 *
 * **Output:** 100
 *
 * **Explanation:** Each ? can be replaced by any digit from 0 to 9, so we have 100 total choices.
 *
 * **Example 3:**
 *
 * **Input:** time = "??:??"
 *
 * **Output:** 1440
 *
 * **Explanation:** There are 24 possible choices for the hours, and 60 possible choices for the minutes. In total, we have 24 \* 60 = 1440 choices.
 *
 * **Constraints:**
 *
 * *   `time` is a valid string of length `5` in the format `"hh:mm"`.
 * *   `"00" <= hh <= "23"`
 * *   `"00" <= mm <= "59"`
 * *   Some of the digits might be replaced with `'?'` and need to be replaced with digits from `0` to `9`.
**/
public class Solution {
    public int countTime(String time) {
        int[] counts = new int[] {3, 10, 0, 6, 10};
        char[] ch = time.toCharArray();
        int result = 1;
        if (ch[0] == '2') {
            counts[1] = 4;
        }
        if ((ch[1] - '0') > 3) {
            counts[0] = 2;
        }
        if (ch[0] == '?' && ch[1] == '?') {
            counts[0] = 1;
            counts[1] = 24;
        }
        for (int i = 0; i < 5; i++) {
            char ch1 = ch[i];
            if (ch1 == '?') {
                result *= counts[i];
            }
        }
        return result;
    }
}