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Formatted question description: https://leetcode.ca/all/2437.html

2437. Number of Valid Clock Times

  • Difficulty: Easy.
  • Related Topics: .
  • Similar Questions: Largest Time for Given Digits, Latest Time by Replacing Hidden Digits.

Problem

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.

Solution (Java, C++, Python)

  • class Solution {
    public int countTime(String time) {
        int ans = 0;
        for (int h = 0; h < 24; ++h) {
            for (int m = 0; m < 60; ++m) {
                String s = String.format("%02d:%02d", h, m);
                int ok = 1;
                for (int i = 0; i < 5; ++i) {
                    if (s.charAt(i) != time.charAt(i) && time.charAt(i) != '?') {
                        ok = 0;
                        break;
                    }
                }
                ans += ok;
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    int countTime(string time) {
        int ans = 0;
        for (int h = 0; h < 24; ++h) {
            for (int m = 0; m < 60; ++m) {
                char s[20];
                sprintf(s, "%02d:%02d", h, m);
                int ok = 1;
                for (int i = 0; i < 5; ++i) {
                    if (s[i] != time[i] && time[i] != '?') {
                        ok = 0;
                        break;
                    }
                }
                ans += ok;
            }
        }
        return ans;
    }
};

  • class Solution:
    def countTime(self, time: str) -> int:
        def check(s, t):
            for a, b in zip(s, t):
                if a != b and b != '?':
                    return 0
            return 1

        return sum(
            check(f'{h:02d}:{m:02d}', time) for h in range(24) for m in range(60)
        )


  • func countTime(time string) int {
	ans := 0
	for h := 0; h < 24; h++ {
		for m := 0; m < 60; m++ {
			s := fmt.Sprintf("%02d:%02d", h, m)
			ok := 1
			for i := 0; i < 5; i++ {
				if s[i] != time[i] && time[i] != '?' {
					ok = 0
					break
				}
			}
			ans += ok
		}
	}
	return ans
}

  • function countTime(time: string): number {
    let [hh, mm] = time.split(':');
    return count(hh, 24) * count(mm, 60);
}

function count(str: string, limit: number): number {
    let [a, b] = str.split('').map(d => Number(d));
    let ans = 0;
    if (isNaN(a) && isNaN(b)) return limit;
    if (isNaN(a)) {
        for (let i = 0; i <= 9; i++) {
            if (i * 10 + b < limit) ans++;
        }
        return ans;
    }
    if (isNaN(b)) {
        for (let i = 0; i <= 9; i++) {
            if (a * 10 + i < limit) ans++;
        }
        return ans;
    }
    return 1;
}


  • impl Solution {
    pub fn count_time(time: String) -> i32 {
        let f = |s: &str, m: usize| -> i32 {
            let mut cnt = 0;
            let first = s.chars().nth(0).unwrap();
            let second = s.chars().nth(1).unwrap();

            for i in 0..m {
                let a = first == '?' || first.to_digit(10).unwrap() as usize == i / 10;

                let b = second == '?' || second.to_digit(10).unwrap() as usize == i % 10;

                if a && b {
                    cnt += 1;
                }
            }

            cnt
        };

        f(&time[..2], 24) * f(&time[3..], 60)
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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