Number of Valid Clock Times

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 integeranswer , the number of valid clock times that can be created by replacing every?_ with a digit from _0 to9.

Examples

Example 1

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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

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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

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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

Method 1 – Brute Force Enumeration

Intuition

Since the number of possible times is small (24 hours × 60 minutes = 1440), we can try all possible hour and minute values, check if they match the pattern, and count the valid ones.

Approach

For each hour from 0 to 23 and each minute from 0 to 59, format as “hh:mm”.
For each candidate, check if it matches the input pattern (where ‘?’ matches any digit).
Count all such valid times.

Code

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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 buf[6];
                sprintf(buf, "%02d:%02d", h, m);
                bool ok = true;
                for (int i = 0; i < 5; ++i) {
                    if (time[i] != '?' && time[i] != buf[i]) ok = false;
                }
                if (ok) ++ans;
            }
        }
        return ans;
    }
};

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func countTime(time string) int {
    ans := 0
    for h := 0; h < 24; h++ {
        for m := 0; m < 60; m++ {
            t := fmt.Sprintf("%02d:%02d", h, m)
            ok := true
            for i := 0; i < 5; i++ {
                if time[i] != '?' && time[i] != t[i] {
                    ok = false
                }
            }
            if ok {
                ans++
            }
        }
    }
    return ans
}

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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 t = String.format("%02d:%02d", h, m);
                boolean ok = true;
                for (int i = 0; i < 5; i++) {
                    if (time.charAt(i) != '?' && time.charAt(i) != t.charAt(i)) ok = false;
                }
                if (ok) ans++;
            }
        }
        return ans;
    }
}

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class Solution {
    fun countTime(time: String): Int {
        var ans = 0
        for (h in 0..23) {
            for (m in 0..59) {
                val t = "%02d:%02d".format(h, m)
                var ok = true
                for (i in 0..4) {
                    if (time[i] != '?' && time[i] != t[i]) ok = false
                }
                if (ok) ans++
            }
        }
        return ans
    }
}

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class Solution:
    def countTime(self, time: str) -> int:
        ans = 0
        for h in range(24):
            for m in range(60):
                t = f"{h:02d}:{m:02d}"
                if all(tc == '?' or tc == sc for tc, sc in zip(time, t)):
                    ans += 1
        return ans

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impl Solution {
    pub fn count_time(time: String) -> i32 {
        let mut ans = 0;
        for h in 0..24 {
            for m in 0..60 {
                let t = format!("{:02}:{:02}", h, m);
                if time.chars().zip(t.chars()).all(|(a,b)| a == '?' || a == b) {
                    ans += 1;
                }
            }
        }
        ans
    }
}

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class Solution {
    countTime(time: string): number {
        let ans = 0;
        for (let h = 0; h < 24; h++) {
            for (let m = 0; m < 60; m++) {
                const t = `${h.toString().padStart(2, '0')}:${m.toString().padStart(2, '0')}`;
                let ok = true;
                for (let i = 0; i < 5; i++) {
                    if (time[i] !== '?' && time[i] !== t[i]) ok = false;
                }
                if (ok) ans++;
            }
        }
        return ans;
    }
}

Complexity

⏰ Time complexity: O(1) (since we check at most 1440 times)
🧺 Space complexity: O(1)

Problem#

Examples#

Example 1#

Example 2#

Example 3#

Constraints#

Solution#

Method 1 – Brute Force Enumeration#

Intuition#

Approach#

Code#

Complexity#