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2437. Number of Valid Clock Times

Description

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.

Solutions

Solution 1: Enumeration

We can directly enumerate all times from $00:00$ to $23:59$, then judge whether each time is valid, if so, increment the answer.

After the enumeration ends, return the answer.

The time complexity is $O(24 \times 60)$, and the space complexity is $O(1)$.

Solution 2: Optimized Enumeration

We can separately enumerate hours and minutes, count how many hours and minutes meet the condition, and then multiply them together.

The time complexity is $O(24 + 60)$, and the space complexity is $O(1)$.

  • 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: str, t: str) -> bool: return all(a == b or b == '?' for a, b in zip(s, t)) 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 ans = 0; for (let h = 0; h < 24; ++h) { for (let m = 0; m < 60; ++m) { const s = `${h}`.padStart(2, '0') + ':' + `${m}`.padStart(2, '0'); let ok = 1; for (let i = 0; i < 5; ++i) { if (s[i] !== time[i] && time[i] !== '?') { ok = 0; break; } } ans += ok; } } return ans; } 
  • impl Solution { pub fn count_time(time: String) -> i32 { let mut ans = 0; for i in 0..24 { for j in 0..60 { let mut ok = true; let t = format!("{:02}:{:02}", i, j); for (k, ch) in time.chars().enumerate() { if ch != '?' && ch != t.chars().nth(k).unwrap() { ok = false; } } if ok { ans += 1; } } } ans } }

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