LeetCode 2437 - Number of Valid Clock Times

The problem asks us to determine the number of valid times that can be formed from a partially known digital clock string of the format "hh:mm", where unknown digits are represented by '?'. Each '?

leetcodeeasystringenumeration

LeetCode Problem 2437

Difficulty: 🟢 Easy
Topics: String, Enumeration

Solution

Problem Understanding

The problem asks us to determine the number of valid times that can be formed from a partially known digital clock string of the format "hh:mm", where unknown digits are represented by '?'. Each '?' can be replaced with any digit from 0 to 9, but the resulting hour (hh) must be in the range 00 to 23, and the minute (mm) must be in the range 00 to 59.

The input is always a string of length 5, including the colon ':' at index 2. The output is an integer representing the total number of valid times obtainable by replacing all '?' characters.

Constraints indicate that the input will always be well-formed, so we do not need to handle malformed strings. Important edge cases include strings where all characters are unknown ("??:??"), strings with a single unknown digit that has limited valid replacements ("?5:00"), and strings where both digits of hours or minutes are unknown but constrained by the maximum valid values (23 for hours, 59 for minutes).

Approaches

The brute-force approach would involve iterating over all possible digits for each '?' and generating all possible time strings, then checking each string for validity. While this would produce correct results, it is inefficient, especially if multiple '?' characters are present. For example, "??:??" would require checking 10,000 combinations, even though we know there are only 24 * 60 = 1440 valid times.

The key insight for an optimal solution is that we can compute the number of valid choices independently for hours and minutes, based on the known and unknown digits. For hours, the first digit dictates the possible range of the second digit. For minutes, the range is always 00 to 59, so each unknown digit can be counted independently. Multiplying the number of valid hour combinations with valid minute combinations gives the total number of valid times.

Approach Time Complexity Space Complexity Notes
Brute Force O(10^n) O(1) Iterate over all replacements for '?' and count valid strings
Optimal O(1) O(1) Directly compute the number of valid hours and minutes based on positions of '?'

Algorithm Walkthrough

  1. Split the string into hour (hh) and minute (mm) components.
  2. Compute the number of valid hour possibilities:

  • If both hour digits are '?', there are 24 possibilities (00 to 23).

  • If the first digit is '?' but the second is known:

  • If the second digit is 0 to 3, the first digit can be 0 or 1 or 2 depending on constraints. Specifically, if the second digit ≤ 3, first digit can be 0, 1, or 2; otherwise only 0 or 1.

  • If the second digit is '?' but the first is known:

  • If the first digit is 0 or 1, the second digit can be 0 to 9.

  • If the first digit is 2, the second digit can be 0 to 3.

  • If neither is '?', there is only 1 valid combination.

  1. Compute the number of valid minute possibilities:
  • If both minute digits are '?', there are 60 possibilities.
  • If the first minute digit is '?', there are 6 possible digits (0 to 5).
  • If the second minute digit is '?', there are 10 possible digits (0 to 9).
  • Multiply possibilities if both are '?'.
  1. Multiply the number of valid hour possibilities with valid minute possibilities to get the final answer.

Why it works: Each '?' can only affect its own position, and the constraints for hours and minutes are independent. By computing the valid combinations for each component separately and multiplying them, we cover all valid permutations without enumerating them.

Python Solution

class Solution:
    def countTime(self, time: str) -> int:
        hour, minute = time.split(':')
        
        # Calculate possible hours
        if hour == '??':
            hour_options = 24
        elif hour[0] == '?':
            hour_options = 2 if hour[1] > '3' else 3
        elif hour[1] == '?':
            hour_options = 10 if hour[0] in '01' else 4
        else:
            hour_options = 1
        
        # Calculate possible minutes
        if minute == '??':
            minute_options = 60
        elif minute[0] == '?':
            minute_options = 6
        elif minute[1] == '?':
            minute_options = 10
        else:
            minute_options = 1
        
        return hour_options * minute_options

Implementation Explanation: We first split the input string into hours and minutes. Then, using conditional logic based on which digits are unknown, we calculate the number of valid hour and minute combinations. Finally, we multiply them to get the total valid times. Special attention is given to the hour constraints because hours cannot exceed 23.

Go Solution

func countTime(time string) int {
    hour := time[:2]
    minute := time[3:]
    
    var hourOptions int
    if hour == "??" {
        hourOptions = 24
    } else if hour[0] == '?' {
        if hour[1] > '3' {
            hourOptions = 2
        } else {
            hourOptions = 3
        }
    } else if hour[1] == '?' {
        if hour[0] == '0' || hour[0] == '1' {
            hourOptions = 10
        } else {
            hourOptions = 4
        }
    } else {
        hourOptions = 1
    }
    
    var minuteOptions int
    if minute == "??" {
        minuteOptions = 60
    } else if minute[0] == '?' {
        minuteOptions = 6
    } else if minute[1] == '?' {
        minuteOptions = 10
    } else {
        minuteOptions = 1
    }
    
    return hourOptions * minuteOptions
}

Go-Specific Notes: The Go implementation mirrors the Python logic. String slicing and indexing are used carefully since Go treats strings as byte arrays. There is no need for additional handling of nil values or overflow since all calculations are well within integer bounds.

Worked Examples

For "?5:00", hour has '?' at first position and second digit is 5. Since 5 > 3, first digit can be 0 or 1, so hour_options = 2. Minutes are fixed, so minute_options = 1. Total = 2 * 1 = 2.

For "0?:0?", hour first digit is 0 and second is '?', so hour_options = 10. Minute first digit is 0 and second is '?', so minute_options = 10. Total = 10 * 10 = 100.

For "??:??", all digits unknown. Hour options = 24, minute options = 60, total = 24 * 60 = 1440.

Step "?5:00" "0?:0?" "??:??"
Hour Options 2 10 24
Minute Options 1 10 60
Total Valid Times 2 100 1440

Complexity Analysis

Measure Complexity Explanation
Time O(1) Constant time computation based on string positions
Space O(1) Only a few integer variables used, no extra data structures

The reasoning is that we only perform a fixed number of checks (at most 4 for hour and minute digits), so time does not depend on the input size. Space is minimal since we store only intermediate counts.

Test Cases

assert Solution().countTime("?5:00") == 2  # single unknown hour digit
assert Solution().countTime("0?:0?") == 100  # multiple unknown digits
assert Solution().countTime("??:??") == 1440  # all unknown digits
assert Solution().countTime("23:59") == 1  # no unknowns
assert Solution().countTime("2?:?9") == 20  # hour first digit fixed, second unknown, minute second digit fixed
assert Solution().countTime("?3:?0") == 24  # hour first unknown, second digit 3; minute first unknown, second digit 0
Test Why
"?5:00" Tests single unknown digit in hours
"0?:0?" Tests unknown digits in both hours and minutes
"??:??" Tests all digits unknown
"23:59" Tests no unknown digits (fixed time)
"2?:?9" Tests upper bound hours with unknowns
"?3:?0" Tests combination of known and unknown digits for hour and minute

Edge Cases

One important edge case is "??:??", where every digit is unknown. This is a source of potential overflow or miscount if one tries to enumerate all combinations instead of using multiplication.

Another edge case is when the first