How do you solve Number of Valid Clock Times on LeetCode?

Number of Valid Clock Times (LeetCode #2437) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding the constraints of valid clock times is crucial for determining valid replacements.

What is the time complexity of Number of Valid Clock Times?

The optimal solution for Number of Valid Clock Times runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only make a constant number of checks (4) regardless of the number of '?' characters.

What is the brute force solution for Number of Valid Clock Times?

The brute force approach for Number of Valid Clock Times is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible combinations of digits for the '?' characters and checking if each generated time is valid. This method is straightforward but can be inefficient as it explores many combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Valid Clock Times?

Number of Valid Clock Times uses the following concepts: String, Enumeration. The recommended patterns to study are: Enumeration, String Manipulation.

What are the interview tips for Number of Valid Clock Times?

Always clarify the constraints and edge cases before diving into coding. Think about how to optimize your solution after implementing a brute-force approach. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Number of Valid Clock Times?

Failing to account for the hour limits when replacing '?' in the hour section. Not considering the impact of '?' in the minutes section on the total count.

#2437

Number of Valid Clock Times

Easy

StringEnumerationEnumerationString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution directly counts valid combinations based on the positions of '?' in the time string without generating all possible times. This reduces unnecessary computations and speeds up the process significantly.

⚙️

Algorithm

6 steps

1Step 1: Initialize a count variable to 1.
2Step 2: Check the first character (hour's tens place): If it's '?', multiply count by 3 if it's the first digit (0-2), otherwise by 2 (0-1).
3Step 3: Check the second character (hour's units place): If it's '?', adjust the count based on the first character's value (0-3).
4Step 4: Check the third character (minute's tens place): If it's '?', multiply count by 6 (0-5).
5Step 5: Check the fourth character (minute's units place): If it's '?', multiply count by 10 (0-9).
6Step 6: Return the final count.

solution.py14 lines

1# Full working Python code
2
3def count_valid_times(time):
4    count = 1
5    if time[0] == '?':
6        count *= 3 if time[1] in '012' else 2
7    if time[1] == '?':
8        count *= 10 if time[0] != '2' else 4
9    if time[3] == '?':
10        count *= 6
11    if time[4] == '?':
12        count *= 10
13    return count
14

ℹ

Complexity note: The time complexity is O(n) because we only make a constant number of checks (4) regardless of the number of '?' characters.

1Understanding the constraints of valid clock times is crucial for determining valid replacements.
2Directly counting valid combinations based on the positions of '?' can significantly reduce computation time.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.