How do you solve Minimum Time Difference on LeetCode?

Minimum Time Difference (LeetCode #539) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting helps in reducing the number of comparisons needed to find the minimum difference.

What is the time complexity of Minimum Time Difference?

The optimal solution for Minimum Time Difference runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step. The space complexity is O(n) because we store the converted minutes in an array.

What is the brute force solution for Minimum Time Difference?

The brute force approach for Minimum Time Difference is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of time points to find the minimum difference. This is straightforward but inefficient for larger lists. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Time Difference?

Minimum Time Difference uses the following concepts: Array, Math, String, Sorting. The recommended patterns to study are: Sorting, Array.

What are the interview tips for Minimum Time Difference?

Always clarify how to handle edge cases, such as the wrap-around in time. Explain your thought process clearly while coding; it shows your understanding. Practice explaining your approach as you code, as this is key in interviews.

What are common mistakes when solving Minimum Time Difference?

Not considering the wrap-around case when calculating differences. Forgetting to convert time points to a consistent format (like total minutes) before comparison.

#539

Minimum Time Difference

Medium

ArrayMathStringSortingSortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

By sorting the time points and calculating differences between consecutive times, we can efficiently find the minimum difference. This takes advantage of the fact that the smallest difference will be between two adjacent times in a sorted list.

⚙️

Algorithm

4 steps

1Step 1: Convert each time point from 'HH:MM' format to total minutes since midnight.
2Step 2: Sort the list of time points in ascending order.
3Step 3: Calculate the difference between each pair of consecutive time points and keep track of the minimum difference.
4Step 4: Also check the difference between the last and first time point to account for the wrap-around.

solution.py14 lines

1# Full working Python code
2
3def minTimeDifference(timePoints):
4    def toMinutes(time):
5        h, m = map(int, time.split(':'))
6        return h * 60 + m
7
8    minutes = sorted(toMinutes(tp) for tp in timePoints)
9    min_diff = float('inf')
10    for i in range(1, len(minutes)):
11        min_diff = min(min_diff, minutes[i] - minutes[i - 1])
12    # Check wrap-around difference
13    min_diff = min(min_diff, 1440 - (minutes[-1] - minutes[0]))
14    return min_diff

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step. The space complexity is O(n) because we store the converted minutes in an array.

1Sorting helps in reducing the number of comparisons needed to find the minimum difference.
2The wrap-around case is crucial when dealing with circular time formats.

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