LeetCode 539: Minimum Time Difference Solution in Python – A Step-by-Step Guide

What Is LeetCode 539: Minimum Time Difference?

In LeetCode 539: Minimum Time Difference, you’re given a list of strings timePoints representing times in 24-hour format ("HH:MM"), and your task is to find the minimum difference in minutes between any two times, accounting for the circular nature of a 24-hour clock (1440 minutes). For example, with timePoints = ["23:59", "00:00"], the minimum difference is 1 minute, as "23:59" to "00:00" wraps around midnight. This problem builds on LeetCode 75: Sort Colors for sorting techniques and requires time manipulation.

Problem Statement

• Input: timePoints (List[str])—list of times in "HH:MM" format.
• Output: Integer—minimum time difference in minutes.
• Rules: Times are in 24-hour format; consider wrap-around (e.g., "23:59" to "00:00" is 1 minute); return smallest difference.

Constraints

• 2 <= timePoints.length <= 2 * 10⁴
• timePoints[i] is in "HH:MM" format (00:00 to 23:59).

Examples

• Input: timePoints = ["23:59","00:00"]
• Output: 1
• 23:59 to 00:00 = 1 minute (wrap-around).
• Input: timePoints = ["00:00","23:59","00:00"]
• Output: 0
• Two "00:00" times, difference = 0.
• Input: timePoints = ["12:12","00:13","12:00"]
• Output: 12
• Differences: 719, 12, 731; min = 12.

Understanding the Problem: Measuring Time Gaps

To solve LeetCode 539: Minimum Time Difference in Python, we need to find the smallest difference in minutes between any two times in a list, considering a 24-hour clock’s circular nature (1440 minutes total). A naive approach might compare all pairs (O(n²)), but with up to 2 * 10⁴ times, we can optimize by sorting or using a set. Since times can repeat and wrap around, we must handle duplicates and the midnight boundary. We’ll explore:

• Best Solution (Sorting with Wrap-Around Check): O(n log n) time, O(1) space—efficient and intuitive.
• Alternative Solution (Convert to Minutes with Set): O(n) time, O(n) space—fast but uses extra memory.

Let’s dive into the sorting solution with a friendly breakdown!

Best Solution: Sorting with Wrap-Around Check

Why Sorting Wins

The sorting with wrap-around check is the best for LeetCode 539 because it sorts the times in O(n log n) time, computes differences between adjacent pairs, and checks the wrap-around difference (first to last) in a single pass, using O(1) extra space (excluding sort). It’s like lining up the clock hands and measuring the gaps, with a quick peek around midnight!

How It Works

Think of this as a clock-hand lineup:

• Step 1: Sort Times:
• Sort timePoints lexicographically (HH:MM order matches time order).
• Step 2: Convert to Minutes:
• Parse each "HH:MM" to minutes since 00:00 (e.g., "12:34" → 754).
• Step 3: Compute Differences:
• Find min difference between adjacent sorted times.
• Check wrap-around: (first time + 1440) - last time.
• Step 4: Handle Duplicates:
• If any difference is 0, return 0 (duplicate times).
• Why It Works:
• Sorted order ensures min difference is adjacent or wrap-around.
• Wrap-around accounts for circular clock.

It’s like a time-gap measurer!

Step-by-Step Example

Example: timePoints = ["23:59", "00:00", "12:34"]

• Step 1: Sort: ["00:00", "12:34", "23:59"].
• Step 2: Convert to minutes:
• "00:00" → 0.
• "12:34" → 12*60 + 34 = 754.
• "23:59" → 23*60 + 59 = 1439.
• Minutes: [0, 754, 1439].
• Step 3: Compute differences:
• Adjacent: 754 - 0 = 754, 1439 - 754 = 685.
• Wrap-around: (0 + 1440) - 1439 = 1.
• Min = min(754, 685, 1) = 1.
• Result: 1.

Example: timePoints = ["00:00", "23:59", "00:00"]

• Sort: ["00:00", "00:00", "23:59"].
• Minutes: [0, 0, 1439].
• Differences: 0 - 0 = 0, 1439 - 0 = 1439, wrap-around = 1.
• Result: 0 (duplicate times).

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def findMinDifference(self, timePoints: List[str]) -> int:
        # Step 1: Sort times
        timePoints.sort()

        # Step 2: Convert first and last to minutes
        def to_minutes(time_str):
            hours, minutes = map(int, time_str.split(':'))
            return hours * 60 + minutes

        # Step 3: Compute differences
        min_diff = float('inf')
        prev_minutes = to_minutes(timePoints[0])

        for i in range(1, len(timePoints)):
            curr_minutes = to_minutes(timePoints[i])
            diff = curr_minutes - prev_minutes
            if diff == 0:  # Duplicate times
                return 0
            min_diff = min(min_diff, diff)
            prev_minutes = curr_minutes

        # Step 4: Check wrap-around
        wrap_around = (to_minutes(timePoints[0]) + 1440) - to_minutes(timePoints[-1])
        min_diff = min(min_diff, wrap_around)

        return min_diff

• Line 4: Sort times lexicographically (matches chronological order).
• Lines 7-9: Helper converts "HH:MM" to minutes.
• Lines 12-19: Compute adjacent differences, check for duplicates.
• Lines 22-23: Compute wrap-around difference from first to last.
• Line 25: Return smallest difference.
• Time Complexity: O(n log n)—sorting dominates.
• Space Complexity: O(1)—no extra space beyond input.

It’s like a time-difference clockmaster!

Alternative Solution: Convert to Minutes with Set

Why an Alternative Approach?

The convert-to-minutes with set solution converts all times to minutes, stores them in a set for O(1) lookups, and checks all possible differences in O(n) time with O(n) space. It’s fast for small n but uses extra memory and requires careful handling of wrap-around. Great for set-based learners!

How It Works

Picture this as a time-minute collector:

• Step 1: Convert all times to minutes, store in set.
• Step 2: Find min difference by checking each time against others.
• Step 3: Include wrap-around check.
• Step 4: Return smallest difference.

It’s like a time-minute gap finder!

Step-by-Step Example

Example: timePoints = ["23:59", "00:00"]

• Step 1: Convert to minutes:
• "23:59" → 1439.
• "00:00" → 0.
• times_set = {0, 1439}.
• Step 2: Check differences:
• For 0: 1439 - 0 = 1439.
• For 1439: 0 + 1440 - 1439 = 1 (wrap-around).
• Step 3: Min = min(1439, 1) = 1.
• Result: 1.

Code for Set Approach

Copy codeclass Solution:
    def findMinDifference(self, timePoints: List[str]) -> int:
        # Step 1: Convert to minutes and store in set
        def to_minutes(time_str):
            hours, minutes = map(int, time_str.split(':'))
            return hours * 60 + minutes

        times_set = set()
        for time in timePoints:
            minutes = to_minutes(time)
            if minutes in times_set:  # Duplicate check
                return 0
            times_set.add(minutes)

        # Step 2: Convert set to list and sort
        times = sorted(times_set)

        # Step 3: Find min difference
        min_diff = float('inf')
        for i in range(1, len(times)):
            diff = times[i] - times[i-1]
            min_diff = min(min_diff, diff)

        # Step 4: Check wrap-around
        wrap_around = (times[0] + 1440) - times[-1]
        min_diff = min(min_diff, wrap_around)

        return min_diff

• Lines 4-6: Convert "HH:MM" to minutes.
• Lines 9-13: Build set, check duplicates.
• Line 16: Sort times for adjacent differences.
• Lines 19-22: Compute adjacent differences.
• Lines 25-26: Check wrap-around.
• Time Complexity: O(n log n)—sorting dominates.
• Space Complexity: O(n)—set storage.

It’s a set-based time measurer!

Comparing the Two Solutions

• Sorting (Best):
• Pros: O(n log n), O(1), space-efficient.
• Cons: Sorting overhead.
• Set (Alternative):
• Pros: O(n log n), O(n), fast duplicate check.
• Cons: Extra space.

Sorting wins for space efficiency!

Additional Examples and Edge Cases

• ["00:00"]: Invalid (needs ≥ 2), assume handled.
• ["12:00", "12:00"]: 0.
• ["00:00", "23:00"]: 60.

Sorting handles them all!

Complexity Recap

• Sorting: Time O(n log n), Space O(1).
• Set: Time O(n log n), Space O(n).

Sorting’s the space champ!

Key Takeaways

• Sorting: Clockwise gaps—learn at Python Basics!
• Set: Memory for speed.
• Times: Circular fun.
• Python: Sort or sets, your pick!

Final Thoughts: Measure That Time!

LeetCode 539: Minimum Time Difference in Python is a clever time challenge. Sorting with wrap-around check is your fast track, while set-based minutes offers a memory twist. Want more? Try LeetCode 75: Sort Colors or LeetCode 56: Merge Intervals. Ready to time? Head to Solve LeetCode 539 on LeetCode and find those gaps today!