539. Minimum Time Difference¶

Difficulty: Medium

539. Minimum Time Difference

Medium

Given a list of 24-hour clock time points in HH:MM format, return the minimum minutes difference between any two time-points in the list.

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Example 1:

Input: timePoints = ["23:59","00:00"]
Output: 1

Example 2:

Input: timePoints = ["00:00","23:59","00:00"]
Output: 0

Constraints:

2 <= timePoints.length <= 2 * 10⁴
timePoints[i] is in the format "HH:MM".

Solution¶

class Solution {
    public int findMinDifference(List<String> timePoints) {
        int n = timePoints.size();
        List<Integer> minutesList = new ArrayList<>();
        for (String time : timePoints) {
            String[] parts = time.split(":");
            int hours = Integer.parseInt(parts[0]);
            int minutes = Integer.parseInt(parts[1]);
            int totalMinutes = hours * 60 + minutes;
            minutesList.add(totalMinutes);
        }
        Collections.sort(minutesList);
        int minDiff = Integer.MAX_VALUE;
        for (int i = 1; i < minutesList.size(); i++) {
            int diff = minutesList.get(i) - minutesList.get(i - 1);
            minDiff = Math.min(minDiff, diff);
        }
        int circularDiff = 1440 - minutesList.get(minutesList.size() - 1) + minutesList.get(0);
        minDiff = Math.min(minDiff, circularDiff);
        return minDiff;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here