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539. Minimum Time Difference

Description

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.

 

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".

Solutions

Solution 1: Sorting

We notice that there can be at most $24 \times 60 = 1440$ distinct time points. Therefore, if the length of $timePoints$ exceeds $1440$, it implies there are duplicate time points, and we can return $0$ early.

Next, we iterate through the list of time points and convert it into a list of minutes $nums$. For example, for the time point 13:14, we convert it into $13 \times 60 + 14$.

Then, we sort the list of minutes in ascending order and append the smallest time $nums[0]$ plus $1440$ to the end of the list. This step is to handle the special case of the difference between the maximum and minimum values.

Finally, we iterate through the list of minutes to find the minimum difference between any two adjacent times.

The time complexity is $O(n \log n)$, and the space complexity is $O(n)$, where $n$ is the number of time points.

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class Solution:
    def findMinDifference(self, timePoints: List[str]) -> int:
        if len(timePoints) > 1440:
            return 0
        nums = sorted(int(x[:2]) * 60 + int(x[3:]) for x in timePoints)
        nums.append(nums[0] + 1440)
        return min(b - a for a, b in pairwise(nums))

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class Solution {
    public int findMinDifference(List<String> timePoints) {
        if (timePoints.size() > 1440) {
            return 0;
        }
        int n = timePoints.size();
        int[] nums = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            String[] t = timePoints.get(i).split(":");
            nums[i] = Integer.parseInt(t[0]) * 60 + Integer.parseInt(t[1]);
        }
        Arrays.sort(nums, 0, n);
        nums[n] = nums[0] + 1440;
        int ans = 1 << 30;
        for (int i = 1; i <= n; ++i) {
            ans = Math.min(ans, nums[i] - nums[i - 1]);
        }
        return ans;
    }
}

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class Solution {
public:
    int findMinDifference(vector<string>& timePoints) {
        if (timePoints.size() > 1440) {
            return 0;
        }
        int n = timePoints.size();
        vector<int> nums(n + 1);
        for (int i = 0; i < n; ++i) {
            int hours = stoi(timePoints[i].substr(0, 2));
            int minutes = stoi(timePoints[i].substr(3, 2));
            nums[i] = hours * 60 + minutes;
        }
        sort(nums.begin(), nums.begin() + n);
        nums[n] = nums[0] + 1440;
        int ans = INT_MAX;
        for (int i = 1; i <= n; ++i) {
            ans = min(ans, nums[i] - nums[i - 1]);
        }
        return ans;
    }
};

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func findMinDifference(timePoints []string) int {
    if len(timePoints) > 1440 {
        return 0
    }

    n := len(timePoints)
    nums := make([]int, n+1)
    for i, time := range timePoints {
        parts := strings.Split(time, ":")
        hours, _ :=