Problem statement

https://leetcode.com/problems/minimum-time-difference/

Solution

One possible way is to transform our times to numbers between 0 and 24*60 - 1 and sort them, then choose the minimum difference between adjacent points and also do not forget about difference between maximum and 24 * 60 minus minimum.

Complexity

Time complexity will be O(n log n), space complexity O(n).

Code

class Solution:
    def findMinDifference(self, timePoints):
        times = sorted([int(x[:2])*60 + int(x[3:]) for x in timePoints])
        return min(min(y-x for x, y in zip(times, times[1:])), times[0] + 1440 - times[-1])

Remark

We can also use optimization, that if we have > 24 * 60 times, by pigeonhole principle we need to return 0. Also we can use bucket sort with O(24 * 60) time and memory, which is more preferable for big data.