Min Time Difference.py

class Solution:
    def findMinDiff(self, times):
        if len(times) < 1:
            return 0
        timelist = []
        # convert all times to minutes
        for time in times:
            minutes = int(time[0:2]) * 60 + int(time[3:5])
            minutes2 = minutes + (24 * 60) # we need to cover for times closer to midnight (00:00)
            timelist.append(minutes2)
            timelist.append(minutes)
            
        timelist.sort()
        diff = (24 * 60) + 1 # set a dummy to estimate minimum
        for i in range(len(timelist) - 1):
            diff = min(diff, timelist[i+1] - timelist[i])
        return diff
      
    # Time O(NlogN)
    # Space O(N)

neat code

Hello @KingAshiru, congratulations 🎉 your solution has been selected as one of the winning solutions in Week 7 of #AlgorithmFridays.

Your solution is clean and readable and really neat.

Out of the many winning solutions, only 3 will be selected for the 20$ prize in a raffle draw. The raffle draw will hold today, Friday May 28 at 3.00pm WAT (7.00 am PST)

If you are interested in being a part of the raffle draw, please send me a DM on Twitter @meekg33k so I can share the event invite with you.

NB: Only solutions of participants who indicated interest in the raffle draw will be considered.

Thanks once again for participating and see you later today for Week 8 of #AlgorithmFridays.