There are only 1440 times. Therefore put each time's occurence into a hashtable, if a same time appear twice than return 0, else sort the key of the map, find time difference between adjacent elements, then find the minimum of the time difference

```
class Solution(object):
def findMinDifference(self, timePoints):
"""
:type timePoints: List[str]
:rtype: int
"""
time=collections.defaultdict(int)
for t in timePoints:
hour=int(t[0:2])
minute=int(t[-2:])
time[hour*60+minute]+=1
if time[hour*60+minute]>1:
return 0
timeList=sorted(time.keys())
deltaM=[timeList[i+1]-timeList[i] for i in range(0, len(timeList)-1)]
deltaM.append(1440-timeList[-1]+timeList[0])
return min(deltaM)
```