### View *Minimum Time Difference* on LeetCode

## Statistics

**Time Spent Coding**

20 minutes

**Time Complexity**

O(n^{2}) - For every element at index i, its difference is calculated with the elements in the range i+1 to n, resulting in the O(n^{2}) time complexity.

**Space Complexity**

O(1) - No extra space is required to run this program since the converted values replace their string counterparts, resulting in the O(1) space complexity.

**Runtime Beats**

90.32% of other submissions

**Memory Beats**

89.41% of other sumbissions

## Explanation

For each element at index i, it calculates the difference between it and the elements in the range i+1 to n.

If the difference is greater than 12 hours, then the shortest time difference is no longer forward in time but backward, e.g. 00:00 to 23:59. Forward, this would be a difference of 23 hours and 59 minutes, but backward, this would only be a difference of 1 minute.

## Solution

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class Solution:
def str_to_int(self,t):
h = t[:2]
m = t[3:]
return int(m) + int(h) * 60
def findMinDifference(self, timePoints: List[str]) -> int:
length = len(timePoints)
day = 1440
min_time = 1441
# Convert the first element into an integer before the for loop to avoid checking its type within the first for loop.
timePoints[0] = self.str_to_int(timePoints[0])
for i in range(length - 1):
t1 = timePoints[i]
for j in range(i+1,length):
# If the element is a str, convert it to an int
if isinstance(timePoints[j],str):
timePoints[j] = self.str_to_int(timePoints[j])
t2 = timePoints[j]
diff = abs(t1-t2)
# Exit early if a difference of 0 is found
if diff == 0:
return 0
# Differences under 12 hours (720 minutes) benefit from going forwards in time
elif diff < 720:
min_time = min(min_time, diff)
# Differences greater than 12 hours benefit from going backward in time
else:
min_time = min(min_time, day - diff)
return min_time