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minimum-average-difference.py
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# 2256. Minimum Average Difference
# 🟠 Medium
#
# https://leetcode.com/problems/minimum-average-difference/
#
# Tags: Array - Prefix Sum
import timeit
from typing import List
# Compute the sum of all array elements in O(1), then iterate over them
# keeping the sum of the left and right side to compute the left and
# right side averages and their differences. Return the index of the
# smallest difference found.
#
# Time complexity: O(n) - We iterate twice over the array elements.
# Space complexity: O(1) - We use constant space.
#
# Runtime: 991 ms, faster than 97.47%
# Memory Usage: 24.9 MB, less than 56.92%
class Solution:
def minimumAverageDifference(self, nums: List[int]) -> int:
res, ls, rs, n = (float("inf"), 0), 0, sum(nums), len(nums)
for i, num in enumerate(nums):
# Add and remove the current value from the sums.
ls += num
rs -= num
# Compute the averages.
la = ls // (i + 1)
ra = rs // (n - i - 1) if i < n - 1 else 0
avg = abs(la - ra)
# Check if we need to update the minimum.
if avg < res[0]:
res = (avg, i)
# Return the index of the best average.
return res[1]
def test():
executors = [Solution]
tests = [
[[0], 0],
[[2, 5, 3, 9, 5, 3], 3],
]
for executor in executors:
start = timeit.default_timer()
for _ in range(1):
for col, t in enumerate(tests):
sol = executor()
result = sol.minimumAverageDifference(t[0])
exp = t[1]
assert result == exp, (
f"\033[93m» {result} <> {exp}\033[91m for"
+ f" test {col} using \033[1m{executor.__name__}"
)
stop = timeit.default_timer()
used = str(round(stop - start, 5))
cols = "{0:20}{1:10}{2:10}"
res = cols.format(executor.__name__, used, "seconds")
print(f"\033[92m» {res}\033[0m")
test()