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running-sum-of-1d-array.py
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# 1480. Running Sum of 1d Array
# 🟢 Easy
#
# https://leetcode.com/problems/running-sum-of-1d-array/
#
# Tags: Array - Prefix Sum
import timeit
from typing import List
# Iterate over the elements in the input computing the prefix sum.
#
# Time complexity: O(n) - We visit each element and do constant work for each.
# Space complexity: O(1) - We use constant extra space.
#
# Runtime 95 ms Beats 6.24%
# Memory 14.1 MB Beats 25.39%
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
sum = 0
for i, num in enumerate(nums):
sum += num
nums[i] = sum
return nums
# Similar logic to the previous solution but mutate the input array.
#
# Time complexity: O(n) - We visit each element and do constant work for each.
# Space complexity: O(1) - We use constant extra space.
#
# Runtime 97 ms Beats5.37%
# Memory 13.9 MB Beats 92.56%
class InPlace:
def runningSum(self, nums: List[int]) -> List[int]:
for i in range(1, len(nums)):
nums[i] += nums[i - 1]
return nums
def test():
executors = [
Solution,
InPlace,
]
tests = [
[[1, 2, 3, 4], [1, 3, 6, 10]],
[[1, 1, 1, 1, 1], [1, 2, 3, 4, 5]],
[[3, 1, 2, 10, 1], [3, 4, 6, 16, 17]],
[[1, 1, 8, -90, 1], [1, 2, 10, -80, -79]],
[[], []],
]
for executor in executors:
start = timeit.default_timer()
for _ in range(1):
for col, t in enumerate(tests):
sol = executor()
result = sol.runningSum([*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()