2021-03-22 09:52:26 +00:00
|
|
|
"""
|
|
|
|
Given an array of integers and another integer target,
|
|
|
|
we are required to find a triplet from the array such that it's sum is equal to
|
|
|
|
the target.
|
|
|
|
"""
|
|
|
|
from __future__ import annotations
|
|
|
|
|
|
|
|
from itertools import permutations
|
|
|
|
from random import randint
|
|
|
|
from timeit import repeat
|
|
|
|
|
|
|
|
|
|
|
|
def make_dataset() -> tuple[list[int], int]:
|
|
|
|
arr = [randint(-1000, 1000) for i in range(10)]
|
|
|
|
r = randint(-5000, 5000)
|
|
|
|
return (arr, r)
|
|
|
|
|
|
|
|
|
|
|
|
dataset = make_dataset()
|
|
|
|
|
|
|
|
|
|
|
|
def triplet_sum1(arr: list[int], target: int) -> tuple[int, int, int]:
|
|
|
|
"""
|
|
|
|
Returns a triplet in the array with sum equal to target,
|
|
|
|
else (0, 0, 0).
|
|
|
|
>>> triplet_sum1([13, 29, 7, 23, 5], 35)
|
|
|
|
(5, 7, 23)
|
|
|
|
>>> triplet_sum1([37, 9, 19, 50, 44], 65)
|
|
|
|
(9, 19, 37)
|
|
|
|
>>> arr = [6, 47, 27, 1, 15]
|
|
|
|
>>> target = 11
|
|
|
|
>>> triplet_sum1(arr, target)
|
|
|
|
(0, 0, 0)
|
|
|
|
"""
|
|
|
|
for triplet in permutations(arr, 3):
|
|
|
|
if sum(triplet) == target:
|
|
|
|
return tuple(sorted(triplet))
|
|
|
|
return (0, 0, 0)
|
|
|
|
|
|
|
|
|
|
|
|
def triplet_sum2(arr: list[int], target: int) -> tuple[int, int, int]:
|
|
|
|
"""
|
|
|
|
Returns a triplet in the array with sum equal to target,
|
|
|
|
else (0, 0, 0).
|
|
|
|
>>> triplet_sum2([13, 29, 7, 23, 5], 35)
|
|
|
|
(5, 7, 23)
|
|
|
|
>>> triplet_sum2([37, 9, 19, 50, 44], 65)
|
|
|
|
(9, 19, 37)
|
|
|
|
>>> arr = [6, 47, 27, 1, 15]
|
|
|
|
>>> target = 11
|
|
|
|
>>> triplet_sum2(arr, target)
|
|
|
|
(0, 0, 0)
|
|
|
|
"""
|
|
|
|
arr.sort()
|
|
|
|
n = len(arr)
|
|
|
|
for i in range(n - 1):
|
|
|
|
left, right = i + 1, n - 1
|
|
|
|
while left < right:
|
|
|
|
if arr[i] + arr[left] + arr[right] == target:
|
|
|
|
return (arr[i], arr[left], arr[right])
|
|
|
|
elif arr[i] + arr[left] + arr[right] < target:
|
|
|
|
left += 1
|
|
|
|
elif arr[i] + arr[left] + arr[right] > target:
|
|
|
|
right -= 1
|
|
|
|
return (0, 0, 0)
|
|
|
|
|
|
|
|
|
|
|
|
def solution_times() -> tuple[float, float]:
|
|
|
|
setup_code = """
|
|
|
|
from __main__ import dataset, triplet_sum1, triplet_sum2
|
|
|
|
"""
|
|
|
|
test_code1 = """
|
|
|
|
triplet_sum1(*dataset)
|
|
|
|
"""
|
|
|
|
test_code2 = """
|
|
|
|
triplet_sum2(*dataset)
|
|
|
|
"""
|
|
|
|
times1 = repeat(setup=setup_code, stmt=test_code1, repeat=5, number=10000)
|
|
|
|
times2 = repeat(setup=setup_code, stmt=test_code2, repeat=5, number=10000)
|
|
|
|
return (min(times1), min(times2))
|
|
|
|
|
|
|
|
|
|
|
|
if __name__ == "__main__":
|
|
|
|
from doctest import testmod
|
|
|
|
|
|
|
|
testmod()
|
|
|
|
times = solution_times()
|
|
|
|
print(f"The time for naive implementation is {times[0]}.")
|
|
|
|
print(f"The time for optimized implementation is {times[1]}.")
|