triplet_sum.py

"""
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, ...]:
    """
    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]}.")