Created triplet_sum in Python/other

[Kush1101]

Describe your change:

• Add an algorithm?
• Fix a bug or typo in an existing algorithm?
• Documentation change?

Checklist:

• I have read CONTRIBUTING.md.
• This pull request is all my own work -- I have not plagiarized.
• I know that pull requests will not be merged if they fail the automated tests.
• This PR only changes one algorithm file. To ease review, please open separate PRs for separate algorithms.
• All new Python files are placed inside an existing directory.
• All filenames are in all lowercase characters with no spaces or dashes.
• All functions and variable names follow Python naming conventions.
• All function parameters and return values are annotated with Python type hints.
• All functions have doctests that pass the automated testing.
• All new algorithms have a URL in its comments that points to Wikipedia or other similar explanation.
• If this pull request resolves one or more open issues then the commit message contains Fixes: #{$ISSUE_NO}.

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[Kush1101]

@cclauss Please review.

[cclauss]

OK... With this one we are going to write two different functions in the same file that solve the same problem and then we are going to write a timeit benchmark to tell us which one is faster. The second function should leverage the itertools module (combinations() or permutations()) and should pass the same doctests.

[Kush1101]

Okk. Working on it

[Kush1101]

OK... With this one we are going to write two different functions in the same file that solve the same problem and then we are going to write a timeit benchmark to tell us which one is faster. The second function should leverage the itertools module (combinations() or permutations()) and should pass the same doctests.

I have written two functions, but I am still not sure about the proper way to include timeit in this code.

def triplet_sum1(arr:List[int],X:int) ->tuple:
    """
    Returns a triplet in in array with sum equal to X,
    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]
    >>> summ = 11
    >>> triplet_sum1(arr,summ)
    (0, 0, 0)
    """        
    triplets = list(permutations(arr,3))
    for triplet in triplets:
        if sum(triplet) == X:
            return tuple(sorted(triplet))
    return (0, 0, 0)    
def triplet_sum2(arr: List[int], X: int) -> tuple:
    """
    Returns a triplet in in array with sum equal to X,
    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]
    >>> summ = 11
    >>> triplet_sum2(arr,summ)
    (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] == X:
                return (arr[i], arr[left], arr[right])
            elif arr[i] + arr[left] + arr[right] < X:
                left += 1
            elif arr[i] + arr[left] + arr[right] > X:
                right -= 1
    else:
        return (0, 0, 0)

Now should I do it like this?

"""
"""
Given an array of integers and another integer X,
we are required to find a triplet from the array such that it's sum is equal to X
"""
from typing import List
from itertools import permutations
from timeit import repeat


def triplet_sum1(arr: List[int], X: int) -> tuple:
    """
    Returns a triplet in in array with sum equal to X,
    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]
    >>> summ = 11
    >>> triplet_sum1(arr,summ)
    (0, 0, 0)
    """
    triplets = list(permutations(arr, 3))
    for triplet in triplets:
        if sum(triplet) == X:
            return tuple(sorted(triplet))
    return (0, 0, 0)


def sol1_time():
    setup_code = """ 
from __main__ import triplet_sum1
from random import randint"""
    test_code = """
arr = [randint(-1000, 1000) for i in range(10)]
r = randint(-5000, 5000)
triplet_sum1(arr,r)
    """
    times = repeat(stmt=test_code, setup=setup_code, repeat=5, number=10000)
    return f"Naive solution time is {min(times)}"


def triplet_sum2(arr: List[int], X: int) -> tuple:
    """
    Returns a triplet in in array with sum equal to X,
    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]
    >>> summ = 11
    >>> triplet_sum2(arr,summ)
    (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] == X:
                return (arr[i], arr[left], arr[right])
            elif arr[i] + arr[left] + arr[right] < X:
                left += 1
            elif arr[i] + arr[left] + arr[right] > X:
                right -= 1
    else:
        return (0, 0, 0)


def sol2_time():
    setup_code = """ 
from __main__ import triplet_sum2
from random import randint"""
    test_code = """
arr = [randint(-1000, 1000) for i in range(10)]
r = randint(-5000, 5000)
triplet_sum2(arr,r)
    """
    times = repeat(stmt=test_code, setup=setup_code, repeat=5, number=10000)
    return f"Optimized solution time is {min(times)}"


if __name__ == "__main__":
    from doctest import testmod

    testmod()
    from random import randint

    arr = [randint(-1000, 1000) for i in range(100)]
    r = randint(-5000, 5000)
    print(f"{triplet_sum1(arr,r)}")
    print(f"{sol1_time()}")
    print(f"{triplet_sum2(arr,r)}")
    print(f"{sol2_time()}")

[Kush1101]

I made a PR, so it will be easier for you to review and propose changes. @cclauss

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[cclauss]

Nice! A few trailing whitespaces to clean up. When benchmarking it is vital that both algorithms are tested with the exact same dataset so please do something like this...

def make_dataset():
     <your code here>
    return arr, target


dataset = make_dataset()

...

from __main__ import dataset, triplet_sum1, triplet_sum2

...

triplet_sum1(*dataset)

...

triplet_sum2(*dataset)

[Kush1101]

Nice! A few trailing whitespaces to clean up. When benchmarking it is vital that both algorithms are tested with the exact same dataset so please do something like this...

def make_dataset():
     <your code here>
    return arr, target


dataset = make_dataset()

...

from __main__ import dataset, triplet_sum1, triplet_sum2

...

triplet_sum1(*dataset)

...

triplet_sum2(*dataset)

I will do it now.

[TravisBuddy]

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[TravisBuddy]

Hey @Kush1101,
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[TravisBuddy]

Hey @Kush1101,
Something went wrong with the build.

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[Kush1101]

I tried to incorporate all the proposed changes. Both the functions are now tested on the same dataset. I also made other small changes. Please review it now. @cclauss

[cclauss]

This is very impressive. Nice work. This shows the power of an algorithm. The Python standard library gives us tools to express complex things simply -- triplet_sum1() short, tidy, and easy to reason about. However, it is no match of the specific algorithm in triplet_sum2() that is longer and more difficult to reason about but delivers correct answers in a fraction of the time.

[cclauss]

EXCELLENT!! The benchmark clearly shows the winner!

[Kush1101]

This is very impressive. Nice work. This shows the power of an algorithm. The Python standard library gives us tools to express complex things simply -- triplet_sum1() short, tidy, and easy to reason about. However, it is no match of the specific algorithm in triplet_sum2() that is longer and more difficult to reason about but delivers correct answers in a fraction of the time.

True.