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move-files-and-2-renames (TheAlgorithms#4285)
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@algobytewise
algobytewise authored Mar 22, 2021
1 parent 14bcb58 commit 8d51c2c
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Showing 11 changed files with 134 additions and 134 deletions.
0 other/binary_exponentiation_2.py → maths/binary_exponentiation_2.py
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0 other/binary_exponentiation.py → maths/binary_exponentiation_3.py
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0 other/euclidean_gcd.py → maths/euclidean_gcd.py
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0 other/integeration_by_simpson_approx.py → maths/integration_by_simpson_approx.py
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0 other/largest_subarray_sum.py → maths/largest_subarray_sum.py
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90 changes: 45 additions & 45 deletions other/max_sum_sliding_window.py → maths/max_sum_sliding_window.py
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@@ -1,45 +1,45 @@
"""
Given an array of integer elements and an integer 'k', we are required to find the
maximum sum of 'k' consecutive elements in the array.
Instead of using a nested for loop, in a Brute force approach we will use a technique
called 'Window sliding technique' where the nested loops can be converted to a single
loop to reduce time complexity.
"""
from typing import List


def max_sum_in_array(array: List[int], k: int) -> int:
"""
Returns the maximum sum of k consecutive elements
>>> arr = [1, 4, 2, 10, 2, 3, 1, 0, 20]
>>> k = 4
>>> max_sum_in_array(arr, k)
24
>>> k = 10
>>> max_sum_in_array(arr,k)
Traceback (most recent call last):
...
ValueError: Invalid Input
>>> arr = [1, 4, 2, 10, 2, 13, 1, 0, 2]
>>> k = 4
>>> max_sum_in_array(arr, k)
27
"""
if len(array) < k or k < 0:
raise ValueError("Invalid Input")
max_sum = current_sum = sum(array[:k])
for i in range(len(array) - k):
current_sum = current_sum - array[i] + array[i + k]
max_sum = max(max_sum, current_sum)
return max_sum


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

testmod()
array = [randint(-1000, 1000) for i in range(100)]
k = randint(0, 110)
print(f"The maximum sum of {k} consecutive elements is {max_sum_in_array(array,k)}")
"""
Given an array of integer elements and an integer 'k', we are required to find the
maximum sum of 'k' consecutive elements in the array.
Instead of using a nested for loop, in a Brute force approach we will use a technique
called 'Window sliding technique' where the nested loops can be converted to a single
loop to reduce time complexity.
"""
from typing import List


def max_sum_in_array(array: List[int], k: int) -> int:
"""
Returns the maximum sum of k consecutive elements
>>> arr = [1, 4, 2, 10, 2, 3, 1, 0, 20]
>>> k = 4
>>> max_sum_in_array(arr, k)
24
>>> k = 10
>>> max_sum_in_array(arr,k)
Traceback (most recent call last):
...
ValueError: Invalid Input
>>> arr = [1, 4, 2, 10, 2, 13, 1, 0, 2]
>>> k = 4
>>> max_sum_in_array(arr, k)
27
"""
if len(array) < k or k < 0:
raise ValueError("Invalid Input")
max_sum = current_sum = sum(array[:k])
for i in range(len(array) - k):
current_sum = current_sum - array[i] + array[i + k]
max_sum = max(max_sum, current_sum)
return max_sum


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

testmod()
array = [randint(-1000, 1000) for i in range(100)]
k = randint(0, 110)
print(f"The maximum sum of {k} consecutive elements is {max_sum_in_array(array,k)}")
0 other/median_of_two_arrays.py → maths/median_of_two_arrays.py
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0 other/primelib.py → maths/primelib.py
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178 changes: 89 additions & 89 deletions other/triplet_sum.py → maths/triplet_sum.py
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@@ -1,89 +1,89 @@
"""
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]}.")
"""
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]}.")
0 other/two_pointer.py → maths/two_pointer.py
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0 other/two_sum.py → maths/two_sum.py
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