Skip to content

Data structures/arrays/triplet sum #11134

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
59 changes: 58 additions & 1 deletion data_structures/arrays/find_triplets_with_0_sum.py
Original file line number Diff line number Diff line change
@@ -1,7 +1,7 @@
from itertools import combinations


def find_triplets_with_0_sum(nums: list[int]) -> list[list[int]]:
def find_triplets_with_0_sum(nums: list) -> list:
Copy link
Member

@cclauss cclauss Nov 5, 2023

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Why?!?

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Since after updating the code in the existing file, "(nums: list[int]) -> list[list[int]]:" was failing in ruff test.
(ruff find_triplets_with_0_sum.py was throwing error so I changed to ( nums: list) -> list: )

"""
Given a list of integers, return elements a, b, c such that a + b + c = 0.
Args:
Expand All @@ -22,3 +22,60 @@ def find_triplets_with_0_sum(nums: list[int]) -> list[list[int]]:
list(x)
for x in sorted({abc for abc in combinations(sorted(nums), 3) if not sum(abc)})
]


def find_triplets_with_0_sum_hashing(arr: list) -> list:
"""
Function for finding the triplets with a given sum in the array using hashing.

Given a list of integers, return elements a, b, c such that a + b + c = 0.

Args:
nums: list of integers
Returns:
list of lists of integers where sum(each_list) == 0
Examples:
>>> find_triplets_with_0_sum_hashing([-1, 0, 1, 2, -1, -4])
[[-1, 0, 1], [-1, -1, 2]]
>>> find_triplets_with_0_sum_hashing([])
[]
>>> find_triplets_with_0_sum_hashing([0, 0, 0])
[[0, 0, 0]]
>>> find_triplets_with_0_sum_hashing([1, 2, 3, 0, -1, -2, -3])
[[-1, 0, 1], [-3, 1, 2], [-2, 0, 2], [-2, -1, 3], [-3, 0, 3]]

Time complexity: O(N^2)
Auxiliary Space: O(N)
Comment on lines +47 to +48
Copy link
Member

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

How do we prove these claims?

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Time complexity: O(N^2)
--> This approach has two nested for loops, outer loop runs from start till end of the array length and inner loop from start+1 to end, that's why it has N^2 time complexity.
Auxiliary Space: O(N)
--> This method Creates a hashmap to store the elements so n extra space has been used.


"""
target_sum = 0

# Initialize the final output array with blank.
output_arr = []

# Set the initial element as arr[i].
for i in range(len(arr) - 2):
Copy link
Member

@cclauss cclauss Nov 5, 2023

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Suggested change
for i in range(len(arr) - 2):
for i, item in enumerate(arr[:-2]):

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

updated as suggested.

# to store second elements that can complement the final sum.
set_initialize = set()

# current sum needed for reaching the target sum
current_sum = target_sum - arr[i]

# Traverse the subarray arr[i+1:].
for j in range(i + 1, len(arr)):
Copy link
Member

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Suggested change
for j in range(i + 1, len(arr)):
for other_item in arr[i+1:]:

Copy link
Contributor Author

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

updated as suggested.

# required value for the second element
required_value = current_sum - arr[j]

# Verify if the desired value exists in the set.
if required_value in set_initialize:
# finding triplet elements combination.
combination_array = sorted([arr[i], arr[j], required_value])
if combination_array not in output_arr:
output_arr.append(combination_array)

# Include the current element in the set
# for subsequent complement verification.
set_initialize.add(arr[j])

# Return all the triplet combinations.
return output_arr