mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-05 02:40:16 +00:00
4700297b3e
* Enable ruff RUF002 rule * Fix --------- Co-authored-by: Christian Clauss <cclauss@me.com>
102 lines
2.7 KiB
Python
102 lines
2.7 KiB
Python
"""
|
|
Question:
|
|
You are given an array of distinct integers and you have to tell how many
|
|
different ways of selecting the elements from the array are there such that
|
|
the sum of chosen elements is equal to the target number tar.
|
|
|
|
Example
|
|
|
|
Input:
|
|
N = 3
|
|
target = 5
|
|
array = [1, 2, 5]
|
|
|
|
Output:
|
|
9
|
|
|
|
Approach:
|
|
The basic idea is to go over recursively to find the way such that the sum
|
|
of chosen elements is “tar”. For every element, we have two choices
|
|
1. Include the element in our set of chosen elements.
|
|
2. Don't include the element in our set of chosen elements.
|
|
"""
|
|
|
|
|
|
def combination_sum_iv(array: list[int], target: int) -> int:
|
|
"""
|
|
Function checks the all possible combinations, and returns the count
|
|
of possible combination in exponential Time Complexity.
|
|
|
|
>>> combination_sum_iv([1,2,5], 5)
|
|
9
|
|
"""
|
|
|
|
def count_of_possible_combinations(target: int) -> int:
|
|
if target < 0:
|
|
return 0
|
|
if target == 0:
|
|
return 1
|
|
return sum(count_of_possible_combinations(target - item) for item in array)
|
|
|
|
return count_of_possible_combinations(target)
|
|
|
|
|
|
def combination_sum_iv_dp_array(array: list[int], target: int) -> int:
|
|
"""
|
|
Function checks the all possible combinations, and returns the count
|
|
of possible combination in O(N^2) Time Complexity as we are using Dynamic
|
|
programming array here.
|
|
|
|
>>> combination_sum_iv_dp_array([1,2,5], 5)
|
|
9
|
|
"""
|
|
|
|
def count_of_possible_combinations_with_dp_array(
|
|
target: int, dp_array: list[int]
|
|
) -> int:
|
|
if target < 0:
|
|
return 0
|
|
if target == 0:
|
|
return 1
|
|
if dp_array[target] != -1:
|
|
return dp_array[target]
|
|
answer = sum(
|
|
count_of_possible_combinations_with_dp_array(target - item, dp_array)
|
|
for item in array
|
|
)
|
|
dp_array[target] = answer
|
|
return answer
|
|
|
|
dp_array = [-1] * (target + 1)
|
|
return count_of_possible_combinations_with_dp_array(target, dp_array)
|
|
|
|
|
|
def combination_sum_iv_bottom_up(n: int, array: list[int], target: int) -> int:
|
|
"""
|
|
Function checks the all possible combinations with using bottom up approach,
|
|
and returns the count of possible combination in O(N^2) Time Complexity
|
|
as we are using Dynamic programming array here.
|
|
|
|
>>> combination_sum_iv_bottom_up(3, [1,2,5], 5)
|
|
9
|
|
"""
|
|
|
|
dp_array = [0] * (target + 1)
|
|
dp_array[0] = 1
|
|
|
|
for i in range(1, target + 1):
|
|
for j in range(n):
|
|
if i - array[j] >= 0:
|
|
dp_array[i] += dp_array[i - array[j]]
|
|
|
|
return dp_array[target]
|
|
|
|
|
|
if __name__ == "__main__":
|
|
import doctest
|
|
|
|
doctest.testmod()
|
|
target = 5
|
|
array = [1, 2, 5]
|
|
print(combination_sum_iv(array, target))
|