"""
In the Combination Sum problem, we are given a list consisting of distinct integers.
We need to find all the combinations whose sum equals to target given.
We can use an element more than one.
Time complexity(Average Case): O(n!)
Constraints:
1 <= candidates.length <= 30
2 <= candidates[i] <= 40
All elements of candidates are distinct.
1 <= target <= 40
"""
def backtrack(
candidates: list, path: list, answer: list, target: int, previous_index: int
) -> None:
"""
A recursive function that searches for possible combinations. Backtracks in case
of a bigger current combination value than the target value.
Parameters
----------
previous_index: Last index from the previous search
target: The value we need to obtain by summing our integers in the path list.
answer: A list of possible combinations
path: Current combination
candidates: A list of integers we can use.
"""
if target == 0:
answer.append(path.copy())
else:
for index in range(previous_index, len(candidates)):
if target >= candidates[index]:
path.append(candidates[index])
backtrack(candidates, path, answer, target - candidates[index], index)
path.pop(len(path) - 1)
def combination_sum(candidates: list, target: int) -> list:
"""
>>> combination_sum([2, 3, 5], 8)
[[2, 2, 2, 2], [2, 3, 3], [3, 5]]
>>> combination_sum([2, 3, 6, 7], 7)
[[2, 2, 3], [7]]
>>> combination_sum([-8, 2.3, 0], 1)
Traceback (most recent call last):
...
RecursionError: maximum recursion depth exceeded in comparison
"""
path = []
answer = []
backtrack(candidates, path, answer, target, 0)
return answer
def main() -> None:
print(combination_sum([-8, 2.3, 0], 1))
if __name__ == "__main__":
import doctest
doctest.testmod()
main()