Combination Sum

Problem

https://leetcode.com/problems/combination-sum/

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1
Output: []

Constraints:

• 1 <= candidates.length <= 30

• 2 <= candidates[i] <= 40

• All elements of candidates are distinct.

• 1 <= target <= 40

Pattern

Array, Backtracking

Approaches

Backtracking

Code

def combinationSum(candidates: list[int], target: int) -> list[list[int]]:
    """Find all unique combinations in ``candidates`` where the candidate
    numbers sum to ``target``, reusing each number in ``candidates`` an
    unlimited number of times.
    """
    result = []

    candidates = sorted(candidates)

    def combination_sum(combination, i):
        s = sum(combination)

        if s == target:
            result.append(combination.copy())
            return

        if i >= len(candidates):
            return

        if s > target:
            return

        for j in range(i, len(candidates)):
            if s + candidates[j] <= target:
                combination_sum(combination + [candidates[j]], j)
            else:
                break

    combination_sum([], 0)
    return result

Test

>>> from combination_sum__backtracking import combinationSum
>>> sorted([sorted(c) for c in combinationSum([2, 3, 6, 7], 7)])
[[2, 2, 3], [7]]
>>> sorted([sorted(c) for c in combinationSum([2, 3, 5], 8)])
[[2, 2, 2, 2], [2, 3, 3], [3, 5]]
>>> combinationSum([2], 1)
[]

combination_sum__backtracking.combinationSum(candidates: list[int], target: int) → list[list[int]]

Find all unique combinations in candidates where the candidate numbers sum to target, reusing each number in candidates an unlimited number of times.