Combination Sum

Because each number can be reused, this is not a simple “include or exclude” subset problem — at every step we choose how many more times to use the current candidate (including zero) before moving on to the next one.

Brute Force: Generate-and-Filter

Time exponentialSpace exponential

Recurse over “use candidate i, or don’t” like Subsets, but allow re-using an index unboundedly, and only check the sum against target once a full combination is built. Without pruning on the running sum, this explores far more branches than necessary — for example it keeps recursing even after the partial sum has already exceeded target.

class Solution:
    def combinationSum(self, candidates: list[int], target: int) -> list[list[int]]:
        result = []
        path = []

        def backtrack(start, path_sum):
            if path_sum == target:
                result.append(path[:])
                return
            if start == len(candidates):
                return
            # option 1: use candidates[start] again
            if path_sum + candidates[start] <= target:
                path.append(candidates[start])
                backtrack(start, path_sum + candidates[start])
                path.pop()
            # option 2: move on to the next candidate
            backtrack(start + 1, path_sum)

        backtrack(0, 0)
        return result

This already has the path_sum <= target check, otherwise it degenerates into infinite recursion (reusing a positive number forever). Even with it, branching into “use again” and “skip” at every node explores overlapping work that the loop-based version below avoids.

Backtracking with a For-Loop and Pruning

OptimalTime O(n^(t/m + 1))Space O(t/m)

Sort the candidates first. At each recursive call, loop over candidates starting from the current index (allowing the same index to be reused on the next call), and stop the loop as soon as a candidate exceeds the remaining budget — since the array is sorted, everything after it is too large as well.

class Solution:
    def combinationSum(self, candidates: list[int], target: int) -> list[list[int]]:
        candidates.sort()
        result = []
        path = []

        def backtrack(start, remaining):
            if remaining == 0:
                result.append(path[:])
                return
            for i in range(start, len(candidates)):
                if candidates[i] > remaining:
                    break  # sorted, so nothing further can fit either
                path.append(candidates[i])
                backtrack(i, remaining - candidates[i])  # i, not i+1: reuse allowed
                path.pop()

        backtrack(0, target)
        return result

Play the trace below to watch the loop reuse the same index and then break the moment a candidate exceeds the remaining budget.

Candidates

loop starts at index 0

Current combination

empty

Remaining

77 target − 0 sum

Recorded combinations

none yet

1 / 10

currentseendiscardedresult

Start: path is empty and 7 remain. The loop begins at index 0, so every candidate is on the table.

Complexity: in the worst case (candidates like [1] with a large target) the recursion depth is t/m where m is the smallest candidate, and each level branches over n candidates, giving O(n^(t/m + 1)) time; the recursion stack and path are bounded by that same depth → O(t/m) space. Sorting first lets the break prune entire subtrees early rather than visiting every combination that overshoots.

Brute Force: Generate-and-Filter

Backtracking with a For-Loop and Pruning

More Backtracking practice