LC 39 - Combination Sum

Question

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

Links

Question here and solution here

Solution

concept

Use backtrack algorithm. One thing to note is that since we allows duplicates in choosing our elements, the idx for the inclusive decision do not need to change (i.e. no need +1 for that branch) since we can choose (or go down that branch again).

code

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        ans = []
        tmp = []

        def dfs(idx):
            if sum(tmp) >= target or idx >= len(candidates):
                if sum(tmp) == target:
                    ans.append(tmp.copy())
                return

            tmp.append(candidates[idx])
            dfs(idx)

            tmp.pop()
            dfs(idx + 1)

        dfs(0)
        return ans
        
class Solution:
	"""
	same solution but pass in the total amount as well
	to avoid summing each time to check if the conditions are met
	"""
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        res = []

        def dfs(i, cur, total):
            if total == target:
                res.append(cur.copy())
                return
            if i >= len(candidates) or total > target:
                return

            cur.append(candidates[i])
            dfs(i, cur, total + candidates[i])
            cur.pop()
            dfs(i + 1, cur, total)

        dfs(0, [], 0)
        return res

Complexity

time: $O(2^{t/m})$
space: $O(t/m)$