Published Jun 12, 2022
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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.
It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.
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.
Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]
Input: candidates = [2], target = 1
Output: []
1 <= candidates.length <= 301 <= candidates[i] <= 200candidates are distinct.1 <= target <= 500candidates, we can either include it or not include it,
this affects the total sum & how we move the pointerfunction combinationSum(candidates: number[], target: number): number[][] {
const LEN: number = candidates.length;
const res: number[][] = []
const candidate: number[] = []
const dfs = (i: number, remain: number) => {
if(remain === 0) {
res.push([...candidate])
return
}
if(i === LEN || remain < 0) return;
// using current number
candidate.push(candidates[i])
// duplicate
dfs(i, remain - candidates[i])
// not using current number, move to next
candidate.pop()
dfs(i + 1, remain)
}
dfs(0, target)
return res;
};