Combination Sum

Problem

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.

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
1 <= candidates[i] <= 200
All elements of candidates are distinct.
1 <= target <= 500

Solution

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum = function(candidates, target) {
    const res = [];
    backtrack(res, [], candidates, target, 0);
    return res;
};

var backtrack = function(res, path, nums, remain, start) {
    if (remain < 0) {
        return;
    } else if (remain === 0) {
        res.push([...path]);
    } else {
        for (let i = start; i < nums.length; i++) {
            path.push(nums[i]);
            backtrack(res, path, nums, remain - nums[i], i)
            path.pop();
        }
    }
};

We will implement a backtracking (DFS) solution. In our backtrack function, we use res to keep track of all valid combinations recorded so far, path to keep track of the current potential combination that adds up to target, nums for all the available numbers, remain for the value required such that the sum of path equals target, and start for the starting index of nums that can be used (ie. nums[:start] cannot be used. Note that since we can reuse values in nums any number of times, when we make our recursive call, start is i instead of i + 1 (ie. the value at nums[i] can be pushed to path again).

Problem​

Solution​

Problem

Solution