[JS] Combination Sum

🔒 문제 (LeetCode 39)

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.

Constraints:

• 1 <= candidates.length <= 30
• 1 <= candidates[i] <= 200
• All elements of candidates are distinct.
• 1 <= target <= 500

 

🌊 입출력

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: []

 

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🔑 해결

🌌 알고리즘 - Backtracking

/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum = function(candidates, target) {
    candidates.sort((a, b) => b-a)
    const result = []
    const backtracking = (comb, sum, idx) => {
        if(sum === target) {
            result.push(comb)
            return
        }
        
        if(sum > target) return
        
        for(let i = idx; i < candidates.length; i++) {
            if(sum + candidates[i] <= target) {
                comb.push(candidates[i])
                backtracking([...comb], sum + candidates[i], i)    
                comb.pop()  
            } else continue
        }
    }
    
    backtracking([], 0, 0)

    
    return result
};