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# 2023 0709 39. Combination Sum

39\. Combination SumMedium16.6K334Companies

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:**

<pre><code><strong>Input: candidates = [2,3,6,7], target = 7
</strong><strong>Output: [[2,2,3],[7]]
</strong><strong>Explanation:
</strong>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.
</code></pre>

**Example 2:**

<pre><code><strong>Input: candidates = [2,3,5], target = 8
</strong><strong>Output: [[2,2,2,2],[2,3,3],[3,5]]
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: candidates = [2], target = 1
</strong><strong>Output: []
</strong></code></pre>

**Constraints:**

* `1 <= candidates.length <= 30`
* `2 <= candidates[i] <= 40`
* All elements of `candidates` are **distinct**.
* `1 <= target <= 40`

{% embed url="<https://leetcode.com/problems/combination-sum/description/>" %}

> time: O(2^n or n^m) ?

> space: O(m^2) ?

```jsx
/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */

let answer = [];
var combinationSum = function(candidates, target) {
    answer = [];
    findCombination(candidates, 0, target);
    return answer;
};

function findCombination(arr, idx, target, currArr = [], currSum = 0) {
    for (let i = idx; i < arr.length; i++) {
        const curr = arr[i];
        if (currSum === target) {
            answer.push(currArr);
            break;
        } else if (currSum > target) {
            continue;
        } else if (currSum < target) {
            findCombination(arr, i, target, [...currArr, curr], currSum + curr);
        }
    }

}

/**
points
1. how to create combination with using duplicate allowed
2. new reference
3. when to stop?
4. time? 2 <= num <= 40, 1 <= target <= 40
    - depth 20 => N
    - time 20 => N
 */
```

Hao

```javascript
var combinationSum = function(candidates, target) {
    const answer = [];
    candidates.forEach((candidate, index) => {
        const nextTarget = target - candidate;
        if (nextTarget === 0) {
            answer.push([target]);
        }
        if (nextTarget > 0) {
            combinationSum(candidates.slice(index), nextTarget).forEach((partialSum) => {
                answer.push([candidate, ...partialSum])
            })
        }
    })

    return answer;
};
```

Choozil

```javascript
/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */

let res;

var combinationSum = function(candidates, target) {
   
    res = []; 
    
    const backtrack = (sum, selected ,idx) => {
        if(sum === target){            
            res.push(selected);
            return;
        }
        
        else if(sum > target){
            return;
        } 
        
        for(let i=idx; i<candidates.length;i++){
            selected.push(candidates[i]);
            sum += candidates[i];
            backtrack(sum, [...selected], i);
            selected.pop();
            sum -= candidates[i];
        }
    }
    
    backtrack(0, [], 0);
    return res;
};
```


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