# Combination Sum

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.
```

## Idea

{% hint style="info" %}
Recursion, in-place update
{% endhint %}

Construct recursive method "comb" with params: int\[] candidates, int target, List\<List>> result, List<>curr, int currIndex

If target==0 and curr not empty, a combination found, add to result

Iterate from currIndex, add currIndex element to curr, recursively call comb on curr with updated target value, if curr result in correct combination, execute base case above, else, return

After all element from a given index iterated, remove the last element from curr and go on iteration

## &#x20;Code

```java
public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> output = new ArrayList<>();
        List<Integer> curr = new ArrayList<>();
        comb(candidates, target, output, curr, 0);
        return output;
}
    private void comb(int[] candidates, int target, List<List<Integer>> output, List<Integer> curr, int currInd){
        if(target==0 && !curr.isEmpty()){
            output.add(new ArrayList<>(curr));
        }
        else{
            for(int i = currInd; i<candidates.length; i++){
                if(target-candidates[i]>=0){
                    curr.add(candidates[i]);
                
                    comb(candidates, target-candidates[i], output, curr, i);
                    curr.remove(curr.size()-1);
                }
            }
        }
    }
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://yingzehou.gitbook.io/docs/leetcode/array/combination-sum.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.
