## The challenge

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.

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: []```

Example 4:

```Input: candidates = [1], target = 1
Output: [[1]]```

Example 5:

```Input: candidates = [1], target = 2
Output: [[1,1]]```

Constraints:

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

## The solution in Java code

This problem is perform for using a `backtracking` algorithm to solve it.

Backtracking is an algorithmic-technique for solving problems recursively by trying to build a solution incrementally, one piece at a time, removing those solutions that fail to satisfy the constraints of the problem at any point of time (by time, here, is referred to the time elapsed till reaching any level of the search tree).

geeksforgeeks.org

Using this technique, we can write the following code to solve our challenge:

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 `````` ``````class Solution { public List> combinationSum(int[] candidates, int target) { // create a return variable List> res = new ArrayList<>(); // use backtracking to attempt to solve // we will pass the following: back(res, new ArrayList<>(), // - return variable target, // - the target number candidates, // - all candidates 0 // - where we should start ); // return our answer return res; } // our backtracking helper method private void back(List> res, List tmp, int remains, int[] nums, int start) { // if our target is less than 0, get out if (remains<0) return; // if the remaining target is 0, return the target if (remains==0) { res.add(new ArrayList<>(tmp)); return; } // loop through all candidates for (int i=start; i