Q20: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 :**

```
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.
Solution:
class Solution {
public:
void search(vector<int>& num, int next, vector<int>& pSol, int target, vector<vector<int> >& result)
{
if(target == 0)
{
result.push_back(pSol);
return;
}
if(next == num.size() || target - num[next] < 0)
return;
pSol.push_back(num[next]);
search(num, next, pSol, target - num[next], result);
pSol.pop_back();
search(num, next + 1, pSol, target, result);
}
vector<vector<int>> combinationSum(vector<int> &num, int target) {
vector<vector<int> > result;
sort(num.begin(), num.end());
vector<int> pSol;
search(num, 0, pSol, target, result);
return result;
}
};
Explantion:
>In this solution we use the concept of backtracking ,we which we call the function
again and again until we get the desire output
>In this code we call the function search again and again for the solution when
we find the solution we push that answer in the result vector and than return the
result answer.
#If anyone have better solution so please comment:)
```