This problem can be solved using a technique called backtracking. Here is a Python solution:

```python
def combinationSum(candidates, target):
def backtrack(remain, comb, start):
if remain == 0:
result.append(list(comb))
return
elif remain

Question

This problem can be solved using a technique called backtracking. Here is a Python solution:

```python
def combinationSum(candidates, target):
    def backtrack(remain, comb, start):
        if remain == 0:
            result.append(list(comb))
            return
        elif remain < 0:
            return
        for i in range(start, len(candidates)):
            comb.append(candidates[i])
            backtrack(remain - candidates[i], comb, i)
            comb.pop()
    result = []
    backtrack(target, [], 0)
    return result
```

Here's how it works:

1. We define a helper function `backtrack` that takes the remaining sum after subtracting the current number, the current combination, and the current start index.

2. If the remaining sum is 0, we have found a valid combination, so we add it to the result list.

3. If the remaining sum is less than 0, we have exceeded the target, so we return without adding the combination to the result list.

4. We iterate over the candidates starting from the current index. For each candidate, we add it to the current combination and recursively call `backtrack` with the updated remaining sum and the current index. We then remove the candidate from the combination to backtrack and try the next candidate.

5. Finally, we call `backtrack` with the initial target, an empty combination, and a start index of 0, and return the result list.

Knowee AI · Accepted Answer

This problem can be solved using a technique called backtracking. Here is a Python solution:

```python
def combinationSum(candidates, target):
    def backtrack(remain, comb, start):
        if remain == 0:
            result.append(list(comb))
            return
        elif remain < 0:
            return
        for i in range(start, len(candidates)):
            comb.append(candidates[i])
            backtrack(remain - candidates[i], comb, i)
            comb.pop()
    result = []
    backtrack(target, [], 0)
    return result
```

Here's how it works:

1. We define a helper function `backtrack` that takes the remaining sum after subtracting the current number, the current combination, and the current start index.

2. If the remaining sum is 0, we have found a valid combination, so we add it to the result list.

3. If the remaining sum is less than 0, we have exceeded the target, so we return without adding the combination to the result list.

4. We iterate over the candidates starting from the current index. For each candidate, we add it to the current combination and recursively call `backtrack` with the updated remaining sum and the current index. We then remove the candidate from the combination to backtrack and try the next candidate.

5. Finally, we call `backtrack` with the initial target, an empty combination, and a start index of 0, and return the result list.

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