Medium

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.

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

To solve the "Combination Sum" problem in Swift with a Solution class, we can follow these steps:

1. Define a Solution class.
2. Define a method named combinationSum that takes an array of integers candidates and an integer target as input and returns a list of lists containing all unique combinations of candidates where the chosen numbers sum to target.
3. Implement backtracking to explore all possible combinations of candidates.
4. Sort the candidates array to ensure that duplicates are grouped together.
5. Create a recursive helper method named backtrack that takes parameters:
   • A list to store the current combination.
   • An integer representing the starting index in the candidates array.
   • The current sum of the combination.
6. In the backtrack method:
   • If the current sum equals the target, add the current combination to the result list.
   • Iterate over the candidates starting from the current index.
   • Add the current candidate to the combination.
   • Recursively call the backtrack method with the updated combination, index, and sum.
   • Remove the last added candidate from the combination to backtrack.
7. Call the backtrack method with an empty combination list, starting index 0, and sum 0.
8. Return the result list containing all unique combinations.

Here's the implementation:

public class Solution {
    public func combinationSum(_ coins: [Int], _ amount: Int) -> [[Int]] {
        var ans = \[\[Int]]()
        var subList = [Int]()
        combinationSumRec(coins.count, coins, amount, &subList, &ans)
        return ans
    }

    private func combinationSumRec(_ n: Int, _ coins: [Int], _ amount: Int, _ subList: inout [Int], _ ans: inout [[Int]]) {
        if amount == 0 || n == 0 {
            if amount == 0 {
                ans.append(subList)
            }
            return
        }
        if amount - coins[n - 1] >= 0 {
            subList.append(coins[n - 1])
            combinationSumRec(n, coins, amount - coins[n - 1], &subList, &ans)
            subList.removeLast()
        }
        combinationSumRec(n - 1, coins, amount, &subList, &ans)
    }
}

This implementation provides a solution to the "Combination Sum" problem in Swift. It explores all possible combinations of candidates using backtracking and returns the unique combinations whose sum equals the target.