39. Combination Sum

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations incandidates where the candidate numbers sums totarget.

The same repeated number may be chosen fromcandidates unlimited number of times.

Note:

• All numbers (includingtarget) will be positive integers.
• The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [2,3,6,7], target = 7,

A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], 
target = 8,

A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

Thoughts:

1. Backtracking: can select distinct number repeated times -> so each time for loop start at the passed-in index

Code: T:O(N^2) S: O(N^2)

class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        def dfs(res, path, start, target):
            if target <= 0 : return 
            for i in range(start, len(candidates)):
                path.append(candidates[i])
                if candidates[i] == target:
                    res.append(list(path))
                    # pop the tail for next candidate
                    path.pop()
                    continue

                dfs(res, path, i, target - candidates[i])
                # pop the tail for next candidate
                path.pop()

            return

        res = []
        dfs(res, [], 0, target)
        return res

Code: java with sorting

class Solution {
    public List<List<Integer>> combinationSum(int[] nums, int target) {
    List<List<Integer>> list = new ArrayList<>();
    Arrays.sort(nums);
    backtrack(list, new ArrayList<>(), nums, target, 0);
    return list;
}

private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
    if(remain < 0) return;
    else if(remain == 0) list.add(new ArrayList<>(tempList));
    else{ 
        for(int i = start; i < nums.length; i++){
            // follow-up: Each number in candidates may only be used once in the combination + there are duplicates.
            // follow-up: if(i > start && nums[i] == nums[i - 1]) continue;
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, remain - nums[i], i); // not i + 1 because we can reuse same elements
            tempList.remove(tempList.size() - 1);
            }
        }
    }
}