Combination Sum

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

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 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: []

Constraints:

• 1 <= candidates.length <= 30
• 2 <= candidates[i] <= 40
• All elements of candidates are distinct.
• 1 <= target <= 40

Approach

• We are maintaining List where we keep adding new number to and check if we reach target if yes then add it to the result otherwise we remove the element anyway next regardless
• We pick an element then we iterate starting from that element or something

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> ans = new ArrayList<>();
        Arrays.sort(candidates);
        d(0,ans,new ArrayList<>(),0,candidates,target);
        return ans;
    }
 
    public void d(int i, List<List<Integer>> a, List<Integer> curr, int t, int[] c, int target) {
        if (t ==  target) {
            a.add(new ArrayList<>(curr));
            return;
        }
 
        for (int j = i; j < c.length; j++) {
            if (t + c[j] > target) {
                return;
            }
            curr.add(c[j]);
            d(j,a,curr,t + c[j],c,target);
            curr.remove(curr.size()-1);
        }
    }
}

• We used sorting so once we know that the total exceeded we can return and no need to check further
• If we want to not use sorting then use continue instead of return
• If we only add c then any later modification would change in the answer as well that is why we create a copy of it and store in the solution

class Solution {
    List<List<Integer>> a;
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        a = new ArrayList<>();
        Arrays.sort(candidates);
        dfs(0,0,new ArrayList<>(),candidates,target);
        return a;
    }
 
    public void dfs(int i, int total, List<Integer> c, int[] n, int target) {
        if (total == target) {
            a.add(new ArrayList<>(c));
            return;
        }
 
        for (int j = i; j < n.length; j++) {
            if (total + n[j] > target)
                return;
 
            c.add(n[j]); //Add element and then check
            dfs(j,total + n[j],c,n,target);
            c.remove(c.size() - 1); // remove element after checking    
        }
    }
}