Combination Sum — Leetcode

Topic: Array
Difficulty: Medium
Question Number: 39

Problem Statement:

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 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]]

Code:

in Java

class Solution {
    List<List<Integer>> combinations;
    int[] candidatesArr;
    int targetSum;

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        combinations = new ArrayList();
        targetSum = target;
        candidatesArr = candidates;

        findCombinations(0, new ArrayList<Integer>(), target);

        return combinations;
    }

    public void findCombinations(int candidate, ArrayList<Integer> current, int total) {
        if (total == 0) {
            combinations.add(new ArrayList<>(current));
            return;
        }

        if (total < 0) {
            return;
        }

        for (int i = candidate; i < candidatesArr.length; i++) {
            current.add(candidatesArr[i]);
            findCombinations(i, current, total - candidatesArr[i]);
            current.remove(current.size() - 1);
        }
    }
}

in Python by Vaishnav Jha

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        ans = []

        def dfs(i, cur, s):
            if s == target:     # Base Case 1
                ans.append(cur[:]) # Append copy of list
                return
            if i>= len(candidates) or s > target:   # Base Case 2
                return

            cur.append(candidates[i])
            dfs(i,cur,s+candidates[i]) # Recirsive call with the ith candidate
            cur.pop()       # Backtrack
            dfs(i+1, cur,s) # Recirsive call without the ith candidate

        dfs(0,[],0)
        return ans

# Contributed By - Vaishnav Jha

Thank You
Oshi Raghav