LeetCode 39. Combination Sum

Description

https://leetcode.com/problems/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 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

Explanation

Combinations are similar to subsets of the candidate array. We can use backtracking approach to find all combinations of the input array.

First, sort candidates array into ascending order.
Second, conduct a backtracking to visit each combination of the input array.

A backtracking function would be:
Whenever adding an element to the combination, we reduce the target.
If the remaining target value is equal to 0, we add the current combination to the result list.
Then, recursively find all combinations for the remaining target value from the remaining elements.

Java Solution

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> results = new ArrayList<>();
        
        if (candidates == null || candidates.length == 0) {
            return results;
        }
        
        Arrays.sort(candidates);
        
        List<Integer> combination = new ArrayList<>();
        toFindCombinationsToTarget(results, combination, candidates, target, 0);
        
        return results;
    }
    
    private void toFindCombinationsToTarget(List<List<Integer>> results, List<Integer> combination, int[] candidates, int target, int startIndex) {
        if (target == 0) {
            results.add(new ArrayList<>(combination));
            return;
        }
        
        for (int i = startIndex; i < candidates.length; i++) {
            if (candidates[i] > target) {
                break;
            }         
            
            combination.add(candidates[i]);
            toFindCombinationsToTarget(results, combination, candidates, target - candidates[i], i);
            combination.remove(combination.size() - 1);
        }        
    }
}

Python Solution

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        results = []
        
        if not candidates:
            return results
        
        candidates = sorted(candidates)
        
        combination = []        
        self.helper(results, candidates, combination, target, 0)
        
        return results
    
    def helper(self, results, candidates, combination, target, start_index):
        if target == 0:            
            results.append(list(combination))            
            return
        
        for i in range(start_index, len(candidates)):
            if candidates[i] > target:
                break
            
            combination.append(candidates[i])
            
            self.helper(results, candidates, combination, target - candidates[i], i)
            
            combination.pop()

• Time Complexity: ~(N^(T/M + 1))
• Space Complexity: ~T/M

N is the number of candidates, T is the target value, and M is the minimal value among the candidates