LeetCode 40. Combination Sum II

Description

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sum to target.

Each number in candidates may only be used once in the combination.

Note: The solution set must not contain duplicate combinations.

Example 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8
Output: 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5
Output: 
[
[1,2,2],
[5]
]

Constraints:

  • 1 <= candidates.length <= 100
  • 1 <= candidates[i] <= 50
  • 1 <= target <= 30

Explanation

Combinations are subsets of the candidate array. We can use depth-first search to find all combinations of the input array. First, sort candidates array into ascending order.
Second, conduct a depth-first search to visit each combination of the input array.

Java Solution

class Solution {
    public List<List<Integer>> combinationSum2(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(candidates, results, combination, 0, target);
        
        return results;
    }
    
    private void toFindCombinationsToTarget(int[] candidates, List<List<Integer>> results, List<Integer> combination, int startIndex, int target) {
        if (target == 0) {
            results.add(new ArrayList<>(combination));
            return;
        }
        
        for (int i = startIndex; i < candidates.length; i++) {
            if (i != startIndex && candidates[i] == candidates[i - 1]) {
                continue;
            }
            
            if (candidates[i] > target) {
                break;
            }         
            
            combination.add(candidates[i]);
            toFindCombinationsToTarget(candidates, results, combination, i + 1, target - candidates[i]);
            combination.remove(combination.size() - 1);
        }
        
    }
}

Python Solution

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        results = []
        
        candidates = sorted(candidates)
        self.helper(results, candidates, target, [], 0)
        
        return results
        
    def helper(self, results, candidates, target, combination, start):
        if target == 0:
            results.append(list(combination))
    
            return
        
        if target < 0:
            return
        
        for i in range(start, len(candidates)):
            if i != start and candidates[i] == candidates[i - 1]:
                continue
            
            combination.append(candidates[i])
            self.helper(results, candidates, target - candidates[i], combination, i + 1)
            combination.pop()
            
        
  • Time Complexity: O(2^N)
  • Space Complexity: O(N)

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