Class Solution
Medium
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 <= 301 <= candidates[i] <= 200- All elements of
candidatesare distinct. 1 <= target <= 500
To solve the “Combination Sum” problem in Java with a Solution class, we can follow these steps:
- Define a
Solutionclass. - Define a method named
combinationSumthat takes an array of integerscandidatesand an integertargetas input and returns a list of lists containing all unique combinations ofcandidateswhere the chosen numbers sum totarget. - Implement backtracking to explore all possible combinations of candidates.
- Sort the
candidatesarray to ensure that duplicates are grouped together. - Create a recursive helper method named
backtrackthat takes parameters:- A list to store the current combination.
- An integer representing the starting index in the
candidatesarray. - The current sum of the combination.
- In the
backtrackmethod:- If the current sum equals the target, add the current combination to the result list.
- Iterate over the candidates starting from the current index.
- Add the current candidate to the combination.
- Recursively call the
backtrackmethod with the updated combination, index, and sum. - Remove the last added candidate from the combination to backtrack.
- Call the
backtrackmethod with an empty combination list, starting index 0, and sum 0. - Return the result list containing all unique combinations.
Here’s the implementation:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(candidates); // Sort the candidates to ensure duplicates are grouped together
backtrack(result, new ArrayList<>(), candidates, target, 0);
return result;
}
private void backtrack(List<List<Integer>> result, List<Integer> combination, int[] candidates, int target, int start) {
if (target == 0) {
result.add(new ArrayList<>(combination));
return;
}
for (int i = start; i < candidates.length && candidates[i] <= target; i++) {
combination.add(candidates[i]);
backtrack(result, combination, candidates, target - candidates[i], i); // Use the same candidate again
combination.remove(combination.size() - 1); // Backtrack by removing the last candidate
}
}
}
This implementation provides a solution to the “Combination Sum” problem in Java. It explores all possible combinations of candidates using backtracking and returns the unique combinations whose sum equals the target.
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Constructor Summary
Constructors -
Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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combinationSum
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