Problem 40 Combination Sum II
Table of Contents
Problem Statement
Link - Problem 40
Question
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`
Solution
/class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
vector<vector<int>> result;
vector<int> current;
backtrack(candidates, target, 0, current, result);
return result;
}
private:
void backtrack(vector<int>& candidates, int target, int start, vector<int>& current, vector<vector<int>>& result) {
if (target == 0) {
result.push_back(current);
return;
}
for (int i = start; i < candidates.size(); i++) {
if (i > start && candidates[i] == candidates[i - 1])
continue;
if (candidates[i] > target)
break;
current.push_back(candidates[i]);
backtrack(candidates, target - candidates[i], i + 1, current, result);
current.pop_back();
}
}
};
Complexity Analysis
| Algorithm | Time Complexity | Space Complexity |
| --------- | --------------- | ---------------- |
| Backtrack | O(2^N) | O(N) |
Explanation
1. Intuition
- To generate all combinations, we can use backtracking.
- But we need to avoid duplicates. Hence we sort the array and skip the duplicates.
- `Backtrack` function is called recursively to generate all possible combinations.
- In backtracking, we add the current element to the `current` vector and call the backtrack function recursively.
- If it reaches the target, we add the current vector to the result.
- If the current element is greater than the target, we break the loop.
- If the current element is a duplicate, we skip it.
- We pop the last element from the current vector after the recursive call.
2. Implementation
- Sort the array to avoid duplicates.
- Initialize the `result` vector to store the final result.
- Initialize the `current` vector to store the current combination.
- Call the `backtrack` function with the `candidates`, `target`, `start`, `current`, and `result`.
- In the `backtrack` function, if the target is 0, add the current vector to the result.
- Iterate over the candidates array from the `start` index.
- If the current element is a duplicate, skip it.
- If the current element is greater than the target, break the loop.
- Add the current element to the current vector and call the backtrack function recursively.
- Pop the last element from the current vector after the recursive call.