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39-combination-sum.cpp
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68 lines (55 loc) · 2.02 KB
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/*
39. Combination Sum
Medium - 75.5%
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 can be used multiple times.
2 + 2 + 3 = 7
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: []
*/
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
vector<vector<int>> result;
vector<int> combination;
// Sort to enable early termination
sort(candidates.begin(), candidates.end());
backtrack(candidates, target, 0, combination, result);
return result;
}
private:
void backtrack(vector<int>& candidates, int target, int start,
vector<int>& combination, vector<vector<int>>& result) {
if (target == 0) {
result.push_back(combination);
return;
}
for (int i = start; i < candidates.size(); i++) {
// Early termination if current candidate exceeds target
if (candidates[i] > target) break;
combination.push_back(candidates[i]);
// Use same index i to allow reuse of same element
backtrack(candidates, target - candidates[i], i, combination, result);
combination.pop_back(); // backtrack
}
}
};