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This approach uses backtracking to explore all possible combinations. Start with an empty combination, then add numbers incrementally and explore subsequent possibilities. If the sum of combination exceeds n or if it reaches the required length k, backtracking occurs to explore other potential combinations.
Time Complexity: O(2^n) because each number has a choice of being included or not.
Space Complexity: O(k) for the recursion stack.
1#include <vector>
2using namespace std;
3
4void backtrack(int k, int n, int start, vector<int>& path, vector<vector<int>>& results) {
5 if (k == 0 && n == 0) {
6 results.push_back(path);
7 return;
8 }
9 for (int i = start; i <= 9; ++i) {
10 if (i > n) break;
11 path.push_back(i);
12 backtrack(k - 1, n - i, i + 1, path, results);
13 path.pop_back();
14 }
15}
16
17vector<vector<int>> combinationSum3(int k, int n) {
18 vector<vector<int>> results;
19 vector<int> path;
20 backtrack(k, n, 1, path, results);
21 return results;
22}
23
24// Example usage:
25// vector<vector<int>> result = combinationSum3(3, 7);
26// for (const auto& seq : result) {
27// for (int num : seq) {
28// cout << num << ' ';
29// }
30// cout << endl;
31// }The C++ code implements backtracking in a similar manner to C. The function modifies the current combination path during exploration. It explores increasingly larger numbers to maintain unique number constraints and uses a vector to store resultant combinations.
This approach leverages bitmasking to generate all subsets of numbers {1,2,...,9} and filters conceivable combinations by size and sum criteria. It's potentially less intuitive, yet eliminates recursion.
Time Complexity: O(2^n), iterating through bitmask combinations for validity checks.
Space Complexity: O(k) due to maintained size of data array.
1
This Java code features non-recursive bitmask exploration, iterating over all subsets as integers 1 through 9 determine sequence membership. Using bitwise operations, subsets producing length k and sum n are collected.