Given two numbers n and k and you have to find all possible combination of k numbers from 1…n.
Examples:
Input : n = 4
k = 2
Output : 1 2
1 3
1 4
2 3
2 4
3 4
Input : n = 5
k = 3
Output : 1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5
We have discussed one approach in below post.
Print all possible combinations of r elements in a given array of size n
In this, we use DFS based approach. We want all numbers from 1 to n. We first push all numbers from 1 to k in tmp_vector and as soon as k is equal to 0, we push all numbers from tmp_vector to ans_vector. After this we remove the last element from tmp_vector and make make all remaining combination.
// C++ program to print all combinations of size // k of elements in set 1..n #include <bits/stdc++.h> using namespace std; void makeCombiUtil(vector<vector<int> >& ans, vector<int>& tmp, int n, int left, int k) { // Pushing this vector to a vector of vector if (k == 0) { ans.push_back(tmp); return; } // i iterates from left to n. First time // left will be 1 for (int i = left; i <= n; ++i) { tmp.push_back(i); makeCombiUtil(ans, tmp, n, i + 1, k - 1); // Popping out last inserted element // from the vector tmp.pop_back(); } } // Prints all combinations of size k of numbers // from 1 to n. vector<vector<int> > makeCombi(int n, int k) { vector<vector<int> > ans; vector<int> tmp; makeCombiUtil(ans, tmp, n, 1, k); return ans; } // Driver code int main() { // given number int n = 5; int k = 3; vector<vector<int> > ans = makeCombi(n, k); for (int i = 0; i < ans.size(); i++) { for (int j = 0; j < ans[i].size(); j++) { cout << ans.at(i).at(j) << " "; } cout << endl; } return 0; } |
Output:
1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5
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