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Combination Sum IIII in C++
Consider we have to generate all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used. Each combination should be a unique set of numbers. All numbers should be positive, and the solution must not contain duplicate combinations. So if k = 3 and n = 9, then the possible combinations are [[1,2,6],[1,3,5],[2,3,4]]
To solve this, we will follow these steps −
- Suppose we will solve this using forming a method called solve. This will be recursive method, this will take k, n, temp array, and start. the start is initially 1
- if n = 0
- if size of temp = k, then insert temp into res and return
- for i := start to minimum of 9 and n
- insert i into temp
- solve(k, n – i, temp, i + 1)
- delete the last element from temp
- The main method will be like
- create one blank vector called temp
- solve(k, n, temp)
- return res
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h> using namespace std; void print_vector(vector<vector<auto> > v){ cout << "["; for(int i = 0; i<v.size(); i++){ cout << "["; for(int j = 0; j <v[i].size(); j++){ cout << v[i][j] << ", "; } cout << "],"; } cout << "]"<<endl; } class Solution { public: vector < vector <int> > res; void solve(int k, int n, vector <int> temp, int start = 1){ if(n == 0){ if(temp.size() == k){ res.push_back(temp); } return; } for(int i = start ; i <= min(9, n); i++){ temp.push_back(i); solve(k, n - i, temp, i + 1); temp.pop_back(); } } vector<vector<int>> combinationSum3(int k, int n) { res.clear(); vector <int> temp; solve(k, n, temp); return res; } }; main(){ Solution ob; print_vector(ob.combinationSum3(2, 9)); }
Input
3 9
Output
[[1, 8, ],[2, 7, ],[3, 6, ],[4, 5, ],]
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