<p>Given two integers <code>n</code> and <code>k</code>, return <em>all possible combinations of</em> <code>k</code> <em>numbers chosen from the range</em> <code>[1, n]</code>.</p>

<p>You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, k = 2
<strong>Output:</strong> [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
<strong>Explanation:</strong> There are 4 choose 2 = 6 total combinations.
Note that combinations are unordered, i.e., [1,2] and [2,1] are considered to be the same combination.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 1, k = 1
<strong>Output:</strong> [[1]]
<strong>Explanation:</strong> There is 1 choose 1 = 1 total combination.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 20</code></li>
	<li><code>1 &lt;= k &lt;= n</code></li>
</ul>

#include <bits/stdc++.h>
using namespace std;

#define pb push_back



class Solution {
public:
    vector<vector<int>> result;
    vector<vector<int>> combine(int n, int k) {
        
        vector<int> nums(n);
        iota(nums.begin(), nums.end(), 1);

        int itr=0;
        vector<int> ans;
        solve(nums, ans, k, itr);
        return result;

    }

    void solve(vector<int> &nums, vector<int> ans , int k, int itr){


        if(ans.size()==k){
            result.pb(ans);
            return;
        }

        // if(1){
        //     result.pb(ans);
        //     // return;
        // }

        int size = nums.size();
        vector<int> temp;
        for(int i=itr; i<size; i++){
            temp = ans;
            temp.pb(nums[i]);
            solve(nums, temp, k, i+1);
            temp.clear();
        }
    }
};
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