<p>Given an integer array <code>nums</code> and a positive integer <code>k</code>, return <em>the most<strong> competitive</strong> subsequence of </em><code>nums</code> <em>of size </em><code>k</code>.</p>

<p>An array&#39;s subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.</p>

<p>We define that a subsequence <code>a</code> is more <strong>competitive</strong> than a subsequence <code>b</code> (of the same length) if in the first position where <code>a</code> and <code>b</code> differ, subsequence <code>a</code> has a number <strong>less</strong> than the corresponding number in <code>b</code>. For example, <code>[1,3,4]</code> is more competitive than <code>[1,3,5]</code> because the first position they differ is at the final number, and <code>4</code> is less than <code>5</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,5,2,6], k = 2
<strong>Output:</strong> [2,6]
<strong>Explanation:</strong> Among the set of every possible subsequence: {[3,5], [3,2], [3,6], [5,2], [5,6], [2,6]}, [2,6] is the most competitive.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,4,3,3,5,4,9,6], k = 4
<strong>Output:</strong> [2,3,3,4]
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

#define ll long long 
class Solution {
public:
    vector<int> mostCompetitive(vector<int>& nums, int k) {
        stack <int> s ;
        for ( ll i = 0 ; i < nums.size() ; i++ ){
            int no_of_next_ele = ( nums.size() - i) ; 
            while ( !s.empty() && (s.size() +  no_of_next_ele ) > k && s.top() > nums[i] ){
                s.pop();
            } 
            s.push(nums[i]) ; 
        }

        while ( s.size() > k ) {
            //cout << " hii " << endl; 
            s.pop() ; 
        }

        vector<int> ans ; 
        while (!s.empty()){
            ans.push_back(s.top() );
            s.pop() ; 
        }
        reverse ( ans.begin() , ans.end() ) ;
        return  ans ; 
    }
};
116300