<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>

class Solution {
public:
    vector<int> mostCompetitive(vector<int>& arr, int k) {
        int n= arr.size();
        
        stack<int> s;  //stores the elements which have a dilemna. In the end, the elements who remain form the answer.
        
        for(int i=0; i<n; i++){
            
            //n-i-1 elements are left in the array after ith index and k-s.size() is the no. of elements still required for our answer
            while(!s.empty() && arr[i]<s.top() && (n-i-1 >= k-s.size())){
                s.pop();
            }
            
            //push elements if stack has less than k elements (these elements are the ones whose dilemna has not been resolved yet) 
            if(s.size()<k){
                s.push(arr[i]);
            }
        }
        
        //stack s contains our answer now, converting to a vector
        vector<int> ans;
        while(!s.empty()){
            ans.push_back(s.top());
            s.pop();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
};
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