Given an integer array nums and a positive integer k, return the most competitive subsequence of nums of size k.

An array's subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.

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

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Example 1:

Input: nums = [3,5,2,6], k = 2
Output: [2,6]
Explanation: 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.


Example 2:

Input: nums = [2,4,3,3,5,4,9,6], k = 4
Output: [2,3,3,4]


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Constraints:


	1 &lt;= nums.length &lt;= 105
	0 &lt;= nums[i] &lt;= 109
	1 &lt;= k &lt;= nums.length


class Solution:
	def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
		end = len(nums) - k
		ans = []
		for num in nums:
			while end and ans and num < ans[-1] :
				ans.pop()
				end -= 1
			ans.append(num)
		
		return ans[:k]
class Solution {
    public int[] mostCompetitive(int[] nums, int k) {
        int[] stack = new int[k];
        for(int i=0,j=0;i<nums.length;i++) {
          while(j>0 && stack[j-1]>nums[i] && j+nums.length-i>k) {
              j--;
          }
          if(j<k) {
              stack[j++]=nums[i];
          }
        }
        return stack;
    }
}
class Solution {
public:
    vector<int> mostCompetitive(vector<int>& nums, int k) {
        vector<int> stack;
        int nums_to_delete = nums.size()-k;
        for (int i = 0; i < nums.size(); i++) {
            while (!stack.empty() && nums[i] < stack.back() && nums_to_delete) {
                stack.pop_back();
                nums_to_delete--;
            }
            stack.push_back(nums[i]);
        }
        return vector<int>(stack.begin(), stack.begin()+k);
    }
};
var mostCompetitive = function(nums, k) {
  let nbToRemove = nums.length - k,stack=[];
  for (const num of nums) {
    while (num<stack[stack.length -1] && stack.length && nbToRemove) {
      stack.pop();
      nbToRemove--;
    }
    stack.push(num)
  }
  while ( nbToRemove ) {
    stack.pop();
    nbToRemove--;
  }
  return stack;
};
Find the Most Competitive Subsequence