# Find the Most Competitive Subsequence

Updated: Jan 21, 2021

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.

**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]

**Constraints:**

1 <= nums.length <= 105

0 <= nums[i] <= 109

1 <= k <= nums.length

**Solution:**

```
class Solution {
public int[] mostCompetitive(int[] nums, int k) {
Stack<Integer> stack = new Stack<Integer>();
int n=nums.length;
for(int i=0;i<n;i++)
{
while(!stack.isEmpty() && nums[i]<stack.peek() && n-i+stack.size()>k)
{
stack.pop();
}
if(stack.size()<k)
stack.push(nums[i]);
}
int[] ans = new int[k];
int index = k-1;
while(!stack.isEmpty())
{
ans[index--] = stack.pop();
}
return ans;
}
}
```