1673. Find the Most Competitive Subsequence

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

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

本质上是求最小的递增序列，不妨先求最长递增序列。t是可以从原始nums里删除的个数，但不一定用到（递减时）。

求出来后先把size确定了再添加到res里。