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

这题，其实挺容易发现规律的，就是要找最小的，而且跟后面距离还有k的数，下一次找k - 1的，再下一次找k - 2的，如此类推。但用heap调了半天每弄对。然后看了答案，发现居然是单调栈。虽然用了deque来实现，但基本操作一样。

public int[] mostCompetitive(int[] nums, int k) {
    if (nums == null || nums.length == 0 || nums.length <= k) {
        return nums;
    }

    Deque<Integer> dq = new ArrayDeque<>();
    // 要从nums里去掉n - k个数，组成这个sequence
    int removeCnt = nums.length - k;
    for (int i = 0; i < nums.length; i++) {
        while (!dq.isEmpty() && dq.peekLast() > nums[i] && removeCnt > 0) {
            dq.pollLast();
            removeCnt--;
        }
        dq.addLast(nums[i]);
    }

    int[] result = new int[k];
    for (int i = 0; i < k; i++) {
        result[i] = dq.pollFirst();
    }

    return result;
}