1673. Find the Most Competitive Subsequence (M)

Find the Most Competitive Subsequence (M)

题目

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 <= 10^5
• 0 <= nums[i] <= 10^9
• 1 <= k <= nums.length

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题意

从数组中删去若干个数，使剩余k个数组成的序列字典序最小。

思路

单调栈。维护一个最大容量为k的栈，遍历所有数字，如果比栈顶元素大则直接压入栈中(栈未满)，如果比栈顶元素小则不断出栈直到栈顶元素小于当前数或栈空。需要注意最后一定要使栈满，所以要记录剩余可丢弃元素的次数。

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代码实现

Java

class Solution {
    public int[] mostCompetitive(int[] nums, int k) {
        Deque<Integer> deque = new ArrayDeque<>();
        int toDrop = nums.length - k;
        int i = 0;

        while (toDrop > 0 && i < nums.length) {
            if (deque.isEmpty() || deque.size() < k && deque.peek() <= nums[i]) {
                deque.push(nums[i]);
            } else if (deque.peek() <= nums[i]) {
                toDrop--;
            } else if (deque.peek() > nums[i]) {
                while (toDrop > 0 && !deque.isEmpty() && deque.peek() > nums[i]) {
                    toDrop--;
                    deque.pop();
                }
                deque.push(nums[i]);
            }
            i++;
        }

        while (deque.size() < k) {
            deque.push(nums[i++]);
        }

        int[] ans = new int[k];
        i = 0;
        while (i < k) {
            ans[i++] = deque.removeLast();
        }

        return ans;
    }
}