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。