Sponsored
Sponsored
This approach uses a stack to build the competitive subsequence. The basic idea is to iterate through the array and maintain a stack where the final size is k
. For each element, we decide whether it should be added to the stack or replace the top element of the stack, which ensures the stack remains a competitive sequence.
We will replace the top element of the stack if the current element is smaller and there are enough elements left to fill the required subsequence size.
This method ensures that the maintained sequence is minimal and competitive, based on the problem constraints.
Time Complexity: O(n)
where n
is the number of elements in nums
.
Space Complexity: O(k)
to store the competitive subsequence.
1import java.util.ArrayDeque;
2
3public class Solution {
4 public int[] mostCompetitive(int[] nums, int k
The Java solution involves using ArrayDeque
as a stack to collect the competitive subsequence. Like the other solutions, it iteratively compares elements to maintain the smallest possible sequence. Excess elements are removed from the stack once all decisions are made.
This approach builds upon the idea that maintaining a competitive subsequence can be achieved by iteratively reducing and replacing elements in a greedy fashion, focusing directly on modifications to an output array.
This strategy is slightly different from a pure stack-based approach as it directly handles result index placement while keeping track of candidates for subsequences and managing the required number of subsequence slots.
Time Complexity: O(n)
.
Space Complexity: O(k)
.
Java implementation ensures through direct index management and placement that we achieve the desired subsequence efficiently. The main addition here is direct selection and placement, indicating inline replacements based on conditions.