1673. Find the Most Competitive Subsequence

Description

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

Solutions

Solution 1

Python Code
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class Solution:
    def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
        stk = []
        n = len(nums)
        for i, v in enumerate(nums):
            while stk and stk[-1] > v and len(stk) + n - i > k:
                stk.pop()
            if len(stk) < k:
                stk.append(v)
        return stk

Java Code
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class Solution {
    public int[] mostCompetitive(int[] nums, int k) {
        Deque<Integer> stk = new ArrayDeque<>();
        int n = nums.length;
        for (int i = 0; i < nums.length; ++i) {
            while (!stk.isEmpty() && stk.peek() > nums[i] && stk.size() + n - i > k) {
                stk.pop();
            }
            if (stk.size() < k) {
                stk.push(nums[i]);
            }
        }
        int[] ans = new int[stk.size()];
        for (int i = ans.length - 1; i >= 0; --i) {
            ans[i] = stk.pop();
        }
        return ans;
    }
}

C++ Code
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class Solution {
public:
    vector<int> mostCompetitive(vector<int>& nums, int k) {
        vector<int> stk;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            while (stk.size() && stk.back() > nums[i] && stk.size() + n - i > k) {
                stk.pop_back();
            }
            if (stk.size() < k) {
                stk.push_back(nums[i]);
            }
        }
        return stk;
    }
};

Go Code
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func mostCompetitive(nums []int, k int) []int {
	stk := []int{}
	n := len(nums)
	for i, v := range nums {
		for len(stk) > 0 && stk[len(stk)-1] > v && len(stk)+n-i > k {
			stk = stk[:len(stk)-1]
		}
		if len(stk) < k {
			stk = append(stk, v)
		}
	}
	return stk
}