Question

1673. Find the Most Competitive Subsequence

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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

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

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;
  }
}

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;
  }
};

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
}

function mostCompetitive(nums: number[], k: number): number[] {
  const stk: number[] = [];
  const n = nums.length;
  for (let i = 0; i < n; ++i) {
    while (stk.length && stk.at(-1) > nums[i] && stk.length + n - i > k) {
      stk.pop();
    }
    if (stk.length < k) {
      stk.push(nums[i]);
    }
  }
  return stk;
}