[LeetCode] 1673. Find the Most Competitive Subsequence

轉自LeetCode

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

Solution

如果這題沒有限制 k，那麼就用一個 stack 來解題就可

但有了k這個限制後，就必須考量當要把數字從stack拿出來的時候

剩下的數字還夠不夠滿足k這個條件

所以必須動態檢查 nums.size - i + stack.size > k 這個條件

kotlin(參考解答)

	class Solution {
	fun mostCompetitive(nums: IntArray, k: Int): IntArray {
	val stack = mutableListOf<Int>()
	for(i in nums.indices) {
	while(stack.isNotEmpty() && nums[i] < stack.last() && nums.size - i + stack.size > k) {
	stack.removeAt(stack.lastIndex)
	}
	if(stack.size < k) {
	stack.add(nums[i])
	}
	}
	return stack.toIntArray()
	}
	}

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