[leetcode] 1608. Special Array With X Elements Greater Than or Equal X

[leetcode] 1608. Special Array With X Elements Greater Than or Equal X

Description

You are given an array nums of non-negative integers. nums is considered special if there exists a number x such that there are exactly x numbers in nums that are greater than or equal to x.

Notice that x does not have to be an element in nums.

Return x if the array is special, otherwise, return -1. It can be proven that if nums is special, the value for x is unique.

Example 1:

Input: nums = [3,5]Output: 2Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.

Example 2:

Input: nums = [0,0]Output: -1Explanation: No numbers fit the criteria for x.If x = 0, there should be 0 numbers >= x, but there are 2.If x = 1, there should be 1 number >= x, but there are 0.If x = 2, there should be 2 numbers >= x, but there are 0.x cannot be greater since there are only 2 numbers in nums.

Example 3:

Input: nums = [0,4,3,0,4]Output: 3Explanation: There are 3 values that are greater than or equal to 3.

Example 4:

Input: nums = [3,6,7,7,0]Output: -1

Constraints:

1 <= nums.length <= 1000 <= nums[i] <= 1000

分析

题目的意思是：给你一个数组，找出其中X个元素大于或者等于X的值，题目给出的这个是唯一的，其中一种暴力破解的方法就是一个一个的判断是否满足要求，只要找出来一个就行了。还有一种更好的方法，首先对数组进行排序，然后从左到右进行判断，如果n-i<=nums[i]说明找到了一个符合条件的值，但是题目中给出x是唯一的，因此不会是重复值，用prev记录一下前面遍历的值，利用n-i>prev可以保证这一点。比如

[1,1,2]

这个反例，应该返回-1，不应该返回1哈。这个想法很难想到，我也没想到哈哈哈哈

代码

class Solution: def specialArray(self, nums: List[int]) -> int: n=len(nums) prev=-1 nums.sort() for i in range(n): if(n-i<=nums[i] and n-i>prev): return n-i prev=nums[i] return -1

参考文献

​​[LeetCode] Simple Python / c solution​​

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