# LeetCode 1608. Special Array With X Elements Greater Than or Equal X

## Description

https://leetcode.com/problems/special-array-with-x-elements-greater-than-or-equal-x/

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: 2
Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.
```

Example 2:

```Input: nums = [0,0]
Output: -1
Explanation: 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: 3
Explanation: 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 <= 100`
• `0 <= nums[i] <= 1000`

## Explanation

Start with x = 0 and increase x to a most len(nums), and check if the total number of elements in list is equal to x.

## Python Solution

``````class Solution:
def specialArray(self, nums: List[int]) -> int:

x = 0

while x <= len(nums):
count = 0
for num in nums:
if num >= x:
count += 1

if count == x:
return x

x += 1

return -1``````
• Time Complexity: O(N^N).
• Space Complexity: O(1).