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.

**Constraints:**

*   `1 <= nums.length <= 100`
*   `0 <= nums[i] <= 1000`

    ```java
public int specialArray(int[] nums) {
    int max_val = Arrays.stream(nums).max().orElse(0);
    for (int x = 1; x <= max_val; ++x) {
        int count = 0;
        for (int num : nums) {
            if (num >= x) count++;
        }
        if (count == x) return x;
    }
    return -1;
}
```
    
    1. Find the maximum value in the input array `nums`.
2. Iterate through the range `[1, max_val]` (both inclusive) and for each value of `x`, count the elements greater than or equal to `x` in the input array.
3. If the count of such elements is equal to `x`, return `x`.
4. If no such `x` is found, return `-1`.
    

    ```cpp
int specialArray(vector<int>& nums) {
    int max_val = *max_element(nums.begin(), nums.end());
    for (int x = 1; x <= max_val; ++x) {
        int count = 0;
        for (int num : nums) {
            if (num >= x) count++;
        }
        if (count == x) return x;
    }
    return -1;
}
```
    
    1. Find the maximum value in the input array `nums`.
2. Iterate through the range `[1, max_val]` (both inclusive) and for each value of `x`, count the elements greater than or equal to `x` in the input array.
3. If the count of such elements is equal to `x`, return `x`.
4. If no such `x` is found, return `-1`.
    

    ```python
def specialArray(nums):
    max_val = max(nums)
    for x in range(1, max_val + 1):
        count = sum([1 for num in nums if num >= x])
        if count == x:
            return x
    return -1
```
    
    1. Find the maximum value in the input array `nums`.
2. Iterate through the range `[1, max_val]` (both inclusive) and for each value of `x`, count the elements greater than or equal to `x` in the input array.
3. If the count of such elements is equal to `x`, return `x`.
4. If no such `x` is found, return `-1`.
    

    ```javascript
function specialArray(nums) {
    let max_val = Math.max(...nums);
    for (let x = 1; x <= max_val; ++x) {
        let count = 0;
        for (let num of nums) {
            if (num >= x) count++;
        }
        if (count === x) return x;
    }
    return -1;
}
```
    
    1. Find the maximum value in the input array `nums`.
2. Iterate through the range `[1, max_val]` (both inclusive) and for each value of `x`, count the elements greater than or equal to `x` in the input array.
3. If the count of such elements is equal to `x`, return `x`.
4. If no such `x` is found, return `-1`.
    
def specialArray(nums):
    max_val = max(nums)
    for x in range(1, max_val + 1):
        count = sum([1 for num in nums if num >= x])
        if count == x:
            return x
    return -1
