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

###### tags: `Leetcode`

https://leetcode.com/problems/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: 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

## Solution in C

### Approach 1 - Brute Force

Time Complexity: $O(n^2)$
Space Complexity: $O(1)$

int specialArray(int* nums, int numsSize){
    for (int i=1; i<=numsSize; i++){
        int count=0;
        for (int j=0; j<numsSize; j++){
            if (nums[j]>=i) count++;
        }
        if (count==i) return i;
    }
    return -1;
}

---

### Approach 2

Time Complexity: $O(n)$
Space Complexity: $O(n)$

int specialArray(int* nums, int numsSize){
    int p[numsSize];
    for (int i=0; i<numsSize; i++) p[i]=0;
    
    for (int i=0; i<numsSize; i++){
        if (nums[i]>numsSize)
            p[numsSize-1]++;
        else if (nums[i]!=0)
            p[nums[i]-1]++;
    }
    
    int sum = 0;
    for (int i=numsSize-1; i>=0; i--){
        sum+=p[i];
        if (sum==i+1) return i+1;
    }
    return -1;
}