Special Array with X Elements Greater Than or Equal X

Problem Statement

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

Solutions

Intuition

To solve this problem, we need to determine if there exists a number x such that there are exactly x elements in the array that are greater than or equal to x. We can achieve this by counting the number of elements for each possible value and checking if the condition holds.

Approach

1. Counting Frequencies:

   • Create an array intArr to count the occurrences of each element in nums.

2. Cumulative Sum:

   • Calculate the cumulative sum from the highest value to the lowest to determine how many elements are greater than or equal to each possible value.

3. Check Condition:

   • Iterate through the possible values and check if there exists a value x where the number of elements greater than or equal to x is exactly x.

Java

class Solution {
    public int specialArray(int[] nums) {
        int val = 0;
        int len = nums.length;
        int[] intArr = new int[1001];
        
        for(int i = 0; i < len; i++) {
            intArr[nums[i]]++;
        }
        
        for(int i = 1000; i >= 0; i--) {
            intArr[i] += val;
            val = intArr[i];
        }
        
        val = -1;
        for(int i = 0; i < 1000; i++) {
            if(intArr[i] == i) {
                val = i;
            }
        }
        
        return val;
    }
    
}

Python

class Solution:
    def specialArray(self, nums: List[int]) -> int:
        n = len(nums)
        intArr = [0] * 1001
        
        for num in nums:
            if num <= 1000:
                intArr[num] += 1
        
        val = 0
        for i in range(1000, -1, -1):
            intArr[i] += val
            val = intArr[i]
        
        val = -1
        for i in range(1001):
            if intArr[i] == i:
                val = i
                break
        
        return val