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

class Solution {
    public int specialArray(int[] nums) {
        int n = nums.length;
        int[] arr = new int[1001];
        for(int i : nums) arr[i]++;
        
        for(int i = 0; i <= 100; i++) {
            if(n == i) return i;
            n -= arr[i];
        }
        return -1;
    }
}

牛逼了我的哥，先用constraint把各个数的频率count一下，然后因为一共不超过100个数，所以实际上nums【i】大于100的时候已经没用了，题目要求的是恰好有x个数大于等于x。

然后从0到100循环，如果剩余数n == i，说明n和i就是x，返回。否则由于这个数不是答案，所以n减去这个数的频率，比如【0，0，3，4，3】，到0的时候不符合，n-2=3，然后到了3直接3 == 3 返回。