Formatted question description: https://leetcode.ca/all/1608.html

# 1608. Special Array With X Elements Greater Than or Equal X (Easy)

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

Related Topics:
Array

## Solution 1.

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

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int specialArray(vector<int>& A) {
sort(begin(A), end(A));
int N = A.size();
for (int i = 0; i <= N; ++i) {
int cnt = N - (lower_bound(begin(A), end(A), i) - begin(A));
if (cnt == i) return i;
}
return -1;
}
};


## Solution 2.

We store the count of numbers in a cnt array. And we just need to iterate from max(A) towards 0 to find the valid value, because for any k > max(A), there are 0 elements >= k and thus must be invalid.

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

// Time: O(N + max(A))
// Space: O(max(A))
class Solution {
public:
int specialArray(vector<int>& A) {
int cnt[1001] = {};
for (int n : A) cnt[n]++;
for (int i = 999; i >= 0; --i) {
cnt[i] += cnt[i + 1];
if (cnt[i] == i) return i;
}
return -1;
}
};


The sorting part takes O(NlogN). But the rest just takes O(logN).

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

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int specialArray(vector<int>& A) {
sort(begin(A), end(A));
int N = A.size(), L = 0, R = N;
while (L <= R) {
int M = (L + R) / 2;
bool cur = M == 0 || A[N - M] >= M, prev = M == N || A[N - M - 1] < M;
if (cur && prev) return M;
if (!cur) R = M - 1;
else L = M + 1;
}
return -1;
}
};


Java

class Solution {
public int specialArray(int[] nums) {
Arrays.sort(nums);
int length = nums.length;
if (length <= nums[0])
return length;
for (int x = 1; x < length; x++) {
if (nums[length - x] >= x && nums[length - x - 1] < x)
return x;
}
return -1;
}
}