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Sort the array first. For each possible count of elements x, check if there are exactly x numbers in the array that are greater than or equal to x.
Time Complexity: O(n log n) due to sorting, followed by O(n) for checking, resulting in O(n log n). Space Complexity: O(1) additional space.
1def special_array(nums):
2 nums.sort()
3 for x in range(len(nums) + 1):
4 count = sum(1 for num in nums if num >= x)
5 if count == x:
6 return x
7 return -1
8
9nums = [0, 4, 3, 0, 4]
10print(special_array(nums))This Python function sorts the number list, and for each possible x, verifies the count of elements greater than or equal to x, returning x if criteria match.
Utilize binary search on the result potential values from 0 to nums.length to efficiently find the special x.
Time Complexity: O(n log n) due to sorting initially, and O(n log n) for the binary search with counting operations. Space Complexity: O(1) additional space.
1
This solution uses binary search over the number of potential special cases, reducing the search space for efficiency.