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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.
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.
1def count_greater_equal(nums, target):
2 return sum(1 for num in nums if num >= target)
3
4def special_array(nums):
5 nums.sort()
6 left, right = 0, len(nums)
7 while left <= right:
8 mid = left + (right - left) // 2
9 count = count_greater_equal(nums, mid)
10 if count == mid:
11 return mid
12 elif count < mid:
13 right = mid - 1
14 else:
15 left = mid + 1
16 return -1
17
18nums = [0, 4, 3, 0, 4]
19print(special_array(nums))This Python approach leverages binary search to locate the special integer x efficiently, checking mid against counts.