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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.
1using System;
2using System.Linq;
3
4class SpecialArray {
5 public int SpecialArrayFunc(int[] nums) {
6 Array.Sort(nums);
7 int n = nums.Length;
8 for (int x = 0; x <= n; x++) {
9 int count = nums.Count(num => num >= x);
10 if (count == x) {
11 return x;
12 }
13 }
14 return -1;
15 }
16
17 static void Main() {
18 SpecialArray sa = new SpecialArray();
19 int[] nums = {0, 4, 3, 0, 4};
20 Console.WriteLine(sa.SpecialArrayFunc(nums));
21 }
22}
This C# version uses Linq to count elements in a sorted array according to the specified criteria for each x.
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 JavaScript method employs binary search to efficiently decide the special x that meets required conditions.