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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.
1function specialArray(nums) {
2 nums.sort((a, b) => a - b);
3 for (let x = 0; x <= nums.length; x++) {
4 let count = nums.filter(num => num >= x).length;
5 if (count === x) {
6 return x;
7 }
8 }
9 return -1;
10}
11
12const nums = [0, 4, 3, 0, 4];
13console.log(specialArray(nums));This JavaScript function sorts the array and filters numbers to check the special condition for each x, returning the x meeting the criteria.
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.