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.
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 1000The goal of this problem is to determine whether there exists a value x such that exactly x elements in the array are greater than or equal to x. A common strategy is to first sort the array and then analyze how many elements satisfy the condition for each possible value of x. After sorting, you can quickly determine how many elements remain to the right of a given index.
Another efficient strategy is to apply binary search on the possible range of x values (typically from 0 to n). For each candidate x, count how many numbers in the array are greater than or equal to it and verify whether the count matches the requirement. Sorting combined with binary search helps reduce repeated scanning.
The sorting-based approach typically runs in O(n log n) time, while the validation step is linear or logarithmic depending on the implementation. Space complexity is usually O(1) if sorting is done in-place.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Sorting + Linear Scan | O(n log n) | O(1) |
| Sorting + Binary Search on X | O(n log n) | O(1) |
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Use these hints if you're stuck. Try solving on your own first.
Count the number of elements greater than or equal to x for each x in the range [0, nums.length].
If for any x, the condition satisfies, return that x. Otherwise, there is no answer.
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.
1#include <stdio.h>
2#include <stdlib.h>
3
4int compare(const void *a, const void *b) {
This C program sorts the input array and then iterates to determine if the number of elements greater than or equal to x is exactly 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
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Problems like this are common in coding interviews because they test sorting, counting logic, and binary search concepts. While the exact question may vary, similar patterns frequently appear in interviews at top tech companies.
The problem mainly uses arrays along with sorting techniques. After sorting, binary search or index-based counting can help determine how many elements meet the condition. No advanced data structures are typically required.
A common optimal approach is to sort the array and then evaluate possible values of x based on element positions. Sorting allows you to quickly determine how many elements are greater than or equal to a given value. This reduces the problem to checking valid candidate values efficiently.
Yes, binary search can be applied on the possible range of x values. For each candidate x, you check how many elements in the array are greater than or equal to it. This approach works well when combined with a sorted array.
This solution uses binary search over the number of potential special cases, reducing the search space for efficiency.