3151. Special Array I - Detailed Explanation

Problem Statement:

You are given an array of non-negative integers nums. An array is considered special if there exists a number x such that there are exactly x elements in nums that are greater than or equal to x. Such an x is called the special number.

Your task is to find the special number if it exists; otherwise, return -1.

Example Inputs and Outputs:

Example 1:
- Input: nums = [3, 5]
- Output: 2
- Explanation: There are exactly 2 elements in the array (3 and 5) that are greater than or equal to 2.
Example 2:
- Input: nums = [0, 0]
- Output: -1
- Explanation: No number x exists such that exactly x elements are greater than or equal to x.
Example 3:
- Input: nums = [0, 4, 3, 0, 4]
- Output: 3
- Explanation: There are exactly 3 elements (the two 4’s and the 3) that are greater than or equal to 3.

Constraints:

The array consists of non-negative integers.
The length of nums is at least 1.
(Typically, the constraints are such that a sorting solution with O(n log n) time or an O(n) counting solution is acceptable.)

Hints for Solving the Problem:

Sorting Insight:
- What happens if you sort the array?
- Can you determine the count of numbers greater than or equal to a candidate value x quickly by knowing the sorted order?
Frequency or Counting:
- If the values in nums have a small range, could you use a frequency count to determine how many numbers are ≥ any candidate value without sorting?
Iterating Through Possibilities:
- Remember that the special number x must be in the range from 0 to n (where n is the number of elements in nums).
- Consider iterating over this range and verifying the condition for each candidate x.

Approach 1: Sorting-Based Method

Step-by-Step Explanation:

Sort the Array:
- Sort the array in non-decreasing order.
- After sorting, if you are checking a candidate value x, you can quickly count how many numbers are ≥ x by looking at the index where numbers become greater than or equal to x.
Iterate Through Candidates:
- For each candidate x (from 0 up to n), determine the number of elements that are ≥ x.
- In a sorted array, if the index of the first number ≥ x is i, then the count is n - i.
- If n - i == x, then x is the special number.
Return Result:
- If you find such an x, return it; if not, return -1.

Python Code (Sorting Approach):

Python3

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Java Code (Sorting Approach):

Java

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Approach 2: Frequency Count (Counting Sort) Method

Step-by-Step Explanation:

Build Frequency Array:
- Since the numbers are non-negative, create an array freq where freq[i] represents the count of the number i in nums.
- Note that any number greater than n can be grouped together (because we only care about counts up to n).
Iterate Over Possible x Values in Reverse:
- For candidate values x from n down to 0, compute the count of numbers ≥ x.
- This can be done by summing frequencies from index x to n (or grouping numbers > n into freq[n]).
Check Condition:
- If the count equals x, return x.
Return -1 if Not Found.

Python Code (Counting Approach):

Python3

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Java Code (Counting Approach):

Java

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Complexity Analysis:

Sorting-Based Approach:
- Time Complexity: O(n log n) due to sorting.
- Space Complexity: O(1) if sorting is done in-place (ignoring the input array).
Frequency Count Approach:
- Time Complexity: O(n) since we iterate through the array and then through the range 0 to n.
- Space Complexity: O(n) for the frequency array.

Edge Cases:

Single Element Array:
- If nums contains only one element, the candidate x is either 1 (if the element is at least 1) or -1 otherwise.
All Zeros:
- An array like [0, 0, 0] should return -1 because there is no candidate x for which exactly x elements are ≥ x.
Elements Larger Than Length:
- If many elements are larger than the length of the array, treat them as if they were equal to n (since they count toward the numbers ≥ any candidate x).

Common Mistakes:

Off-by-One Errors:
- Ensure that the candidate x is checked for all values from 0 to n.
Incorrect Counting:
- When using the counting approach, make sure to correctly group numbers larger than n into the freq[n] bucket.
Not Sorting (or Mis-sorting):
- In the sorting approach, careful handling of indices when counting elements ≥ x is essential.

Alternative Variations:

Special Array II:
- Variations might ask for different conditions (e.g., at most or at least a given count) or for multiple such numbers.
Related Problems:
- Problems involving threshold counts or frequency analysis can be seen as variations of this theme.

Majority Element (LeetCode 169):
- Find the element that appears more than ⌊n/2⌋ times.
Find the Difference of Two Arrays:
- Involves counting frequencies to determine differences.
Kth Largest Element in an Array (LeetCode 215):
- Though not directly related, it also uses sorting or counting methods.