3151. Special Array I - Detailed Explanation

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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:

  1. 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.

  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.

  3. 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:

  1. 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?
  2. 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?
  3. 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:

  1. 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.

  2. 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.

  3. Return Result:

    • If you find such an x, return it; if not, return -1.

Python Code (Sorting Approach):

Python3
Python3

. . . .

Java Code (Sorting Approach):

Java
Java

. . . .

Approach 2: Frequency Count (Counting Sort) Method

Step-by-Step Explanation:

  1. 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).

  2. 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]).
  3. Check Condition:

    • If the count equals x, return x.
  4. Return -1 if Not Found.

Python Code (Counting Approach):

Python3
Python3

. . . .

Java Code (Counting Approach):

Java
Java

. . . .

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:

  1. 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.
  2. All Zeros:

    • An array like [0, 0, 0] should return -1 because there is no candidate x for which exactly x elements are ≥ x.
  3. 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:

  1. Off-by-One Errors:

    • Ensure that the candidate x is checked for all values from 0 to n.
  2. Incorrect Counting:

    • When using the counting approach, make sure to correctly group numbers larger than n into the freq[n] bucket.
  3. Not Sorting (or Mis-sorting):

    • In the sorting approach, careful handling of indices when counting elements ≥ x is essential.

Alternative Variations:

  1. Special Array II:

    • Variations might ask for different conditions (e.g., at most or at least a given count) or for multiple such numbers.
  2. Related Problems:

    • Problems involving threshold counts or frequency analysis can be seen as variations of this theme.

Related Problems for Further Practice:

  1. Majority Element (LeetCode 169):

    • Find the element that appears more than ⌊n/2⌋ times.
  2. Find the Difference of Two Arrays:

    • Involves counting frequencies to determine differences.
  3. Kth Largest Element in an Array (LeetCode 215):

    • Though not directly related, it also uses sorting or counting methods.
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