easy Special Array With X Elements Greater Than or Equal X

Problem Statement

You are given an array nums containing positive integers. A nums array is special if there exists a number x such that there are exactly x elements of the array greater than or equal to x.

Return the value of x if the array is special. Otherwise, return -1.

Note: The array may or may not contain the x.

Examples

Example 1:

• Input: [3, 5, 3, 4]
• Expected Output: -1
• Justification: There's no value X from 1 to 4 such that there are exactly X elements greater than or equal to X.

Example 2:

• Input: [0, 5, 2, 4, 3]
• Expected Output: 3
• Justification: There are 3 elements (5, 4, 3), which are greater than or equal to 3 in the nums array.

Example 3:

• Input: [5, 6, 7, 8, 9]
• Expected Output: 5
• Justification: Each number in the array is greater than or equal to 5, and since there are 5 elements in total, 5 is the special value.

Solution

To solve this problem, we employ a strategy that involves sorting the array and then methodically checking each element to identify a special value. This special value, X, is defined as the number of elements in the array that are greater than or equal to X itself. By sorting the array in non-decreasing order, we position ourselves to efficiently determine the number of elements meeting the criteria for any candidate X. The sorted arrangement allows for a straightforward calculation of elements greater than or equal to each index, facilitating a direct comparison against the potential special value.

In the second phase of our approach, we iterate through the sorted array, assessing each element's eligibility to be the special value. For each element, we calculate the count of elements greater than or equal to it and verify if this count satisfies the unique condition to be considered special. This iterative process continues until we either find a value that meets the criteria or exhaust all possibilities, concluding with a -1 to indicate the absence of such a special value. This method is effective because it combines the logical simplicity of sorted data with a precise criterion check, ensuring a thorough and efficient search for the special value.

Step-by-Step Algorithm

1. Sort the Array: First, sort the given array in ascending order. This makes it easier to figure out how many elements are greater than or equal to each number.

2. Loop Through the Array: Go through each element of the sorted array one by one.

3. Calculate Greater or Equal Elements: For every element at position i, calculate how many elements are greater than or equal to it. You do this by subtracting i from the total number of elements in the array (let's call this total n). So, the calculation is n - i.

4. Check Conditions: For the element at position i, check two things:

   • If the number of elements greater than or equal to it is less than or equal to the value of the element itself.
   • And if i is not the first position, make sure this number is greater than the value of the element just before i.

5. Find Special Value: If you find an element that meets these conditions, that's your special value. Return it.

6. Return -1 If Not Found: If you finish checking all elements and don't find one that meets the conditions, return -1. This means there's no special value in the array.

Step-by-Step Algorithm Walkthrough

Let's consider the input: nums = [0,5,2,4,3].

1. Sort the Array: First, sort the array to get [0, 2, 3, 4, 5].

2. Iterate Through the Sorted Array:

   • Start with the first element (0 at index 0):
     • elementsGreaterOrEqual = 5 - 0 = 5
     • 5 is not less than or equal to 0. Continue to the next element.
   • Next element (2 at index 1):
     • elementsGreaterOrEqual = 5 - 1 = 4
     • 4 is not less than or equal to 2. Continue to the next element.
   • Next element (3 at index 2):
     • elementsGreaterOrEqual = 5 - 2 = 3
     • 3 is equal to 3. However, we need to check if it's also greater than the previous element (which was 2). Here 3 > 2, which statisfies the condition. So, return the value of elementsGreaterOrEqual, which is 3.

Code

import java.util.Arrays;

public class Solution {

  public int specialArray(int[] nums) {
    int size = nums.length; // Total number of elements in nums
    Arrays.sort(nums); // Sort nums in non-decreasing order

    for (int index = 0; index < size; index++) {
      int elementsGreaterOrEqual = size - index; // Number of elements greater than or equal to nums[index]
      if (
        elementsGreaterOrEqual <= nums[index] &&
        (index == 0 || elementsGreaterOrEqual > nums[index - 1])
      ) {
        return elementsGreaterOrEqual; // Return the special value
      }
    }

    return -1; // If no special value is found
  }

  public static void main(String[] args) {
    Solution solution = new Solution();

    // Example 1
    System.out.println(
      "Example 1: " + solution.specialArray(new int[] { 3, 5, 3, 4 })
    );
    // Example 2
    System.out.println(
      "Example 2: " + solution.specialArray(new int[] { 0, 5, 2, 4, 3 })
    );
    // Example 3
    System.out.println(
      "Example 3: " + solution.specialArray(new int[] { 5, 6, 7, 8, 9 })
    );
  }
}

Complexity Analysis

Time Complexity

• Sorting: The initial sorting of the array takes time, where N is the number of elements in the array.
• Linear Scan: After sorting, the algorithm performs a linear scan of the sorted array to find the special value. This step takes time.

Overall, the time complexity of the algorithm is dominated by the sorting step, resulting in a total time complexity of .

Space Complexity

The overall space complexity of the algorithm is , which is used to sort the array.

🎯 STRICT STANDOUT — Solution Special Array With X Elements ≥ X

1. Why / judgment

You need an integer x such that exactly x values are ≥ x. After sorting ascending, for index i the count of elements ≥ nums[i] is n−i; candidates for x also include every integer in [0..n] (x need not appear in the array). Binary search works because the function f(x)=|{a: a≥x}| is nonincreasing in x — classic binary search on answer / sorted-count hybrid. Naive O(n²) recount per x is correct but slow; uniqueness of special x (when it exists) is a problem property worth citing.

2. Big-O derivation (K11)

Sort: O(n log n). Then either:
  (A) scan i=0..n with candidate x = n−i, check nums[i]≥x and (i==0 or nums[i−1]<x) → O(n);
  (B) BS x in [0,n], each mid counts ≥mid in O(n) or O(log n) after sort → O(n log n).
Dominant: O(n log n). Space O(1)/O(n) sort.
Trace [0,5,2,4,3] sorted [0,2,3,4,5]:
  x=3: elements ≥3 → {3,4,5} count=3 ✓
Trace [3,5,3,4] sorted [3,3,4,5]: no x with count==x → -1 ✓
Trace [5,6,7,8,9]: x=5, all five ≥5 ✓

3. Pattern + when-NOT (K12)

Name: SORT + COUNT / BINARY SEARCH ON ANSWER (SPECIAL-X)

Recognition: “exactly x elements ≥ x”; x may or may not be in the array; return unique x or -1.

When-NOT: Need count of elements in arbitrary ranges many times → Fenwick/segment tree, not one-shot sort. Unsorted online stream with updates → maintain order statistic tree. If x must be an array element only (different problem) → restrict candidates to values present.

4. Edge hand-run (K13)

empty []: x=0 has 0 elements ≥0 → special 0 (if allowed by constraints).
all zeros [0,0]: x=0 → all ≥0 count 2 ≠0; x=1 count 0; x=2 count 0 → -1 or check carefully.
single [1]: x=1 → one element ≥1 ✓
duplicates don’t break count of ≥x.

5. Interviewer follow-ups (model answers)

Q1. Why is f(x)=count(a≥x) monotone?
A: Raising x can only drop or keep the count; never increase. Monotonicity licenses binary search on x.

Q2. Can there be two special x?
A: At most one: if two distinct x1<x2 both special, then f(x1)=x1 and f(x2)=x2 but f nonincreasing implies contradiction for strict inequalities in typical proofs — problem guarantees uniqueness or asks any/-1 framing.

Q3. Sort vs frequency array?
A: If values in tiny range, counting sort / bucket can drop sort to O(n+U); general case sort is fine.

6. Short drills

Drill: [0,4,3,0,4] → special? count for x=3: {4,3,4}=3 → 3.
Drill: prove O(n log n) not O(n²).
Drill: BS x∈[0,n] with sorted lower_bound for count ≥x.