1608. Special Array With X Elements Greater Than or Equal X

1608. Special Array With X Elements Greater Than or Equal X

Difficulty: Easy

Topics: Array, Binary Search, Sorting

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 <= 100
• 0 <= nums[i] <= 1000

Solution:

We need to find a number x such that there are exactly x numbers in the array that are greater than or equal to x. The solution involves sorting the array and then efficiently checking possible values of x using the sorted properties of the array.

Approach

1. Sort the Array: Sorting helps in efficiently determining the number of elements greater than or equal to a given value x.
2. Check for Special Case (x = n): If the smallest element in the sorted array is greater than or equal to the length of the array (n), then x = n is the solution. This is because all elements in the array are at least n, so there are exactly n elements greater than or equal to n.
3. Iterate Over Possible x Values: For each possible x from 1 to n-1:
   • Determine the Critical Index: The critical index is n - x, which is the position in the sorted array where the x-th element from the end resides.
   • Check Conditions:
     • The element at the critical index must be greater than or equal to x.
     • The element immediately before the critical index must be less than x. This ensures that exactly x elements are greater than or equal to x.
4. Return Result: If such an x is found during the iteration, return it. If no valid x is found after checking all possibilities, return -1.

Let's implement this solution in PHP: 1608. Special Array With X Elements Greater Than or Equal X

<?php
/**
 * @param Integer[] $nums
 * @return Integer
 */
function specialArray($nums) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
print_r(specialArray([3, 5]));        // Output: 2
print_r(specialArray([0, 0]));        // Output: -1
print_r(specialArray([0, 4, 3, 0, 4])); // Output: 3
?>

Explanation:

1. Sorting the Array: The array is sorted in non-decreasing order to facilitate efficient checking of elements relative to x.
2. Special Case Handling (x = n): If the smallest element (first element after sorting) is greater than or equal to the array length n, then x = n is immediately returned because all elements will satisfy the condition of being at least n.
3. Iterating for Valid x: For each x from 1 to n-1:
   • The element at position n - x must be at least x. This ensures that there are at least x elements from this position onwards that are greater than or equal to x.
   • The element immediately before (n - x - 1) must be less than x. This ensures that there are not more than x elements that are greater than or equal to x.
4. Result Handling: If the conditions are met for any x during the iteration, that x is returned. If no such x is found after checking all possibilities, the function returns -1.

This approach efficiently checks possible values of x by leveraging the sorted properties of the array, ensuring optimal performance with a time complexity dominated by the sorting step, which is O(n log n). The subsequent iteration is O(n), making the overall complexity O(n log n).

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