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Special Array With X Elements Greater Than or Equal X

Number: 1730

Difficulty: Easy

Paid? No

Companies: Google, Bloomberg, Meta, Amazon

Problem Description

Given an array of non-negative integers, determine if the array is "special". An array is special if there exists a number x such that exactly x elements in the array are greater than or equal to x. Return x if it exists; otherwise, return -1.

Key Insights

  • Sorting the array simplifies counting elements that meet the condition.
  • The candidate x can range from 0 to the length of the array.
  • Iterating through each candidate x and counting elements >= x is a straightforward approach.
  • Although binary search might optimize counting in a sorted array, a simple iteration suffices given the small constraints.

Space and Time Complexity

Time Complexity: O(n log n) due to sorting, with an additional O(n) for each candidate check leading to O(n^2) in the worst-case scenario, though n is at most 100. Space Complexity: O(1) additional space if sorting in-place.

Solution

  1. Sort the array to leverage the order during counting.
  2. For each possible candidate x from 0 to n (where n is the length of the array):
    • Count the number of elements in the array that are greater than or equal to x.
    • If the count is exactly equal to x, return x as the answer.
  3. If no candidate satisfies the condition, return -1. The algorithm uses the sorted array to efficiently determine the count of elements and iterates over possible x values to find the unique solution.

Code Solutions

# Python solution with line-by-line comments
class Solution:
    def specialArray(self, nums: list[int]) -> int:
        # Sort the array to help with counting elements efficiently.
        nums.sort()
        n = len(nums)
        # Try every candidate x from 0 up to n.
        for x in range(n + 1):
            count = 0
            # Count how many numbers are greater than or equal to x.
            for num in nums:
                if num >= x:
                    count += 1
            # Check if the count matches the candidate x.
            if count == x:
                return x
        return -1

# Example usage:
# sol = Solution()
# print(sol.specialArray([3, 5]))
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