Special Array With X Elements Greater Than or Equal X

Problem Description

Welcome to the world of arrays, where numbers play hide and seek! The problem at hand is like trying to find out how many friends you need to invite to a party so that at least X of them can say, “Hey, I have at least X dollars in my pocket!”

In more technical terms, the problem is to determine if there exists an integer X such that exactly X elements in the array are greater than or equal to X. Sounds simple, right? Well, it’s like trying to figure out how many people in a room can actually afford to buy a round of drinks!

Code Solution

Copy Code

class Solution:
    def specialArray(self, nums: list[int]) -> int:
        nums.sort()

        if nums[0] >= len(nums):
            return len(nums)

        for i, (a, b) in enumerate(itertools.pairwise(nums)):
            count = len(nums) - i - 1
            if a < count and b >= count:
                return count

        return -1

Approach Explanation

The approach taken in this solution is quite straightforward. First, we sort the array to make it easier to count how many elements are greater than or equal to a certain value. Then, we check for each possible value of X (from 0 to the length of the array) whether there are exactly X elements that meet the criteria. If we find such an X, we return it; otherwise, we return -1.

Time and Space Complexity

Time Complexity: O(n log n) due to the sorting step, where n is the number of elements in the array.

Space Complexity: O(1) if we ignore the space used by the input array, as we are only using a few extra variables.

Real-World Example

Imagine you’re at a job interview, and the interviewer asks, “How many of you can code in Python?” If you’re in a room of 10 people and only 3 can code in Python, you might think, “Well, I guess I’m not getting this job!” This scenario is akin to our problem: we need to find out if there’s a number of people (or elements) that can meet the criteria of being able to code (or being greater than or equal to X).