Leetcode – Special Array With X Elements Greater Than or Equal X Solution

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Leetcode is a goldmine of programming challenges that range from simple algorithmic problems to intricate system designs. Among these is the “Special Array With X Elements Greater Than or Equal X” problem, a problem that captures the essence of array manipulation and computational logic. Though it may appear straightforward at first glance, it provides an excellent opportunity to delve into sorting algorithms, array manipulation, and binary search.

In this comprehensive article, we will dissect the problem in detail, walk through a Pythonic solution, discuss the time and space complexities, and explore potential optimizations and enhancements.

Problem Statement

The original problem statement from Leetcode is as follows:

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:

Input: nums = [3,5,6,7]
Output: 4

In this example, there are exactly 4 numbers in nums that are greater than or equal to 4 (the numbers are [5,6,7,4]). Therefore, the array is special.

Understanding the Problem

Before rushing to code, it’s crucial to understand what the problem asks for. We need to find a value x such that there are exactly x numbers in the array that are greater than or equal to x. If we find such an x, the array is termed ‘special’.

Algorithmic Approach

There are several ways to approach this problem, but a sorted array would make our task much easier. The main steps to solve the problem can be as follows:

  1. Sort the Array: Sort the given array in ascending order.
  2. Check for Special Condition: Traverse the sorted array and check if it satisfies the special condition.

Python Code Implementation

Let’s implement our algorithm in Python:

def specialArray(nums):
    nums.sort()
    n = len(nums)
    
    for i in range(n):
        x = n - i
        if nums[i] >= x:
            if i == 0 or nums[i-1] < x:
                return x
    return -1

Code Explanation

  1. Sorting: We start by sorting the array in ascending order using Python’s built-in sort() method.
  2. Checking for Special Condition: We then traverse the sorted array and perform the following checks:
    • x = n - i: This is the number of elements in nums that are greater than or equal to x (since the array is sorted).
    • nums[i] >= x: This checks if the element nums[i] is greater than or equal to x.
    • nums[i-1] < x: This ensures that we have exactly x numbers greater than or equal to x.

If these conditions are satisfied, we return x; otherwise, we return -1.

Time and Space Complexity

  1. Time Complexity:
    • Sorting the array takes O(n log⁡ n) time.
    • The loop traversal takes O(n) time.
    • Therefore, the overall time complexity is O(n log⁡ n).
  2. Space Complexity:
    • The space complexity is O(1) since we do the sorting in-place and only use a few extra variables.

Possible Optimizations and Variants

  1. Binary Search: You could potentially employ a binary search mechanism to speed up the checking process, though this might complicate the code.
  2. Counting Sort: Since the input is non-negative integers, you could use counting sort to bring the sorting complexity down to O(n)O(n) in certain cases.
  3. Pre-calculation of Sums: Another potential optimization could involve pre-calculating the sums or counts for quicker checks.

Conclusion

The “Special Array With X Elements Greater Than or Equal X” problem is a compelling programming challenge that combines array manipulation with some logical ingenuity. It provides an excellent platform to practice sorting, looping, and condition-checking in Python. Additionally, it opens up avenues for thinking about optimization techniques.

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