Solution Explanations for Finding the Kth Largest Element in an Array

This problem asks to find the kth largest element in an array. Several approaches exist, each with different time and space complexity trade-offs. We'll explore three: QuickSelect, using a Priority Queue (Min Heap), and Counting Sort.

1. QuickSelect

Approach: QuickSelect is a variation of the QuickSort algorithm. Instead of sorting the entire array, it recursively partitions the array around a pivot element. The goal is to find the position of the kth largest element without fully sorting.

Algorithm:

1. Choose a pivot: Select a pivot element (the code uses the middle element).
2. Partition: Rearrange the array such that elements smaller than the pivot are placed before it, and elements larger are placed after.
3. Recurse: Based on the pivot's position, determine which subarray contains the kth largest element and recurse on that subarray.

Time Complexity: Average case: O(n), Worst case: O(n²). The average case is linear because the partitioning process effectively reduces the problem size by half in each step. The worst-case occurs when the pivot selection consistently leads to highly unbalanced partitions (e.g., always selecting the smallest or largest element).

Space Complexity: O(log n) due to the recursive calls on the call stack. In the worst case, the recursion depth could be O(n).

Code (Python):

class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        def quickselect(l, r):
            if l == r:
                return nums[l]
            i, j = l - 1, r + 1
            pivot = nums[(l + r) // 2]
            while i < j:
                while True:
                    i += 1
                    if nums[i] >= pivot:
                        break
                while True:
                    j -= 1
                    if nums[j] <= pivot:
                        break
                if i < j:
                    nums[i], nums[j] = nums[j], nums[i]
            if j < len(nums) - k:
                return quickselect(j + 1, r)
            return quickselect(l, j)
 
        return quickselect(0, len(nums) - 1)

The code in other languages (Java, C++, Go, TypeScript, Rust) follows a similar approach; the key differences lie in syntax and how they handle array manipulation and recursion.

2. Priority Queue (Min Heap)

Approach: A min-heap data structure is used to efficiently track the k largest elements encountered so far.

Algorithm:

1. Initialize a min-heap of size k.
2. Iterate through the input array:
   • If the heap size is less than k, add the element to the heap.
   • If the heap size is equal to k, and the current element is larger than the smallest element in the heap (heap's root), remove the root and insert the current element.
3. After iterating through all elements, the root of the min-heap will be the kth largest element.

Time Complexity: O(n log k). Building the heap takes O(k log k) initially. Then, for each of the remaining n-k elements, we perform a heap operation (O(log k)).

Space Complexity: O(k) to store the min-heap.

Code (Python):

import heapq
 
class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        return heapq.nlargest(k, nums)[-1] # Efficient Python library function

Other language implementations will involve creating a min-heap manually using their respective priority queue libraries.

3. Counting Sort (If applicable)

Approach: Counting sort works well when the input numbers are within a relatively small range. It counts the occurrences of each number.

Algorithm:

1. Create a hash map (or array if the range is known) to store the frequency of each number.
2. Iterate through the numbers in descending order. Accumulate counts until you reach the kth largest.

Time Complexity: O(n + m), where 'n' is the number of elements, and 'm' is the range of numbers (maximum - minimum). This is linear if 'm' is relatively small compared to 'n'.

Space Complexity: O(m) to store the frequency counts.

Code (Python):

from collections import Counter
 
class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        count = Counter(nums)
        for i in range(max(nums), -1, -1):
            k -= count[i]
            if k <= 0:
                return i
 

This approach is only efficient if the range of numbers is limited; otherwise, it's less efficient than QuickSelect or using a min-heap. The other language versions will have similar structures, using their respective hash map or frequency array implementations.

Choosing the Right Solution:

• QuickSelect: Generally the most efficient for a wide range of inputs. The average case linear time is a significant advantage.
• Min-Heap: A good choice if k is much smaller than n. The logarithmic time complexity for heap operations is beneficial in this scenario.
• Counting Sort: Only use this if the input numbers are within a small, known range. Otherwise, the space complexity can be high.