Kth Largest Element in an Array

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given an array of integers and an integer k.
  • We need to find the kth largest element in the array.

2. Analyze the problem: To solve this problem, we can use the concept of partitioning in the QuickSelect algorithm. We'll choose a pivot element, partition the array around it, and recursively apply the partitioning process on the appropriate subarray until we find the kth largest element.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Define a helper function called partition that takes the array, a start index, and an end index as parameters. This function will choose a pivot element and partition the array around it.
    • Choose the last element as the pivot.
    • Initialize a variable partitionIndex to keep track of the partition index.
    • Iterate through the array from the start index to the end index:
      • If the current element is less than or equal to the pivot, swap it with the element at the partitionIndex and increment partitionIndex.
    • Finally, swap the pivot element with the element at the partitionIndex.
    • Return the partitionIndex.
  2. Define the main function findKthLargest that takes the array and k as parameters.
    • Initialize variables start and end to 0 and len(array) - 1 respectively.
    • Use a while loop to find the kth largest element:
      • Call the partition function on the array with start and end as arguments.
      • If the partition index is equal to k - 1, return the element at that index.
      • If the partition index is greater than k - 1, update end to partitionIndex - 1.
      • If the partition index is less than k - 1, update start to partitionIndex + 1.
  3. Return -1 if k is out of range (less than 1 or greater than the array size).

4. Explain your approach: The approach involves using the QuickSelect algorithm to find the kth largest element in the array. We choose a pivot element, partition the array around it, and recursively apply the partitioning process on the appropriate subarray until we find the kth largest element.

5. Write clean and readable code:

python
def partition(array, start, end):
    pivot = array[end]
    partitionIndex = start

    for i in range(start, end):
        if array[i] <= pivot:
            array[i], array[partitionIndex] = array[partitionIndex], array[i]
            partitionIndex += 1

    array[partitionIndex], array[end] = array[end], array[partitionIndex]
    return partitionIndex

def findKthLargest(nums, k):
    start, end = 0, len(nums) - 1

    while True:
        partitionIndex = partition(nums, start, end)

        if partitionIndex == k - 1:
            return nums[partitionIndex]
        elif partitionIndex > k - 1:
            end = partitionIndex - 1
        else:
            start = partitionIndex + 1

    return -1

6. Test your code: Let's test the code with the following test case:

python
nums = [3, 2, 1, 5, 6, 4]
k = 2

result = findKthLargest(nums, k)
print(result)

The expected output is:

5

7. Optimize if necessary: The provided solution using the QuickSelect algorithm has an average time complexity of O(n) and a worst-case time complexity of O(n^2) if the pivot selection is not balanced. To optimize the algorithm, we can use the Randomized QuickSelect algorithm that randomly selects the pivot, reducing the chances of worst-case time complexity.

8. Handle error cases: The code already handles the case where k is out of range (less than 1 or greater than the array size) by returning -1.

9. Discuss complexity analysis: The average time complexity of the QuickSelect algorithm is O(n), where n is the length of the array. This is because, on average, the partition process eliminates a constant fraction of the remaining elements at each step. However, in the worst-case scenario, the time complexity can be O(n^2) if the pivot selection is not balanced. The space complexity of the algorithm is O(1) since we are using a constant amount of extra space.

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