0215. Kth Largest in Array

Medium | Array + Sort + QuickSelect | 40 ms (98.02%), 14.1 MB (94.93%)

❓ Problem Statement

Soure: LeetCode - Kth Largest Element in an Array GitHub: Solution / Performance

Given an integer array nums and an integer k, return the kth largest element in the array.

Note that it is the kth largest element in the sorted order, not the kth distinct element.

✍🏻 Constraints & Example

💡 Ideas

Do quick sort to put larger elements on the left of the selected pivot.

For quick sorting, we need a partition function based on the randomly selected pivot index to divide the array into two parts:

1. The left part contains larger elements (> randomly-selected pivot)

2. The right part contains smaller elements (< randomly-selected pivot)

Since the partition function returns n-th largest index, the quick soring ends whenn == k. Otherwise, we modify the range between left and right for the next partition.

🤖 Python3

class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:        
        # (base case)
        if len(nums) == 1: return nums[0]
        
        # ==================================================
        #  Quickselect                                     =
        # ==================================================
        # time  : O(n)
        # space : O(1)
        
        self.nums = nums
        self.quickSelect(0, len(nums) - 1, k - 1)
        return self.nums[k-1]
        
        # (Return top-k largest elements)
        # return self.nums[:k]
    
    def quickSelect(self, left, right, kLargest) -> None:
        if not (left < right): return 
        
        pivot = self.partition(left, right)

        if pivot == kLargest: return
        elif pivot > kLargest: self.quickSelect(left, pivot-1, kLargest)
        else: self.quickSelect(pivot+1, right, kLargest)
                
    def partition(self, left, right) -> int:
        randomNum = (left + right) // 2
        
        # move pivot to the end/right
        pivot = right
        self.nums[randomNum], self.nums[pivot] = self.nums[pivot], self.nums[randomNum]
        
        # move larger elements to the left
        placeP = left
        for i in range(left, right):
            if self.nums[i] >= self.nums[pivot]:
                self.nums[placeP], self.nums[i] = self.nums[i], self.nums[placeP]
                placeP += 1
            
        # move back the pivot
        self.nums[placeP], self.nums[pivot] = self.nums[pivot], self.nums[placeP]
        return placeP

🤖 Python3 (Heap)

🤖 Java