LC-0215 Medium LeetCode

215. Kth Largest Element in an Array

Array Divide and Conquer Sorting Heap (Priority Queue) Quickselect

Read the full problem statement on LeetCode.

Difficulty: medium Acceptance: 68% Topics: Array, Divide and Conquer, Sorting, Heap (Priority Queue), Quickselect

Reading material

Arrays & Hashing 12 min Divide & Conquer 11 min Sorting 13 min

Reference solution (spoiler · python)

# Time:  O(n) on average, using Median of Medians could achieve O(n) (Intro Select)
# Space: O(1)

from random import randint


# optimized for duplicated nums
class Solution(object):
    def findKthLargest(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        def nth_element(nums, n, compare=lambda a, b: a < b):
            def tri_partition(nums, left, right, target, compare):
                mid = left
                while mid <= right:
                    if nums[mid] == target:
                        mid += 1
                    elif compare(nums[mid], target):
                        nums[left], nums[mid] = nums[mid], nums[left]
                        left += 1
                        mid += 1
                    else:
                        nums[mid], nums[right] = nums[right], nums[mid]
                        right -= 1
                return left, right

            left, right = 0, len(nums)-1
            while left <= right:
                pivot_idx = randint(left, right)
                pivot_left, pivot_right = tri_partition(nums, left, right, nums[pivot_idx], compare)
                if pivot_left <= n <= pivot_right:
                    return
                elif pivot_left > n:
                    right = pivot_left-1
                else:  # pivot_right < n.
                    left = pivot_right+1

        nth_element(nums, k-1, compare=lambda a, b: a > b)
        return nums[k-1]


# Time:  O(n) on average, using Median of Medians could achieve O(n) (Intro Select)
# Space: O(1)
class Solution2(object):
    # @param {integer[]} nums
    # @param {integer} k
    # @return {integer}
    def findKthLargest(self, nums, k):
        left, right = 0, len(nums) - 1
        while left <= right:
            pivot_idx = randint(left, right)
            new_pivot_idx = self.PartitionAroundPivot(left, right, pivot_idx, nums)
            if new_pivot_idx == k - 1:
                return nums[new_pivot_idx]
            elif new_pivot_idx > k - 1:
                right = new_pivot_idx - 1
            else:  # new_pivot_idx < k - 1.
                left = new_pivot_idx + 1

    def PartitionAroundPivot(self, left, right, pivot_idx, nums):
        pivot_value = nums[pivot_idx]
        new_pivot_idx = left
        nums[pivot_idx], nums[right] = nums[right], nums[pivot_idx]
        for i in xrange(left, right):
            if nums[i] > pivot_value:
                nums[i], nums[new_pivot_idx] = nums[new_pivot_idx], nums[i]
                new_pivot_idx += 1

        nums[right], nums[new_pivot_idx] = nums[new_pivot_idx], nums[right]
        return new_pivot_idx

Solution from kamyu104/LeetCode-Solutions · MIT