# Time:  O(n)
# Space: O(1)

from random import randint

class Solution:
    # @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
        

@@ -6,16 +6,25 @@
 e:  O(n)
+ ~ O(n%5E2)
 %0A# Space

Update kth-largest-element-in-an-array.py