Oct 1, 2020 · Oct 1, 2020 · Oct 1, 2020 · Oct 2, 2020
diff --git a/Heap/P03_FindKthLargestElementInArray.py b/Heap/P03_FindKthLargestElementInArray.py
 import Heap

 def findKthLargestElement(nums:list,k:int):
    '''
    The problem here is to find the Kth Largest Element from the array.
    to solve this we could've used sort and given the answer but it would've cost us O(N*logN) where N is the number of elements in the array

    to even optimize that solution we can use heaps.
    the idea here is:
    1. create a min-heap with adding all the values as negative to it
    2. deleting till second last step
    3. returning the element at the last step as the answer

    T - O(N +K*log(N)) where K is the position of Kth largest element and N is the number of elements in the array

    '''
    h=Heap.BinaryHeap()
    [h.insert(-num) for num in nums]
    [h.delete() for _ in range(k-1)]

    return -h.delete()


 if __name__ == '__main__':

    nums=[2,3,5,6,4]
    k=2
    kthLargestElement = findKthLargestElement(nums,k)

    print("the Kth("+str(k)+") largest element is",kthLargestElement)
diff --git a/Linked Lists/P03_FindingMidOfLinkedList.py b/Linked Lists/P03_FindingMidOfLinkedList.py
 #  Author: VARNIT SHARMA (LunaticProgrammer)

 import SinglyLinkedList

 def findMid(myLinkedList):
    '''
    The approch is simple:
    1. creating a fast pointer that is moving and skipping one element of the linked-list
    2. creating a slow pointer that goes through every next element in the linked-list
    3. when the fast pointer has reached the end the slower one is supposed to be at mid
    4. return the element pointed by slow pointer

    Time-O(logN) where N is the number of elements in linked-list

    '''
    if not myLinkedList or not myLinkedList.head.next: return myLinkedList.head

    slow = myLinkedList.head
    fast = myLinkedList.head

    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next

    return slow


 if __name__ == '__main__':
    myLinkedList = SinglyLinkedList.LinkedList()
    for i in range(10, 0, -1):
        myLinkedList.insertAtStart(i)
    print("LinkedList",end=" ")
    myLinkedList.printLinkedList()
    print()
    midleElement = findMid(myLinkedList)
    print("The middle element of the LinkedList is",midleElement.data)

    # OUTPUT:
    # The middle element of the LinkedList is 6
Original file line number	Diff line number	Diff line change
		@@ -0,0 +1,30 @@
		import Heap

		def findKthLargestElement(nums:list,k:int):
		'''
		The problem here is to find the Kth Largest Element from the array.
		to solve this we could've used sort and given the answer but it would've cost us O(N*logN) where N is the number of elements in the array

		to even optimize that solution we can use heaps.
		the idea here is:
		1. create a min-heap with adding all the values as negative to it
		2. deleting till second last step
		3. returning the element at the last step as the answer

		T - O(N +K*log(N)) where K is the position of Kth largest element and N is the number of elements in the array

		'''
		h=Heap.BinaryHeap()
		[h.insert(-num) for num in nums]
		[h.delete() for _ in range(k-1)]

		return -h.delete()


		if __name__ == '__main__':

		nums=[2,3,5,6,4]
		k=2
		kthLargestElement = findKthLargestElement(nums,k)

		print("the Kth("+str(k)+") largest element is",kthLargestElement)
Original file line number	Diff line number	Diff line change
		@@ -0,0 +1,39 @@
		# Author: VARNIT SHARMA (LunaticProgrammer)

		import SinglyLinkedList

		def findMid(myLinkedList):
		'''
		The approch is simple:
		1. creating a fast pointer that is moving and skipping one element of the linked-list
		2. creating a slow pointer that goes through every next element in the linked-list
		3. when the fast pointer has reached the end the slower one is supposed to be at mid
		4. return the element pointed by slow pointer

		Time-O(logN) where N is the number of elements in linked-list

		'''
		if not myLinkedList or not myLinkedList.head.next: return myLinkedList.head

		slow = myLinkedList.head
		fast = myLinkedList.head

		while fast and fast.next:
		slow = slow.next
		fast = fast.next.next

		return slow


		if __name__ == '__main__':
		myLinkedList = SinglyLinkedList.LinkedList()
		for i in range(10, 0, -1):
		myLinkedList.insertAtStart(i)
		print("LinkedList",end=" ")
		myLinkedList.printLinkedList()
		print()
		midleElement = findMid(myLinkedList)
		print("The middle element of the LinkedList is",midleElement.data)

		# OUTPUT:
		# The middle element of the LinkedList is 6