diff --git a/Heap/P03_FindKthLargestElementInArray.py b/Heap/P03_FindKthLargestElementInArray.py
new file mode 100644
index 0000000..12738ac
--- /dev/null
+++ b/Heap/P03_FindKthLargestElementInArray.py
@@ -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)
\ No newline at end of file
diff --git a/Linked Lists/P03_FindingMidOfLinkedList.py b/Linked Lists/P03_FindingMidOfLinkedList.py
new file mode 100644
index 0000000..faad4d6
--- /dev/null
+++ b/Linked Lists/P03_FindingMidOfLinkedList.py	
@@ -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