Explain retrieving an element with a given rank.

Retrieving an element with a given rank is a common operation in data structures, particularly in trees like Binary Search Trees (BSTs) and Balanced Binary Search Trees (AVL, Red-Black Trees, etc.). The rank of an element in a set of elements is the position it would occupy in a list if all the elements were sorted. For example, in the set {7, 10, 3, 20, 15}, the rank of 10 is 2 because it is the second smallest element.

Here is a step-by-step approach to retrieve an element with a given rank in a Binary Search Tree:

1. Add a size field to tree nodes: To support rank operations efficiently, we need to maintain size information in every node. The size of a node is the total number of nodes in the subtree rooted at that node. For example, if we have a node with key 15, and it has 5 nodes in its left subtree and 3 nodes in its right subtree, the size of this node will be 5 (left subtree) + 1 (itself) + 3 (right subtree) = 9.

2. Update the size field during insertion/deletion: Whenever we insert or delete a node, we need to update the size field of all ancestor nodes. This can be done by incrementing/decrementing the size of all ancestors of the inserted/deleted node.

3. Retrieve an element with a given rank: To retrieve an element with a given rank, we can use the following algorithm:

- Start from the root node. - If the size of the left subtree of the current node (let's denote it as leftSize) is equal to the rank - 1, then the current node is the node with the given rank. - If leftSize > rank - 1, move to the left child of the current node. - If leftSize < rank - 1, subtract leftSize + 1 from the rank and move to the right child of the current node. - Repeat the above steps until you find the node with the given rank.

Here is a Python example of the algorithm:

class Node:
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
        self.size = 1

def insert(node, key):
    if node is None:
        return Node(key)
    if key < node.key:
        node.left = insert(node.left, key)
    else:
        node.right = insert(node.right, key)
    node.size = 1 + size(node.left) + size(node.right)
    return node

def size(node):
    if node is None:
        return 0
    else:
        return node.size

def getRank(node, rank):
    if node is None:
        return None
    leftSize = size(node.left)
    if leftSize == rank - 1:
        return node.key
    elif leftSize > rank - 1:
        return getRank(node.left, rank)
    else:
        return getRank(node.right, rank - leftSize - 1)

root = None
keys = [20, 8, 22, 4, 12, 10, 14]
for key in keys:
    root = insert(root, key)
print(getRank(root, 3))  # Output: 10

In this example, we first insert keys into the BST while maintaining the size of each node. Then we retrieve the element with rank 3, which is 10.

The time complexity of the getRank operation is O(h), where h is the height of the tree. If the tree is balanced, the height is logarithmic in the number of elements, so the operation is efficient.