The following key values are inserted into a B+ tree in which order of the internal nodes is 3, and that of the leaf nodes is 2, in the sequence given below. The order of internal nodes is the maximum number of tree pointers in each node, and the order of leaf nodes is the maximum number of data items that can be stored in it. The B+ tree is initially empty.10, 3, 6, 8, 4, 2, 1The maximum number of times leaf nodes would get split up as a result of these insertions is___(Assume right biasing).Correct answer is '3'. Can you explain this answer?

Question

Answer

Explanation:

Given sequence of keys: 10, 3, 6, 8, 4, 2, 1

Step 1: Insert 10
- Insert 10 as the root of the tree

Step 2: Insert 3
- Insert 3 as a child of the root
- Since the order of internal nodes is 3, there is no need for split

Step 3: Insert 6
- Insert 6 as a child of the root
- Since the order of internal nodes is 3, there is no need for split

Step 4: Insert 8
- Insert 8 as a child of the root
- Since the order of internal nodes is 3, there is no need for split

Step 5: Insert 4
- Insert 4 as a child of the node containing 3 and 6
- Since the order of leaf nodes is 2 and this node already contains 2 keys, it needs to be split
- Split the node and promote the middle key (4) to the parent node

Step 6: Insert 2
- Insert 2 as a child of the node containing 1 and 4
- Since the order of leaf nodes is 2 and this node already contains 2 keys, it needs to be split
- Split the node and promote the middle key (2) to the parent node

Step 7: Insert 1
- Insert 1 as a child of the node containing 1 and 2
- Since the order of leaf nodes is 2 and this node already contains 2 keys, it needs to be split
- Split the node and promote the middle key (1) to the parent node

Conclusion:
- Leaf nodes got split 3 times while inserting the given sequence of keys.

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