Answer is 2
Given:
- B+ tree
- Max no. of keys in the internal node is 5
- Max no. of keys in the leaf node is 4
Solution:
We know that the max child pointer in the internal node of the B+ tree
= max no of keys in internal node + 1
= 5 + 1 i.e 6
Hence the max order of B+ tree internal node = 6
Now for a given internal node of the B+ tree, the min order of any internal node
= ceil (order / 2)
= ceil (6/2)
= 3
Now, if the min order is 3, then min number of keys would be 3-1 = 2