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The following key values are inserted into B+

tree in which the order of internal nodes is 4 and that of the leaf node is 5 in the sequence given below. The order of internal node is maximum number of keys in each node and the order of leaf node is the maximum number of pointers that can be stored in it. The B+

tree initially empty.

 

 2, 7, 3, 11, 17, 5, 19, 31, 29, 23

 

Answer is given as :- 17+31+23 = 71

 

But my doubt is since leaf can carry 5 pointer then till 5 element we can  insert in one leaf. So answer should be 5+ 19+23 = 47.

AM i Right?
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if there are x pointer in the leaf node then key will be x-1 as one extra pointer is used in b+ tree to point to the other leaf node in a linked list manner,so if there are 5 pointer you can insert 4 keys upto maximum

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