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The best algorithm to solve a given problem is one that requires less space in
memory and takes less time to complete its execution. But in practice it is not always possible to achieve both of these objectives. There may be more than one approach to solve a problem. One approach may require more space but less time to complete its execution. The 2nd approach may require less space but takes more time to complete execution. We choose 1st approach if time is a constraint and 2nd approach if space is a constraint. Thus we may have to sacrifice one at cost of the other. That is what we can say that there exists a time space trade among algorithm.
Q. Draw a B-tree of order 3 for the sequence of keys written below: 2, 4, 9, 8, 7, 6, 3, 1, 5, 10
Ask queConsider the following functional dependencies: Applicant_ID -> Applicant_Name Applicant_ID -> Applicant_Address Position_ID -> Positoin_Title Position_ID -> Date_Position_O
A set s is conveniently shown in a computer store by its characteristic function C(s). This is an array of logical numbers whose ith element has the meaning "i is present in s". As
Explain State Space Tree If it is convenient to execute backtracking by constructing a tree of choices being made, the tree is known as a state space tree. Its root indicates a
Q. Define a method for keeping two stacks within a single linear array S in such a way that neither stack overflows until entire array is used and a whole stack is never shifted to
the above title please send give for the pdf file and word file
H o w can you r ot a t e a B i n a r y Tr e e? E x pl a i n r i g h t a n d l eft r ot a tion s by taking an e x a mpl e. If after
The insertion procedure in a red-black tree is similar to a binary search tree i.e., the insertion proceeds in a similar manner but after insertion of nodes x into the tree T, we c
Draw the process flow diagram: Anand Dairy (AD) sources 150,000 litres of milk daily from large number of local villagers .The milk is collected from 4:00 AM to 6:00 am and
Q. Construct a binary tree whose nodes in inorder and preorder are written as follows: Inorder : 10, 15, 17, 18, 20, 25, 30, 35, 38, 40, 50 Preorder: 20, 15, 10
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