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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.
Step-1: For the current node, verify whether it contain a left child. If it has, then go to step-2 or else go to step-3 Step-2: Repeat step-1 for left child Step-3: Visit (th
A representation of an array structure is a mapping of the (abstract) array with elements of type T onto the store which is an array with elements of type BYTE. The array could be
This notation bounds a function to in constant factors. We say f(n) = Θ(g(n)) if there presents positive constants n 0 , c 1 and c 2 such that to the right of n 0 the value of f
A graph with n vertices will absolutely have a parallel edge or self loop if the total number of edges is greater than n-1
how to implement multiple stack using single dimension array in c
Preorder traversal of a binary tree struct NODE { struct NODE *left; int value; /* can take any data type */ struct NODE *right; }; preorder(struct N
give any example of page replacement using fifo and use your own reference string
A driver takes shortest possible route to attain destination. The problem which we will discuss here is similar to this type of finding shortest route in any specific graph. The gr
For preorder traversal, in the worst case, the stack will rise to size n/2, where n refer to number of nodes in the tree. Another method of traversing binary tree non-recursively t
Write a function that performs integer division. The function should take the large number in memory location 1 and divide it by the large number in memory location 2 disregarding
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