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A Stack has an ordered list of elements & an array is also utilized to store ordered list of elements. Therefore, it would be very simple to manage a stack by using an array. Though, the problem along with an array is that we are needed to declare the size of the array before using it in a program. Thus, the size of stack would be fixed. However, an array & a stack are completely different data structures, an array may be utilized to store the elements of a stack. We may declare the array along a maximum size large sufficient to manage a stack. Program1 implements a stack by using an array.
The advantage of list over Arrays is flexibility. Over flood is not a problem until the computer memory is bushed. When the individual record are quite large, it may be difficult t
After learning this, you will be able to: understand the concept of algorithm; understand mathematical foundation underlying the analysis of algorithm; to understand se
Q . Write down the non-recursive algorithm to traverse a tree in preorder. Ans: T he Non- Recursive algorithm for preorder traversal is written below: Initially i
Q. Explain w hat are the stacks? How can we use the stacks to check whether an expression is correctly parentheses or not. For example (()) is well formed but (() or )()( is not w
What are the different ways of representing a graph? The different ways of representing a graph is: Adjacency list representation: This representation of graph having of an
Indexed Sequential Files An index is inserted to the sequential file to provide random access. An overflow area required to be maintained to permit insertion in sequence. I
Q. Draw the expression tree of the infix expression written below and then convert it intoPrefix and Postfix expressions. ((a + b) + c * (d + e) + f )* (g + h )
Q. State the difference between a grounded header link list and a circular header link list? Ans: A header linked list is a linked list which all the time c
#questionalgorithm for implementing multiple\e queues in a single dimensional array
Insertion & deletion of target key requires splaying of the tree. In case of insertion, the tree is splayed to find the target. If, target key is found out, then we have a duplicat
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