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Q. How do we represent a max-heap sequentially? Explain by taking a valid example.
Ans: A max heap is also called as a descending heap, of size n is an almost complete binary tree of n nodes such as the content of each node is less than or equal to the content or matter of its father. If sequential representation of an almost whole binary tree is used, this condition reduces to the condition of inequality. Info[j] <= info[(j-1)/2] for 0 <= ((j-1)/2) < j <= n-1 It is very much clear from this definition that root of the tree contains the largest element in the heap in the descending heap. Any of the paths from the root to a leaf is an ordered list in the descending order.
Ans:
A max heap is also called as a descending heap, of size n is an almost complete binary tree of n nodes such as the content of each node is less than or equal to the content or matter of its father. If sequential representation of an almost whole binary tree is used, this condition reduces to the condition of inequality.
Info[j] <= info[(j-1)/2] for 0 <= ((j-1)/2) < j <= n-1
It is very much clear from this definition that root of the tree contains the largest element in the heap in the descending heap. Any of the paths from the root to a leaf is an ordered list in the descending order.
for(int i = 0; i for (int j = n - 1; j >= i ; j--){ System.out.println(i+ " " + j);
Ans: A procedure to reverse the singly linked list: reverse(struct node **st) { struct node *p, *q, *r; p = *st; q = NULL; while(p != NULL) { r =q;
Data array A has data series from 1,000,000 to 1 with step size 1, which is in perfect decreasing order. Data array B has data series from 1 to 1,000,000, which is in random order.
Program: Program segment for insertion of an element into the queue add(int value) { struct queue *new; new = (struct queue*)malloc(sizeof(queue)); new->value = val
MID SQUARE METHOD
Implement multiple queues in a single dimensional array. Write algorithms for various queue operations for them.
How do collisions happen during hashing? Usually the key space is much larger than the address space, thus, many keys are mapped to the same address. Assume that two keys K1 an
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Algo rithm to Insert a Node p at the End of a Linked List is explained below Step1: [check for space] If new1= NULL output "OVERFLOW" And exit Step2: [Allocate fr
Q. Suggest a method of implementing two stacks in one array such that as long as space is there in an array, you should be capable to add an element in either stack. Using proposed
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