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- Change the following code for Depth First Search Strategy so that the graph (graph values in dictionary) generates randomly. # Using a Python dictionary to act as an adjacency listgraph = { '5' : ['3','7'], '3' : ['2', '4'], '7' : ['8'], '2' : [], '4' : ['8'], '8' : []} visited = set() # Set to keep track of visited nodes of graph. def dfs(visited, graph, node): #function for dfs if node not in visited: print (node) visited.add(node) for neighbour in graph[node]: dfs(visited, graph, neighbour) # Driver Codeprint("Following is the Depth-First Search")dfs(visited, graph, '5')Suppose that the height of a Binary Search Tree is ℎ, then the time complexty to delete a node in the BST will be ?(ℎ). True or false Please explain whyTHIS IS MY CODE HELP ME ACHIEVE POINTS OUTLINED BELOW : #include <stdio.h>#include <stdlib.h>#include <string.h>#include <float.h>#include "graph.h"#include "dijkstra.h" #define INFINITY DBL_MAX /* find shortest paths between source node id and all other nodes in graph. *//* upon success, returns an array containing a table of shortest paths. *//* return NULL if *graph is uninitialised or an error occurs. *//* each entry of the table array should be a Path *//* structure containing the path information for the shortest path between *//* the source node and every node in the graph. If no path exists to a *//* particular desination node, then next should be set to -1 and weight *//* to DBL_MAX in the Path structure for this node */Path *dijkstra(Graph *graph, int id, int *pnEntries){ int n; int i, j; int* nv = get_vertices(graph, &n); int *S = malloc(n * sizeof(int)); double *D = malloc(n * sizeof(double)); int *R = malloc(n…
- Given the following binary search tree, what would be the sequence of nodes, if traversed using inorder traversal after deleting 8. Image 2.png 3 1 4 6 7 13 14 10 1 3 4 6 7 10 13 14 3 1 6 4 7 10 14 13 1 4 6 3 13 14 10 7implement the algorithm in C++ program. Algorithm to Linked List Creation Algorithm: (ITEM), [ Here the initial value of FRONT & REAR are NULL ] Step 1. Create dynamic NODE to store (Info and Next) for the next node and store its address into NewNode pointer [Insert ITEM in newly created Node for QUEUE.] Step 2. Set NewNode -> Info = ITEM and NewNode -> Next = NULL Step 3. If REAR = NULL then: Set REAR and FRONT both = NewNode [ First Node Enqueued ] Else Set REAR -> Next = NewNode Set REAR = NewNode [ End of If Structure ] Step 4. End.What advantages does a hash table have over a linear list? What presumably is the issue?putting a hashtable in place
- Computer Science Create a circular link list with at least 7 nodes, and explain in words that how insertion, deletion of a node, and traverse will work. No need for codes and algorithm, just explain in words and design the circular link of 7 nodes.09.Given a linked list of N nodes such that it may contain a loop. A loop here means that the last node of the link list is connected to the node at position X(1-based index). If the link list does not have any loop, X=0. Remove the loop from the linked list, if it is present, i.e. unlink the last node which is forming the loop. Example 1: Input: N = 3 value[] = {1,3,4} X = 2 Output: 1 Explanation: The link list looks like 1 -> 3 -> 4 ^ | |____|Suppose there are two singly linked lists both of which intersect at some point and become a single linked list. The head or start pointers of both the lists are known, but the intersecting node is unknown. Also, the number of nodes in each of the list before they intersect are unknown and both the list may have it different. List1 may have n nodes before it reaches intersection point and List2 might have m nodes before it reaches intersection point where m and n may be m = n, m > n or m < n. Give an algorithm for finding the merging point. Hints: A brute force approach would be to compare every pointer in one list with every pointer in another list. But in this case the complexity would be O(mn)
- Develop a linked-list processing function, IsolateTargetSoloAsTail, to process a linked list as follows. ● If a target cannot be found on the given list, a new node containing the target is created and added to the list's end (made the new tail node). ► This includes the case where the given list is empty, in which the new tail node added is also the new head node. (This is so because the only node in a 1-node list is the list's head and tail node.) ● If the target appears only once on the given list, the target-matching node is moved to the list's end (made the new tail node). ► Nothing needs to be done if the target-matching node is already the tail node (of the given list). ● If the target appears multiple times on the given list, the first target-matching node is moved to the list's end (made the new tail node), and all other target-matching nodes are to be deleted from the list. ► Note that although the target-matching node to be…ANSWER!!Is it possible to implement a search algorithm that runs in logarithmic time for a linked list? Why or why not?Given the following search tree, state the order in which the nodes will be searched for breadth first and depth first, until a solution B is reached.