Given a linked list of co-ordinates where adjacent points either form a vertical line or a horizontal line. Delete points from the linked list which are in the middle of a horizontal or vertical line.

Examples:

Input: (0,10)->(1,10)->(5,10)->(7,10) | (7,5)->(20,5)->(40,5) Output: Linked List should be changed to following (0,10)->(7,10) | (7,5)->(40,5) The given linked list represents a horizontal line from (0,10) to (7, 10) followed by a vertical line from (7, 10) to (7, 5), followed by a horizontal line from (7, 5) to (40, 5). Input: (2,3)->(4,3)->(6,3)->(10,3)->(12,3) Output: Linked List should be changed to following (2,3)->(12,3) There is only one vertical line, so all middle points are removed.

Source: Microsoft Interview Experience

The idea is to keep track of current node, next node and next-next node. While the next node is same as next-next node, keep deleting the next node. In this complete procedure we need to keep an eye on shifting of pointers and checking for NULL values.

Following are C/C++ and Java implementations of above idea.

## C/C++

// C program to remove intermediate points in a linked list // that represents horizontal and vertical line segments #include <stdio.h> #include <stdlib.h> // Node has 3 fields including x, y coordinates and a pointer // to next node struct Node { int x, y; struct Node *next; }; /* Function to insert a node at the beginning */ void push(struct Node ** head_ref, int x,int y) { struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); new_node->x = x; new_node->y = y; new_node->next = (*head_ref); (*head_ref) = new_node; } /* Utility function to print a singly linked list */ void printList(struct Node *head) { struct Node *temp = head; while (temp != NULL) { printf("(%d,%d)-> ", temp->x,temp->y); temp = temp->next; } printf("\n"); } // Utility function to remove Next from linked list // and link nodes after it to head void deleteNode(struct Node *head, struct Node *Next) { head->next = Next->next; Next->next = NULL; free(Next); } // This function deletes middle nodes in a sequence of // horizontal and vertical line segments represented by // linked list. struct Node* deleteMiddle(struct Node *head) { // If only one node or no node...Return back if (head==NULL || head->next ==NULL || head->next->next==NULL) return head; struct Node* Next = head->next; struct Node *NextNext = Next->next ; // Check if this is a vertical line or horizontal line if (head->x == Next->x) { // Find middle nodes with same x value, and delete them while (NextNext !=NULL && Next->x==NextNext->x) { deleteNode(head, Next); // Update Next and NextNext for next iteration Next = NextNext; NextNext = NextNext->next; } } else if (head->y==Next->y) // If horizontal line { // Find middle nodes with same y value, and delete them while (NextNext !=NULL && Next->y==NextNext->y) { deleteNode(head, Next); // Update Next and NextNext for next iteration Next = NextNext; NextNext = NextNext->next; } } else // Adjacent points must have either same x or same y { puts("Given linked list is not valid"); return NULL; } // Recur for next segment deleteMiddle(head->next); return head; } // Driver program to tsst above functions int main() { struct Node *head = NULL; push(&head, 40,5); push(&head, 20,5); push(&head, 10,5); push(&head, 10,8); push(&head, 10,10); push(&head, 3,10); push(&head, 1,10); push(&head, 0,10); printf("Given Linked List: \n"); printList(head); if (deleteMiddle(head) != NULL); { printf("Modified Linked List: \n"); printList(head); } return 0; }

## Java

// Java program to remove middle points in a linked list of // line segments, class LinkedList { Node head; // head of list /* Linked list Node*/ class Node { int x,y; Node next; Node(int x, int y) { this.x = x; this.y = y; next = null; } } // This function deletes middle nodes in a sequence of // horizontal and vertical line segments represented // by linked list. Node deleteMiddle() { // If only one node or no node...Return back if (head == null || head.next == null || head.next.next == null) return head; Node Next = head.next; Node NextNext = Next.next; // check if this is vertical or horizontal line if (head.x == Next.x) { // Find middle nodes with same value as x and // delete them. while (NextNext != null && Next.x == NextNext.x) { head.next = Next.next; Next.next = null; // Update NextNext for the next iteration Next = NextNext; NextNext = NextNext.next; } } // if horizontal else if (head.y == Next.y) { // find middle nodes with same value as y and // delete them while (NextNext != null && Next.y == NextNext.y) { head.next = Next.next; Next.next = null; // Update NextNext for the next iteration Next = NextNext; NextNext = NextNext.next; } } // Adjacent points should have same x or same y else { System.out.println("Given list is not valid"); return null; } // recur for other segment // temporarily store the head and move head forward. Node temp = head; head = head.next; // call deleteMiddle() for next segment this.deleteMiddle(); // restore head head = temp; // return the head return head; } /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void push(int x, int y) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(x,y); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } void printList() { Node temp = head; while (temp != null) { System.out.print("("+temp.x+","+temp.y+")->"); temp = temp.next; } System.out.println(); } /* Drier program to test above functions */ public static void main(String args[]) { LinkedList llist = new LinkedList(); llist.push(40,5); llist.push(20,5); llist.push(10,5); llist.push(10,8); llist.push(10,10); llist.push(3,10); llist.push(1,10); llist.push(0,10); System.out.println("Given list"); llist.printList(); if (llist.deleteMiddle() != null) { System.out.println("Modified Linked List is"); llist.printList(); } } } /* This code is contributed by Rajat Mishra */

## Python

# Python program to remove middle points in a linked list of # line segments, class LinkedList(object): def __init__(self): self.head = None # Linked list Node class Node(object): def __init__(self, x, y): self.x = x self.y = y self.next = None # This function deletes middle nodes in a sequence of # horizontal and vertical line segments represented # by linked list. def deleteMiddle(self): # If only one node or no node...Return back if self.head == None or self.head.next == None or self.head.next.next == None: return self.head Next = self.head.next NextNext = Next.next # check if this is vertical or horizontal line if self.head.x == Next.x: # Find middle nodes with same value as x and # delete them. while NextNext != None and Next.x == NextNext.x: self.head.next = Next.next Next.next = None # Update NextNext for the next iteration Next = NextNext NextNext = NextNext.next elif self.head.y == Next.y: # find middle nodes with same value as y and # delete them while NextNext != None and Next.y == NextNext.y: self.head.next = Next.next Next.next = None # Update NextNext for the next iteration Next = NextNext NextNext = NextNext.next else: # Adjacent points should have same x or same y print "Given list is not valid" return None # recur for other segment # temporarily store the head and move head forward. temp = self.head self.head = self.head.next # call deleteMiddle() for next segment self.deleteMiddle() # restore head self.head = temp # return the head return self.head # Given a reference (pointer to pointer) to the head # of a list and an int, push a new node on the front # of the list. def push(self, x, y): # 1 & 2: Allocate the Node & # Put in the data new_node = self.Node(x, y) # 3. Make next of new Node as head new_node.next = self.head # 4. Move the head to point to new Node self.head = new_node def printList(self): temp = self.head while temp != None: print "(" + str(temp.x) + "," + str(temp.y) + ")->", temp = temp.next print '' # Driver program llist = LinkedList() llist.push(40,5) llist.push(20,5) llist.push(10,5) llist.push(10,8) llist.push(10,10) llist.push(3,10) llist.push(1,10) llist.push(0,10) print "Given list" llist.printList() if llist.deleteMiddle() != None: print "Modified Linked List is" llist.printList() # This code is contributed by BHAVYA JAIN

Output:

Given Linked List: (0,10)-> (1,10)-> (3,10)-> (10,10)-> (10,8)-> (10,5)-> (20,5)-> (40,5)-> Modified Linked List: (0,10)-> (10,10)-> (10,5)-> (40,5)->

Time Complexity of the above solution is O(n) where n is number of nodes in given linked list.

**Exercise: **

The above code is recursive, write an iterative code for the same problem.

This article is contributed by Sanket Jain. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above