**Find Merge Point of Two Lists**

### Problem Statement :

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share a common node, return that node's data value. Note: After the merge point, both lists will share the same node pointers. Example In the diagram below, the two lists converge at Node x: [List #1] a--->b--->c \ x--->y--->z--->NULL / [List #2] p--->q Function Description Complete the findMergeNode function in the editor below. findMergeNode has the following parameters: SinglyLinkedListNode pointer head1: a reference to the head of the first list SinglyLinkedListNode pointer head2: a reference to the head of the second list Returns int: the data value of the node where the lists merge Input Format Do not read any input from stdin/console. The first line contains an integer t, the number of test cases.

### Solution :

` ````
Solution in C :
In C++ :
/*
Find merge point of two linked lists
Node is defined as
struct Node
{
int data;
Node* next;
}
*/
int getCount(Node* head)
{
Node* current = head;
int count = 0;
while (current != NULL)
{
count++;
current = current->next;
}
return count;
}
int getNode(int d, Node* head1, Node* head2)
{
int i;
Node* current1 = head1;
Node* current2 = head2;
for(i = 0; i < d; i++)
{
if(current1 == NULL)
{ return -1; }
current1 = current1->next;
}
while(current1 != NULL && current2 != NULL)
{
if(current1 == current2)
return current1->data;
current1= current1->next;
current2= current2->next;
}
return -1;
}
int FindMergeNode(Node *headA, Node *headB)
{
// Complete this function
// Do not write the main method.
int c1 = getCount(headA);
int c2 = getCount(headB);
int d;
if(c1 > c2)
{
d = c1 - c2;
return getNode(d, headA, headB);
}
else
{
d = c2 - c1;
return getNode(d, headB, headA);
}
}
In Java :
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
}
*/
int FindMergeNode(Node headA, Node headB) {
// Complete this function
// Do not write the main method.
int countA=0;
int countB=0;
Node tempA=headA;
Node tempB=headB;
while(tempA!=null)
{
countA++;
tempA=tempA.next;
}
while(tempB!=null)
{
countB++;
tempB=tempB.next;
}
int diff=0;
if(countA>countB)
diff=countA-countB;
else
diff=countB-countA;
tempA=headA;
tempB=headB;
if(countA>countB)
{
while(diff >0)
{tempA=tempA.next;
diff--;}
}
else
{while(diff >0)
{ tempB=tempB.next;
diff--;}
}
while(tempA!=null && tempB!=null)
{
tempA=tempA.next;
tempB=tempB.next;
if(tempA==tempB)
return tempA.data;
}
return 0;
}
In python3 :
"""
Find the node at which both lists merge and return the data of that node.
head could be None as well for empty list
Node is defined as
class Node(object):
def __init__(self, data=None, next_node=None):
self.data = data
self.next = next_node
"""
def FindMergeNode(a, b):
h = {}
while a != None:
h[a.data] = a.data
a = a.next
while b != None:
if b.data in h:
return h[b.data]
b = b.next
return None
In C :
// Complete the findMergeNode function below.
/*
* For your reference:
*
* SinglyLinkedListNode {
* int data;
* SinglyLinkedListNode* next;
* };
*
*/
#include <math.h>
int findMergeNode(SinglyLinkedListNode *headA, SinglyLinkedListNode *headB)
{
// Complete this function
// Do not write the main method.
int countA = 0;
int countB = 0;
struct SinglyLinkedListNode *tempHeadA, *tempHeadB;
tempHeadA = headA;
tempHeadB = headB;
while(tempHeadA != NULL){
countA++;
tempHeadA = tempHeadA->next;
}
while(tempHeadB != NULL){
countB++;
tempHeadB = tempHeadB->next;
}
// printf("%d %d", countA,countB);
int biggerNodeExtraNodes = abs(countA - countB);
// printf("%d", biggerNodeExtraNodes);
while(biggerNodeExtraNodes--){
printf("33333333333");
if(countA > countB){
headA = headA->next;
} else{
headB = headB->next;
}
}
while(headA != NULL){
printf("asdfghjklpoiuytredxdcfgh");
if(headA == headB){
return headA->data;
}else{
headA = headA->next;
headB = headB->next;
}
}
return 1;
}
```

## View More Similar Problems

## Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →## Pair Sums

Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v

View Solution →## Lazy White Falcon

White Falcon just solved the data structure problem below using heavy-light decomposition. Can you help her find a new solution that doesn't require implementing any fancy techniques? There are 2 types of query operations that can be performed on a tree: 1 u x: Assign x as the value of node u. 2 u v: Print the sum of the node values in the unique path from node u to node v. Given a tree wi

View Solution →## Ticket to Ride

Simon received the board game Ticket to Ride as a birthday present. After playing it with his friends, he decides to come up with a strategy for the game. There are n cities on the map and n - 1 road plans. Each road plan consists of the following: Two cities which can be directly connected by a road. The length of the proposed road. The entire road plan is designed in such a way that if o

View Solution →## Heavy Light White Falcon

Our lazy white falcon finally decided to learn heavy-light decomposition. Her teacher gave an assignment for her to practice this new technique. Please help her by solving this problem. You are given a tree with N nodes and each node's value is initially 0. The problem asks you to operate the following two types of queries: "1 u x" assign x to the value of the node . "2 u v" print the maxim

View Solution →## Number Game on a Tree

Andy and Lily love playing games with numbers and trees. Today they have a tree consisting of n nodes and n -1 edges. Each edge i has an integer weight, wi. Before the game starts, Andy chooses an unordered pair of distinct nodes, ( u , v ), and uses all the edge weights present on the unique path from node u to node v to construct a list of numbers. For example, in the diagram below, Andy

View Solution →