# Find Merge Point of Two Lists Hackerrank problem solution

##### Find Merge Point of Two Lists Hackerrank

Given pointers to the head nodes of 2 linked lists that merge together at some point, find the Node where the two lists merge. It is guaranteed that the two head Nodes will be different, and neither will be NULL.

In the diagram below, the two lists converge at Node x:

[List #1] a—>b—>c
\
x—>y—>z—>NULL
/
[List #2] p—>q
Complete the int FindMergeNode(Node* headA, Node* headB) method so that it finds and returns the data value of the Node where the two lists merge.

Input Format

The FindMergeNode(Node*,Node*) method has two parameters, headA and headB, which are the non-null head Nodes of two separate linked lists that are guaranteed to converge.
Do not read any input from stdin/console.

Output Format

Each Node has a data field containing an integer; return the integer data for the Node where the two lists converge.
Do not write any output to stdout/console.

Sample Input

The diagrams below are graphical representations of the lists that input Nodes headA and headB are connected to. Recall that this is a method-only challenge; the method only has initial visibility to those 2 Nodes and must explore the rest of the Nodes using some algorithm of your own design.

Test Case 0

1
\
2—>3—>NULL
/
1
Test Case 1

1—>2
\
3—>Null
/
1
Sample Output

2
3
Explanation

Test Case 0: As demonstrated in the diagram above, the merge Node’s data field contains the integer 2 (so our method should return 2).
Test Case 1: As demonstrated in the diagram above, the merge Node’s data field contains the integer 3 (so our method should return 3).

##### Solution

To calculate the merge point, first calculate the difference in the sizes of the linked lists. Move the pointer of the smaller linked list by this difference. Increment both pointers till you reach the merge point.

```int getCount(Node* head)
{
int count = 0;

while (current != NULL)
{
count++;
current = current->next;
}

return count;
}

{
int i;

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;
}

{
// Complete this function
// Do not write the main method.
int d;

if(c1 > c2)
{
d = c1 - c2;
}
else
{
d = c2 - c1;
}
}```
``````/*
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;
}
*/

//Do till the two nodes are the same
while(currentA != currentB){
//If you reached the end of one list start at the beginning of the other one
//currentA
if(currentA.next == null){
}else{
currentA = currentA.next;
}
//currentB
if(currentB.next == null){
}else{
currentB = currentB.next;
}
}
return currentB.data;
}
``````
``````"""
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 next(ptr):
if ptr is None:
return None
else:
return ptr.next

def ahit(d, key):
if key is None:
return 0
elif d.has_key(key):
return 1
else:
d[key] = 1
return 0

result = None
d = {}
while True:
break
break
break
else:

return result``````
```/*
Find merge point of two linked lists
Node is defined as
struct Node
{
int data;
Node* next;
}
*/
{
int count = 0;

while (current != NULL)
{
count++;
current = current->next;
}

return count;
}

{
int i;

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;
}

{
// Complete this function
// Do not write the main method.
int d;

if(c1 > c2)
{
d = c1 - c2;