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Last Updated: 20 Oct, 2020

Moderate

```
If the first linked list is 1 -> 2 -> 3 -> 4 -> 5 -> NULL and the second linked list is 4 -> 5 -> NULL.
The two numbers represented by these two lists are 12345 and 45, respectively. So, adding these two numbers gives 12390.
So, the linked list representation of this number is 1 -> 2 -> 3 -> 9 -> 0 -> NULL.
```

```
The first line of input contains an integer 'T' representing the number of test cases.
The first line of each test case contains the elements of the first linked list separated by a single space and terminated by -1. Hence, -1 would never be a list element.
The second line of each test case contains the elements of the second linked list separated by a single space and terminated by -1. Hence, -1 would never be a list element.
```

```
For each test case, return the head of linked list after summation. The elements of the linked list must be terminated by -1.
```

```
You don't need to print anything, it has already been taken care of. Just implement the given function.
```

```
Try to solve this problem in linear time complexity and constant space complexity.
```

```
1 <= T <= 100
1 <= L <= 5000
0 <= data <= 9 and data != -1
Where 'L' is the number of nodes in either of the two Linked List and 'data' is the element value in a node of the linked list.
Time limit: 1 sec
```

We will use the naive method to add two linked lists. While adding two numbers, we always add digits in the right to left manner. So first, we will reverse both linked lists so that we can easily process the number in this manner.

We will be storing the result of the addition in a different linked list. We will start adding the nodes of both linked lists and use another variable ‘carry’ to keep track of the carry generated in each addition.

In the end, if we are left with a ‘carry’ greater than 0, we will just add a new node in our answer linked list.

We will use the same naive method to add two linked lists. Like the previous approach, we will reverse both linked lists so that we can easily process the number in this manner.

Now, we will start adding the nodes of both linked lists and use another variable ‘carry’ to keep track of the carry generated in each addition. We will be updating either of the linked lists to their sum. Hence, we will not be using any extra space for storing the result.

In the end, if we are left with a ‘carry’ greater than 0, we will just add a new node in our resultant linked list.