### Problem Statement :

```Given a singly linked list l0 and another linked list l1, each representing a number with least significant digits first, return the summed linked list.

Note: Each value in the linked list is guaranteed to be between 0 and 9.

Constraints

n ≤ 100,000 where n is the number of nodes in l0
m ≤ 100,000 where m is the number of nodes in l1

Example 1

Input

l0 = [6, 4]
l1 = [4, 7]

Output

[0, 2, 1]

Explanation

This is 46 + 74 = 120```

### Solution :

```                        ```Solution in C++ :

/**
* class LLNode {
*     public:
*         int val;
*         LLNode *next;
* };
*/
LLNode* solve(LLNode* l0, LLNode* l1) {
int carry = 0;
// add the numbers, then send over the carry

while (l0 or l1 or carry) {
res->next = new LLNode();
res = res->next;
int a, b;

if (!l0)
a = 0;
else
a = l0->val, l0 = l0->next;

if (!l1)
b = 0;
else
b = l1->val, l1 = l1->next;

// now just add these two to each other!
int digit = (a + b + carry) % 10;
carry = (a + b + carry) / 10;

res->val = digit;
// cout << " added " << digit << " to the list " << endl;
}

}```
```

```                        ```Solution in Python :

# class LLNode:
#     def __init__(self, val, next=None):
#         self.val = val
#         self.next = next
class Solution:
def solve(self, l0, l1):
def finish_remaining(l, c):
s, head, tail = 0, None, None
while l is not None:  # Need to do extra addition only if there is a leftover carry
entry = l.val + c
s = entry % 10
c = entry // 10
l.val = s
else:
tail.next = l
tail = l
prev, l = l, l.next
if c == 1:  # Try l =[9,9] and c=1 what should happen?
prev.next = LLNode(1)

if l0 is None and l1 is not None:
return l1
elif l1 is None and l0 is not None:
return l0
elif l0 is None and l1 is None:
return None

s, c, prev, head = 0, 0, None, l0
while l0 is not None and l1 is not None:
entry = l0.val + l1.val + c
s, c = entry % 10, entry // 10
l0.val = s
prev, l0, l1 = l0, l0.next, l1.next

prev.next = None
if l0 is None and l1 is not None:
prev.next = finish_remaining(l1, c)
elif l1 is None and l0 is not None:
prev.next = finish_remaining(l0, c)
elif c == 1:  # l1 and l0 are both None But there is a leftover carry. Try for Example 1
prev.next = LLNode(1)
```

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