**Difficulty:** Easy, **Asked-In:** Microsoft, Amazon, Adobe, Morgan Stanley, Qualcomm.

**Key takeaway:** An excellent problem to learn problem-solving using fast and slow pointers.

Given a singly linked list, write a program to find the middle node of the linked list. If the number of nodes is even, we need to return the second middle node.

**Example 1:** Input: 5->4->**3**->2->1, Output: 3

Explanation: Here the number of nodes is 5, so there is one middle node which is 3.

**Example 2:** Input: 6->5->4->**3**->2->1, Output: 3

Explanation: The number of nodes is 6, where the first middle node is 4 and the second middle node is 3. So we need to return the pointer to the node 3.

- Brute force approach using extra memory
- By counting nodes: Double traversal
- By counting nodes: Single traversal
- Efficient solution using slow and fast pointers

The basic idea is to store each linked list node, in the same order, into an extra array with a size equal to the length of the linked list. Then, we can get the middle node by accessing the middle index of the array.

```
int listLength(ListNode* head)
{
int length = 0;
while (head != NULL)
{
length = length + 1;
head = head->next;
}
return length;
}
ListNode* findMiddleLinkedList(ListNode* head)
{
int length = listLength(head);
ListNode* temp[length];
int count = 0;
while (head != NULL)
{
temp[count] = head;
count = count + 1;
head = head->next;
}
return temp[count / 2];
}
```

```
def listLength(head):
length = 0
while head:
length = length + 1
head = head.next
return length
def findMiddleLinkedList(head):
length = listLength(head)
temp = [None] * length
count = 0
while head:
temp[count] = head
count = count + 1
head = head.next
return temp[count // 2]
```

Suppose the length of the linked list is n. So the time complexity = Time complexity of finding the length of the linked list + Time complexity of storing nodes in extra memory = O(n) + O(n) = O(n). The space complexity is O(n) for storing nodes in an n-size array.

In the above approach, we are using extra space to find the middle node of the linked list. Now, the critical question is: Can we reduce the space complexity to O(1)? One idea is to first traverse the linked list to count the number of nodes and then traverse the list again count/2 number of times. So, by the end of the second traversal, we will be present at the middle node.

- We initialize the variable
**count**to 0. - We also initialize a list pointer
**middle**to track the middle node, i.e., middle = head. - Now we run a loop until
**head != NULL**and count the number of nodes. - Next, we run another loop until
**i < count/2**and increment middle pointer in each iteration. - By the end of the second loop, we return the middle pointer.

```
ListNode findMiddleLinkedList(ListNode* head)
{
int count = 0;
ListNode middle = *head;
// Count the number of nodes in the linked list
while (head != NULL)
{
count = count + 1;
head = head->next;
}
// Find the middle node
int i = 0;
while (i < count / 2)
{
middle = *(middle.next);
i = i + 1;
}
return middle;
}
```

```
ListNode findMiddleLinkedList(ListNode head)
{
int count = 0;
ListNode middle = head;
while (head != null)
{
count = count + 1;
head = head.next;
}
int i = 0;
while (i < count / 2)
{
middle = middle.next;
i = i + 1;
}
return middle;
}
```

```
def findMiddleLinkedList(head):
count = 0
middle = head
while head:
count = count + 1
head = head.next
i = 0
while i < count//2:
middle = middle.next
i = i + 1
return middle
```

Suppose the length of the linked list is n. So the time complexity = Time complexity of finding the node count + Time complexity of finding the middle node = O(n) + O(n) = O(n). Space complexity = O(1), as we are using constant extra space.

Here is another idea: we can track the count of nodes and the middle node simultaneously in a single traversal.

- We initialize a variable
**count**to track the node count. - Also, we initialize a pointer
**middle**to track the middle node. - Now we run a loop until head->next is not equal to NULL. At each iteration, we increment the count value by 1. When the count value is even, we increment the middle pointer by one.
- By the end of the loop, the middle pointer will be at the middle node.

```
ListNode* findMiddleLinkedList(ListNode* head)
{
int count = 0;
ListNode* middle = head;
while (head->next != NULL)
{
if (count % 2 == 0)
middle = middle->next;
count = count + 1;
head = head->next;
}
return middle;
}
```

```
def findMiddleLinkedList(head):
count = 0
middle = head
while head.next is not None:
if count % 2 == 0:
middle = middle.next
count = count + 1
head = head.next
return middle
```

```
ListNode findMiddle(ListNode head)
{
int count = 0;
ListNode middle = head;
while (head.next != null)
{
if (count % 2 == 0)
middle = middle.next;
count = count + 1;
head = head.next;
}
return middle;
}
```

We are running a single loop and performing constant operations at each iteration. So the time complexity is O(n). The space complexity is O(1), as we are using constant extra space.

Now, the critical question is: Can we come up with another approach to finding the middle node in a single traversal? Let's think! Suppose we use two pointers to traverse the linked list, where one pointer is moving with double the speed of the other pointer. So when the **fast** pointer reaches the end of the linked list, then the **slow** pointer must be present at the middle node. The idea looks straightforward, and we can get the middle node in a single scan of the linked list.

- We initialize
**slow**and**fast**pointers with the head. - Now run a loop until
**fast**or**fast->next**becomes NULL. - At each iteration of the loop, we move the slow pointer by one and the fast pointer by two steps forward.
- By the end of the loop, the slow pointer will point to the middle node, and we return it.

```
ListNode* findMiddleLinkedList(ListNode* head)
{
ListNode* fast = head;
ListNode* slow = head;
while (fast != NULL && fast->next != NULL)
{
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
```

```
def findMiddleLinkedList(head):
fast = head
slow = head
while fast is not None and fast.next is not None:
slow = slow.next
fast = fast.next.next
return slow
```

```
ListNode findMiddleLinkedList(ListNode head)
{
ListNode fast = head;
ListNode slow = head;
while (fast != null && fast.next != null)
{
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
```

If the number of nodes is n, then the fast pointer will move n/2 steps to reach the end. Similarly, the slow pointer will also move n/2 steps to reach the middle node. So we are running a single loop n/2 number of times and doing O(1) operation at each iteration. Time complexity = n/2 * O(1) = O(n). The space complexity is O(1), as we are using constant extra space.

- Using extra memory: Time = O(n), Space = O(n).
- By counting nodes (Double traversal): Time = O(n), Space = O(1).
- By counting nodes (Single traversal): Time = O(n), Space = O(1).
- Using slow and fast pointers: Time = O(n), Space = O(1).

- Can the above approaches work if the linked list has a loop?
- In terms of pointer movement, which approach seems more efficient?
- How do we modify the above code to delete the middle node?
- In the case of an even nodes, how do we modify the above code to return the first middle node?
- In the case of the fast and slow pointers approach, can we optimize it further by using the head as the slow pointer? If yes, then how do we modify the above code?
- Can this problem be solved using other approaches?

- Reversing a linked list
- Remove the Nth node from the list end
- Detect loop in a linked list
- Swap list nodes in pairs
- Merge two sorted list
- Reverse a linked list in groups of a given size
- Intersection points in Y shaped linked lists

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