# Javascript Program For Partitioning A Linked List Around A Given Value And Keeping The Original Order

Last Updated : 22 Jun, 2022

Given a linked list and a value x, partition it such that all nodes less than x come first, then all nodes with a value equal to x, and finally nodes with a value greater than or equal to x. The original relative order of the nodes in each of the three partitions should be preserved. The partition must work in place.
Examples:

```Input: 1->4->3->2->5->2->3,
x = 3
Output: 1->2->2->3->3->4->5

Input: 1->4->2->10
x = 3
Output: 1->2->4->10

Input: 10->4->20->10->3
x = 3
Output: 3->10->4->20->10 ```

To solve this problem we can use partition method of Quick Sort but this would not preserve the original relative order of the nodes in each of the two partitions.
Below is the algorithm to solve this problem :

1. Initialize first and last nodes of below three linked lists as NULL.
• Linked list of values smaller than x.
• Linked list of values equal to x.
• Linked list of values greater than x.
2. Now iterate through the original linked list. If a node’s value is less than x then append it at the end of the smaller list. If the value is equal to x, then at the end of the equal list. And if a value is greater, then at the end of the greater list.
3. Now concatenate three lists.

Below is the implementation of the above idea.

## Javascript

 ``

Output:

`2 10 4 5 30 50`

Time Complexity: O(n) where n is the size of the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Please refer complete article on Partitioning a linked list around a given value and keeping the original order for more details!

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