# Partition a Linked List into 3 parts such that the maximum difference between their sizes is minimum

Given a singly linked list, the task is to split the given linked list into exactly three parts such that the maximum difference between the length of the split linked lists is minimum.

**Examples:**

Input:1->2->3->4->5Output:

1->2

3->4

5Explanation:

Consider the splitting of the linked list as:

1->2:The size is 1.3->4:The size is 1.5:The size is 1.The maximum difference between the length of any two splitted linked lists is 1, which is minimum.

Input:7 -> 2 -> 1Output:

7

2

1

**Approach:** Follow the steps below to solve the given problem:

- Initialize a vector, say
**ans[]**that stores the split linked list - If the
**size**of the given linked list is less than**3**, then create**size**time**3 – size**linked list with**null**nodes and add it to the**ans**vector and**return**. - Initialize a variable, say
**minSize**as**size / 3**that will be the minimum size of the linked list to be divided and**rem**as**size % 3**. - Iterate over the linked list until size becomes
**0**and perform the following steps:- In each iteration, if
**rem**equals**0**, then iterate again**minSize**times into the linked list and add that linked list to the**ans**and decrement**rem**by**1.** - Otherwise, iterate
**(minSize + 1)**a number of times into the linked list and add that linked list to the**ans**.

- In each iteration, if
- After completing the above steps, print all the Linked List stored in the vector
**ans[]**.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Structure of a Node` `class` `Node {` `public` `:` ` ` `int` `data;` ` ` `Node* next;` `};` `// Function to find the length of` `// the Linked List` `int` `sizeOfLL(Node* head)` `{` ` ` `int` `size = 0;` ` ` `// While head is not null` ` ` `while` `(head != NULL) {` ` ` `++size;` ` ` `head = head->next;` ` ` `}` ` ` `return` `size;` `}` `// Function to partition list into` `// 3 parts such that maximum size` `// difference between parts is minimum` `vector<Node*> Partition_of_list(Node* head)` `{` ` ` `int` `size = sizeOfLL(head);` ` ` `Node* temp = head;` ` ` `vector<Node*> ans;` ` ` `// If size is less than 3` ` ` `if` `(3 >= size) {` ` ` `// Partition linked list` ` ` `// into one node each` ` ` `while` `(temp != NULL) {` ` ` `Node* next = temp->next;` ` ` `temp->next = NULL;` ` ` `ans.push_back(temp);` ` ` `temp = next;` ` ` `}` ` ` `// The remaining parts (3-size)` ` ` `// will be filled by empty` ` ` `// the linkded list` ` ` `int` `y = 3 - size;` ` ` `while` `(y != 0) {` ` ` `ans.push_back(NULL);` ` ` `y--;` ` ` `}` ` ` `}` ` ` `else` `{` ` ` `// Minimum size` ` ` `int` `minSize = size / 3;` ` ` `int` `rem = size % 3;` ` ` `// While size is positive` ` ` `// and temp is not null` ` ` `while` `(size > 0 && temp != NULL) {` ` ` `int` `m = 0;` ` ` `// If remainder > 0, then` ` ` `// partition list on the` ` ` `// basis of minSize + 1` ` ` `if` `(rem != 0) {` ` ` `m = minSize + 1;` ` ` `rem--;` ` ` `}` ` ` `// Otherwise, partition` ` ` `// on the basis of minSize` ` ` `else` `{` ` ` `m = minSize;` ` ` `}` ` ` `Node* curr = temp;` ` ` `// Iterate for m-1 steps` ` ` `// in the list` ` ` `for` `(` `int` `j = 1; j < m` ` ` `&& temp->next != NULL;` ` ` `j++) {` ` ` `temp = temp->next;` ` ` `}` ` ` `// Change the next of the` ` ` `// current node to NULL` ` ` `// and add it to the ans` ` ` `if` `(temp->next != NULL) {` ` ` `Node* x = temp->next;` ` ` `temp->next = NULL;` ` ` `temp = x;` ` ` `ans.push_back(curr);` ` ` `}` ` ` `// Otherwise` ` ` `else` `{` ` ` `// Pushing to ans` ` ` `ans.push_back(curr);` ` ` `break` `;` ` ` `}` ` ` `size -= m;` ` ` `}` ` ` `}` ` ` `// Return the resultant lists` ` ` `return` `ans;` `}` `// Function to insert elements in list` `void` `push(Node** head, ` `int` `d)` `{` ` ` `Node* temp = ` `new` `Node();` ` ` `temp->data = d;` ` ` `temp->next = NULL;` ` ` `// If the head is NULL` ` ` `if` `((*head) == NULL)` ` ` `(*head) = temp;` ` ` `// Otherwise` ` ` `else` `{` ` ` `Node* curr = (*head);` ` ` `// While curr->next is not NULL` ` ` `while` `(curr->next != NULL) {` ` ` `curr = curr->next;` ` ` `}` ` ` `curr->next = temp;` ` ` `}` `}` `// Function to display the Linked list` `void` `display(Node* head)` `{` ` ` `// While head is not null` ` ` `while` `(head->next != NULL) {` ` ` `// Print the data` ` ` `cout << head->data << ` `"->"` `;` ` ` `head = head->next;` ` ` `}` ` ` `cout << head->data << ` `"\n"` `;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given Input` ` ` `Node* head = NULL;` ` ` `push(&head, 1);` ` ` `push(&head, 2);` ` ` `push(&head, 3);` ` ` `push(&head, 4);` ` ` `push(&head, 5);` ` ` `// Function Call` ` ` `vector<Node*> v = Partition_of_list(head);` ` ` `for` `(` `int` `i = 0; i < v.size(); i++) {` ` ` `display(v[i]);` ` ` `}` ` ` `return` `0;` `}` |

**Output:**

1->2 3->4 5

**Time Complexity:** O(N)**Auxiliary Space:** O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.