Given a matrix. Convert it into a linked list matrix such that each node is connected to its next right and down node.
Input : 2D matrix
1 2 3
4 5 6
7 8 9
Output :
1 -> 2 -> 3 -> NULL
| | |
v v v
4 -> 5 -> 6 -> NULL
| | |
v v v
7 -> 8 -> 9 -> NULL
| | |
v v v
NULL NULL NULL
Question Source: Factset Interview Experience | Set 9
Approach: The problem can be solved based on the following idea:
Connect every cell to its right cell of the same row and to its bottom cell in the same column and also for each cell and keep track of those created node.
Follow the steps mentioned below to solve this problem:
- First create a variable of Node type, which will store address of its right and bottom Node corresponding to cell in the matrix.
- Recursively do the following steps for any cell in the matrix:
- If Node is not created for any corresponding cell in the matrix, then create a new Node and store it.
- Else we reach at some cell which has already been created for its corresponding cell in the matrix then return that stored Node.
- Attach Node to its right and bottom cell which is created and return the current Node.
- Finally return the root Node.
Below is the implementation of the above approach:
// CPP program to construct a linked list // from given 2D matrix #include <bits/stdc++.h> using namespace std;
// struct node of linked list struct Node {
int data;
Node *right, *down;
}; // returns head pointer of linked list // constructed from 2D matrix Node* construct( int arr[][4], int i, int j, int m, int n,
vector<vector<Node*> >& visited)
{ // return if i or j is out of bounds
if (i > m - 1 || j > n - 1)
return NULL;
// Check if node is previously created then,
// don't need to create new/
if (visited[i][j]) {
return visited[i][j];
}
// create a new node for current i and j
// and recursively allocate its down and
// right pointers
Node* temp = new Node();
visited[i][j] = temp;
temp->data = arr[i][j];
temp->right = construct(arr, i, j + 1, m, n, visited);
temp->down = construct(arr, i + 1, j, m, n, visited);
return temp;
} // utility function for displaying // linked list data void display(Node* head)
{ // pointer to move right
Node* Rp;
// pointer to move down
Node* Dp = head;
// loop till node->down is not NULL
while (Dp) {
Rp = Dp;
// loop till node->right is not NULL
while (Rp) {
cout << Rp->data << " " ;
Rp = Rp->right;
}
cout << "\n" ;
Dp = Dp->down;
}
} // driver program int main()
{ // 2D matrix
int arr[][4] = { { 1, 2, 3, 0 },
{ 4, 5, 6, 1 },
{ 7, 8, 9, 2 },
{ 7, 8, 9, 2 } };
int m = 4, n = 4;
vector<vector<Node*> > visited(m, vector<Node*>(n));
Node* head = construct(arr, 0, 0, m, n, visited);
display(head);
return 0;
} |
// Java program to construct a linked list // from given 2D matrix public class Linked_list_2D_Matrix {
// node of linked list
static class Node {
int data;
Node right;
Node down;
};
// returns head pointer of linked list
// constructed from 2D matrix
static Node construct( int arr[][], int i, int j, int m,
int n, Node[][] visited)
{
// return if i or j is out of bounds
if (i > m - 1 || j > n - 1 )
return null ;
// Check if node is previously created then,
// don't need to create new/
if (visited[i][j] != null ) {
return visited[i][j];
}
// create a new node for current i and j
// and recursively allocate its down and
// right pointers
Node temp = new Node();
visited[i][j] = temp;
temp.data = arr[i][j];
temp.right
= construct(arr, i, j + 1 , m, n, visited);
temp.down = construct(arr, i + 1 , j, m, n, visited);
return temp;
}
// utility function for displaying
// linked list data
static void display(Node head)
{
// pointer to move right
Node Rp;
// pointer to move down
Node Dp = head;
// loop till node->down is not NULL
while (Dp != null ) {
Rp = Dp;
// loop till node->right is not NULL
while (Rp != null ) {
System.out.print(Rp.data + " " );
Rp = Rp.right;
}
System.out.println();
Dp = Dp.down;
}
}
// driver program
public static void main(String args[])
{
// 2D matrix
int arr[][] = { { 1 , 2 , 3 , 0 },
{ 4 , 5 , 6 , 1 },
{ 7 , 8 , 9 , 2 },
{ 7 , 8 , 9 , 2 } };
int m = 4 , n = 4 ;
// List<List<Node>> arr = new
// ArrayList<List<Node>>();
Node[][] visited = new Node[m][n];
Node head = construct(arr, 0 , 0 , m, n, visited);
display(head);
}
} // This code is contributed by Sumit Ghosh |
# Python program to construct a linked list # from given 2D matrix # Node class to represent a node in the linked list class Node:
def __init__( self , data = None , right = None , down = None ):
self .data = data
self .right = right
self .down = down
# Function to construct the linked list from the given 2D matrix def construct(arr, i, j, m, n, visited):
# Return if i or j is out of bounds
if i > m - 1 or j > n - 1 :
return None
# Check if node is previously created, then don't create new
if visited[i][j]:
return visited[i][j]
# Create a new node for current i and j and
# recursively allocate its down and right pointers
node = Node(arr[i][j])
visited[i][j] = node
node.right = construct(arr, i, j + 1 , m, n, visited)
node.down = construct(arr, i + 1 , j, m, n, visited)
return node
# Utility function for displaying linked list data def display(head):
# Pointer to move right
Rp = head
# Pointer to move down
Dp = head
# Loop till node.down is not None
while Dp:
Rp = Dp
# Loop till node.right is not None
while Rp:
print (Rp.data, end = " " )
Rp = Rp.right
print ()
Dp = Dp.down
# Driver program if __name__ = = "__main__" :
# 2D matrix
arr = [[ 1 , 2 , 3 , 0 ],
[ 4 , 5 , 6 , 1 ],
[ 7 , 8 , 9 , 2 ],
[ 7 , 8 , 9 , 2 ]]
m = 4
n = 4
visited = [[ None ] * n for _ in range (m)]
head = construct(arr, 0 , 0 , m, n, visited)
display(head)
|
// C# program to construct a linked list // from given 2D matrix using System;
using System.Collections.Generic;
// node of linked list class Node {
public int data;
public Node right;
public Node down;
} class Linked_list_2D_Matrix
{ // returns head pointer of linked list
// constructed from 2D matrix
static Node construct( int [, ] arr, int i, int j, int m,
int n, Node[, ] visited)
{
// return if i or j is out of bounds
if (i > m - 1 || j > n - 1)
return null ;
// Check if node is previously created then,
// don't need to create new/
if (visited[i, j] != null ) {
return visited[i, j];
}
// create a new node for current i and j
// and recursively allocate its down and
// right pointers
Node temp = new Node();
visited[i, j] = temp;
temp.data = arr[i, j];
temp.right
= construct(arr, i, j + 1, m, n, visited);
temp.down = construct(arr, i + 1, j, m, n, visited);
return temp;
}
// utility function for displaying
// linked list data
static void display(Node head)
{
// pointer to move right
Node Rp;
// pointer to move down
Node Dp = head;
// loop till node->down is not NULL
while (Dp != null ) {
Rp = Dp;
// loop till node->right is not NULL
while (Rp != null ) {
Console.Write(Rp.data + " " );
Rp = Rp.right;
}
Console.WriteLine();
Dp = Dp.down;
}
}
// driver program
public static void Main( string [] args)
{
// 2D matrix
int [, ] arr = { { 1, 2, 3, 0 },
{ 4, 5, 6, 1 },
{ 7, 8, 9, 2 },
{ 7, 8, 9, 2 } };
int m = 4, n = 4;
// List<List<Node>> arr = new
// ArrayList<List<Node>>();
Node[, ] visited = new Node[m, n];
Node head = construct(arr, 0, 0, m, n, visited);
display(head);
}
} // This code is contributed by phasing17 |
// Javascript program to construct a linked list // from given 2D matrix
//node of linked list
class Node {
constructor() {
this .data;
this .right;
this .down;
}
}
// returns head pointer of linked list
// constructed from 2D matrix
function construct(arr, i, j, m, n, visited) {
// return if i or j is out of bounds
if (i > m - 1 || j > n - 1) return null ;
// Check if node is previously created then,
// don't need to create new/
if (visited[i][j]) {
return visited[i][j];
}
// create a new node for current i and j
// and recursively allocate its down and
// right pointers
let temp = new Node();
visited[i][j] = temp;
temp.data = arr[i][j];
temp.right = construct(arr, i, j + 1, m, n, visited);
temp.down = construct(arr, i + 1, j, m, n, visited);
return temp;
}
// utility function for displaying
// linked list data
function display(head) {
// pointer to move right
let Rp;
// pointer to move down
let Dp = head;
// loop till node.down is not NULL
while (Dp) {
Rp = Dp;
// loop till node.right is not NULL
while (Rp) {
console.log(Rp.data + " " );
Rp = Rp.right;
}
Dp = Dp.down;
}
}
// driver program
// 2D matrix
var arr = [
[1, 2, 3, 0],
[4, 5, 6, 1],
[7, 8, 9, 2],
[7, 8, 9, 2],
];
let m = 4,
n = 4;
let visited = Array.from(Array(m), () => new Array(n));
var head = construct(arr, 0, 0, m, n, visited);
display(head);
|
1 2 3 0 4 5 6 1 7 8 9 2 7 8 9 2
Time complexity: O(N*M), where N is the number of row and M is the number of column.
Auxiliary space: O(N*M)
Use a pre-allocated array of nodes instead of allocating new nodes:
This will reduce the overhead of dynamic memory allocation and deallocation, which can be significant in some cases. The nodeCounter variable keeps track of the next available node in the array. A sentinel node is used to simplify the linked list construction logic, and eliminates the need for special cases when creating the first node.
Here is an example implementation that incorporates these optimizations:
#include <bits/stdc++.h> using namespace std;
const int MAXN = 100;
struct Node
{ int data;
Node *next;
}; Node nodes[MAXN]; int nodeCounter = 0;
Node *constructLinkedList( int matrix[][4], int rows, int columns)
{ Node *sentinel = &nodes[nodeCounter++];
sentinel->next = NULL;
Node *current = sentinel;
for ( int i = 0; i < rows; i++)
{
for ( int j = 0; j < columns; j++)
{
Node *newNode = &nodes[nodeCounter++];
newNode->data = matrix[i][j];
newNode->next = NULL;
current->next = newNode;
current = current->next;
}
}
return sentinel->next;
} void printLinkedList(Node *head)
{ while (head != NULL)
{
cout << head->data << " " ;
head = head->next;
}
cout << endl;
} int main()
{ int matrix[3][4] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12}};
Node *head = constructLinkedList(matrix, 3, 4);
printLinkedList(head);
return 0;
} |
// Java code to implement the approach import java.util.*;
// Node class definition class Node {
public int data;
public Node next;
} // LinkedListFromMatrix class definition class LinkedListFromMatrix {
// This method constructs a linked list from a 2D array
static Node constructLinkedList( int [][] matrix,
int rows, int columns)
{
Node sentinel = new Node();
sentinel.next = null ;
Node current = sentinel;
// Iterating over the rows
for ( int i = 0 ; i < rows; i++) {
// Iterating over the columns
for ( int j = 0 ; j < columns; j++) {
// Adding node to linked list
Node newNode = new Node();
newNode.data = matrix[i][j];
newNode.next = null ;
current.next = newNode;
current = current.next;
}
}
return sentinel.next;
}
// This function displays the linked list
static void printLinkedList(Node head)
{
while (head != null ) {
System.out.print(head.data + " " );
head = head.next;
}
System.out.println( " " );
}
// Driver code
public static void main(String[] args)
{
int [][] matrix = new int [][] { { 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 },
{ 9 , 10 , 11 , 12 } };
// Function calls
Node head = constructLinkedList(matrix, 3 , 4 );
printLinkedList(head);
}
} |
// C# code to implement the approach using System;
// Node class definition class Node
{ public int data;
public Node next;
} class LinkedListFromMatrix
{ // This function constructs a linked list from a 2D array
static Node ConstructLinkedList( int [,] matrix, int rows, int columns)
{
Node sentinel = new Node();
sentinel.next = null ;
Node current = sentinel;
// Iterating over the rows
for ( int i = 0; i < rows; i++)
{
// Iterating over the columns
for ( int j = 0; j < columns; j++)
{
// Adding node to linked list
Node newNode = new Node();
newNode.data = matrix[i, j];
newNode.next = null ;
current.next = newNode;
current = current.next;
}
}
return sentinel.next;
}
// This function prints the linked list
static void PrintLinkedList(Node head)
{
while (head != null )
{
Console.Write(head.data + " " );
head = head.next;
}
Console.WriteLine( " " );
}
// Driver code
static void Main( string [] args)
{
int [,] matrix = new int [3, 4] {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12}};
// Function calls
Node head = ConstructLinkedList(matrix, 3, 4);
PrintLinkedList(head);
}
} |
# GFG # Python program to construct a linked list # from given 2D matrix # Pre-allocated array of nodes instead of allocating new nodes class Node:
def __init__( self , data = None ):
self .data = data
self . next = None
def construct_linked_list(matrix):
sentinel = Node()
current = sentinel
for row in matrix:
for element in row:
new_node = Node(element)
current. next = new_node
current = current. next
return sentinel. next
def print_linked_list(head):
while head is not None :
print (head.data, end = " " )
head = head. next
print ()
if __name__ = = "__main__" :
matrix = [[ 1 , 2 , 3 , 4 ],
[ 5 , 6 , 7 , 8 ],
[ 9 , 10 , 11 , 12 ]]
head = construct_linked_list(matrix)
print_linked_list(head)
# This code is written by Sundaram |
// Javascript program to construct a linked list // from given 2D matrix // Pre-allocated array of nodes instead of allocating new nodes class Node { constructor(data) {
this .data = data;
this .next = null ;
}
} function constructLinkedList(matrix) {
const sentinel = new Node();
let current = sentinel;
for (const row of matrix) {
for (const element of row) {
const newNode = new Node(element);
current.next = newNode;
current = current.next;
}
}
return sentinel.next;
} function printLinkedList(head) {
let currentNode = head;
let temp = []
while (currentNode !== null ) {
temp.push(currentNode.data);
currentNode = currentNode.next;
}
console.log(temp.join( ' ' ));
} const matrix = [ [1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]
]; const head = constructLinkedList(matrix); printLinkedList(head); // Contributed by adityasharmadev01 |
1 2 3 4 5 6 7 8 9 10 11 12
Time complexity: O(N*M)
Auxiliary space: O(N*M)