Form a Rectangle from boundary elements of Matrix using Linked List
Last Updated :
22 Aug, 2022
Given a Matrix grid[][] of size NxM where N is number of rows and M is number of columns. The task is to form a rectangle from boundary elements of grid[][] using linked list having four pointers namely prev, next, top and bottom. Print the final linked list.
Examples:
Input: A = [[13, 42, 93, 88],
[26, 38, 66, 42],
[75, 63, 78, 12]]
Output: 13 42 93 88 42 12 78 63 75 26
Explanation:
1. Make A[0][0] and head node
2. Traverse through 0th row and create node for each element and connect them through next pointer.
3. Traverse through (m-1)th column and create node for each element and connect them through bottom pointer.
4. Traverse through (n-1)th row and create node for each element and connect them through prev pointer.
5. Traverse through 0th column and create node for each element and connect them through top pointer.
6. Step 2, 3, 4, 5 is repeated till temp. The top become equal to head.
Input: A = [[1, 2, 3]
[8, 9, 4]
[7, 6, 5]]
Output: 1 2 3 4 5 6 7 8
Approach: This problem can be solved by Performing boundary traversal of matrix and creating nodes for each element and link them using next, prev, bottom or top and create a linked list.
Follow the steps below:
Step 1: Make grid[0][0] as the head of the Linked list and initialize temp as the head.
Step 2: Traverse through the first row from j=1 to j=m-1 where i=0 and create a node for each element and link them through the next pointer.
Step 3: Traverse through the last column from i=0 to i=n-1 where j=m-1 and create a node for each element and link them through a bottom pointer.
Step 4: Traverse through the last row from j=m-1 to j=0 where i=n-1 and create a node for each element and link them through the prev pointer.
Step 5: Traverse through the first column from i=n-1 to i=0 where j=0 and create a node for each element and link them through the top pointer.
Step 6: Step 2, 3, 4, 5 is repeated till temp.top becomes equal to head.
Step 7: Print the required Linked List.
Below is the implementation of the above algorithm.
C++
// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
// Node Class
struct Node {
int data;
Node* next;
Node* prev;
Node* top;
Node* bottom;
// Constructor to initialize the node object
Node(int data)
{
this->data = data;
next = NULL;
prev = NULL;
top = NULL;
bottom = NULL;
}
};
// Linked List class
struct LinkedList {
// initialize head
Node* head = NULL;
// function to form square
// linked list of matrix.
void Quad(vector<vector<int> > grid, int n, int m)
{
// initialising A[0][0] as head.
head = new Node(grid[0][0]);
// head is assigned to head.
Node* temp = head;
// i is row index, j is column index
int i = 0;
int j = 1;
// loop till temp.top become equal to head.
while (temp != NULL and temp->top != head) {
// as we iterating over boundary
// of matrix so we will iterate
// over first(0) and last(n-1) row
// and first(0) and last(m-1) column.
// iterating over first i.e 0th row
// and connecting node.
if (j < m && i == 0) {
temp->next = new Node(grid[i][j]);
temp = temp->next;
j += 1;
}
// iterating over last i.e (m-1)th
// column and connecting Node.
else if (j == m && i < n - 1) {
i = i + 1;
temp->bottom = new Node(grid[i][j - 1]);
temp = temp->bottom;
}
// iterating over last i.e (n-1)th row
// and connecting Node.
else if (i == n - 1 && j <= m && j >= 1) {
if (j == m)
j = j - 1;
j = j - 1;
temp->prev = new Node(grid[i][j]);
temp = temp->prev;
}
// iterating over first i.e 0th column
// and connecting Node.
else if (i <= n - 1 && j == 0) {
i = i - 1;
temp->top = new Node(grid[i][j]);
temp = temp->top;
if (i == 1)
temp->top = head;
}
}
}
// function to print Linked list.
void printList(Node* root)
{
Node* temp = root;
// printing head of linked list
cout << temp->data << " ";
// loop till temp.top
// become equal to head
while (temp->top != root) {
// printing the node
if (temp->next) {
cout << temp->next->data << " ";
temp = temp->next;
}
if (temp->prev) {
cout << temp->prev->data << " ";
temp = temp->prev;
}
if (temp->bottom) {
cout << temp->bottom->data << " ";
temp = temp->bottom;
}
if (temp->top) {
cout << temp->top->data << " ";
temp = temp->top;
}
}
}
};
// Driver Code
int main()
{
vector<vector<int> > grid = { { 13, 42, 93, 88 },
{ 26, 38, 66, 42 },
{ 75, 63, 78, 12 } };
// n is number of rows
int n = grid.size();
// m is number of columns
int m = grid[0].size();
// creation of object
LinkedList* l = new LinkedList();
// Call Quad method to create Linked List.
l->Quad(grid, n, m);
// Call Quad method to create Linked List.
l->printList(l->head);
return 0;
}
// The code is contributed by Gautam goel (gautamgoel962)
// Java program for above approach
// Node Class
class Node {
int data;
Node next;
Node prev;
Node top;
Node bottom;
// Constructor to initialize the node object
Node(int data)
{
this.data = data;
next = null;
prev = null;
top = null;
bottom = null;
}
}
class GFG {
// initialize head
public Node head = null;
// function to form square linked list of matrix.
public void Quad(int[][] grid, int n, int m)
{
// initialising A[0][0] as head.
head = new Node(grid[0][0]);
// head is assigned to head.
Node temp = head;
// i is row index, j is column index
int i = 0;
int j = 1;
// loop till temp.top become equal to head.
while (temp.top != head) {
// as we iterating over boundary
// of matrix so we will iterate
// over first(0) and last(n-1) row
// and first(0) and last(m-1) column.
// iterating over first i.e 0th row
// and connecting node.
if (j < m && i == 0) {
temp.next = new Node(grid[i][j]);
temp = temp.next;
j += 1;
}
// iterating over last i.e (m-1)th
// column and connecting Node.
else if (j == m && i < n - 1) {
i = i + 1;
temp.bottom = new Node(grid[i][j - 1]);
temp = temp.bottom;
}
// iterating over last i.e (n-1)th row
// and connecting Node.
else if (i == n - 1 && j <= m && j >= 1) {
if (j == m) {
j = j - 1;
}
j = j - 1;
temp.prev = new Node(grid[i][j]);
temp = temp.prev;
}
// iterating over first i.e 0th column
// and connecting Node.
else if (i <= n - 1 && j == 0) {
i = i - 1;
temp.top = new Node(grid[i][j]);
temp = temp.top;
if (i == 1) {
temp.top = head;
}
}
}
}
// function to print Linked list.
public void printList(Node root)
{
Node temp = root;
// printing head of linked list
System.out.print(temp.data + " ");
// loop till temp.top
// become equal to head
while (temp.top != root) {
// printing the node
if (temp.next != null) {
System.out.print(temp.next.data + " ");
temp = temp.next;
}
if (temp.prev != null) {
System.out.print(temp.prev.data + " ");
temp = temp.prev;
}
if (temp.bottom != null) {
System.out.print(temp.bottom.data + " ");
temp = temp.bottom;
}
if (temp.top != null) {
System.out.print(temp.top.data + " ");
temp = temp.top;
}
}
}
public static void main(String[] args)
{
int[][] grid = new int[][] { { 13, 42, 93, 88 },
{ 26, 38, 66, 42 },
{ 75, 63, 78, 12 } };
// n is number of rows
int n = grid.length;
// m is number of columns
int m = grid[0].length;
GFG l = new GFG();
// Call Quad method to create Linked List.
l.Quad(grid, n, m);
// Call Quad method to create Linked List.
l.printList(l.head);
}
}
// This code is contributed by lokesh (lokeshmvs21).
# Python program for above approach
# Node Class
class Node:
# Constructor to initialize the node object
def __init__(self, val):
self.data = val
self.next = None
self.prev = None
self.top = None
self.bottom = None
# Linked List class
class LinkedList:
# Constructor to initialize head
def __init__(self):
self.head = None
# function to form square
# linked list of matrix.
def Quad(self, grid, n, m):
# initialising A[0][0] as head.
self.head = Node(grid[0][0])
# head is assigned to head.
temp = self.head
# i is row index, j is column index
i = 0
j = 1
# loop till temp.top become equal to head.
while temp.top != self.head:
# as we iterating over boundary
# of matrix so we will iterate
# over first(0) and last(n-1) row
# and first(0) and last(m-1) column.
# iterating over first i.e 0th row
# and connecting node.
if j < m and i == 0:
temp.next = Node(grid[i][j])
temp = temp.next
j += 1
# iterating over last i.e (m-1)th
# column and connecting Node.
elif j == m and i < n - 1:
i = i + 1
temp.bottom = Node(grid[i][j - 1])
temp = temp.bottom
# iterating over last i.e (n-1)th row
# and connecting Node.
elif i == n - 1 and j <= m and j >= 1:
if j == m: j = j - 1
j = j - 1
temp.prev = Node(grid[i][j])
temp = temp.prev
# iterating over first i.e 0th column
# and connecting Node.
elif i <= n - 1 and j == 0:
i = i - 1
temp.top = Node(grid[i][j])
temp = temp.top
if i == 1:
temp.top = self.head
# function to print Linked list.
def printList(self, root):
temp = root
# printing head of linked list
print(temp.data, end =" ")
# loop till temp.top
# become equal to head
while temp.top != root:
# printing the node
if temp.next:
print(temp.next.data, end =" ")
temp = temp.next
if temp.prev:
print(temp.prev.data, end =" ")
temp = temp.prev
if temp.bottom:
print(temp.bottom.data, end =" ")
temp = temp.bottom
if temp.top:
print(temp.top.data, end =" ")
temp = temp.top
# Driver Code
grid = [[13, 42, 93, 88],
[26, 38, 66, 42],
[75, 63, 78, 12]]
# n is number of rows
n = len(grid)
# m is number of column
m = len(grid[0])
# creation of object
l = LinkedList()
# Call Quad method to create Linked List.
l.Quad(grid, n, m)
# Call printList method to print list.
l.printList(l.head)
// C# program for above approach
using System;
// Node Class
class Node {
public int data;
public Node next;
public Node prev;
public Node top;
public Node bottom;
// Constructor to initialize the node object
public Node(int data)
{
this.data = data;
this.next = null;
this.prev = null;
this.top = null;
this.bottom = null;
}
}
public class GFG {
// initialize head
Node head = null;
// function to form square linked list of matrix.
public void Quad(int[, ] grid, int n, int m)
{
// initialising A[0][0] as head.
head = new Node(grid[0, 0]);
// head is assigned to head.
Node temp = head;
// i is row index, j is column index
int i = 0;
int j = 1;
// loop till temp.top become equal to head.
while (temp.top != head)
{
// as we iterating over boundary
// of matrix so we will iterate
// over first(0) and last(n-1) row
// and first(0) and last(m-1) column.
// iterating over first i.e 0th row
// and connecting node.
if (j < m && i == 0) {
temp.next = new Node(grid[i, j]);
temp = temp.next;
j += 1;
}
// iterating over last i.e (m-1)th
// column and connecting Node.
else if (j == m && i < n - 1) {
i = i + 1;
temp.bottom = new Node(grid[i, j - 1]);
temp = temp.bottom;
}
// iterating over last i.e (n-1)th row
// and connecting Node.
else if (i == n - 1 && j <= m && j >= 1) {
if (j == m) {
j = j - 1;
}
j = j - 1;
temp.prev = new Node(grid[i, j]);
temp = temp.prev;
}
// iterating over first i.e 0th column
// and connecting Node.
else if (i <= n - 1 && j == 0) {
i = i - 1;
temp.top = new Node(grid[i, j]);
temp = temp.top;
if (i == 1) {
temp.top = head;
}
}
}
}
// function to print Linked list.
void printList(Node root)
{
Node temp = root;
// printing head of linked list
Console.Write(temp.data + " ");
// loop till temp.top
// become equal to head
while (temp.top != root) {
// printing the node
if (temp.next != null) {
Console.Write(temp.next.data + " ");
temp = temp.next;
}
if (temp.prev != null) {
Console.Write(temp.prev.data + " ");
temp = temp.prev;
}
if (temp.bottom != null) {
Console.Write(temp.bottom.data + " ");
temp = temp.bottom;
}
if (temp.top != null) {
Console.Write(temp.top.data + " ");
temp = temp.top;
}
}
}
static public void Main()
{
// Code
int[, ] grid = { { 13, 42, 93, 88 },
{ 26, 38, 66, 42 },
{ 75, 63, 78, 12 } };
// n is number of rows
int n = grid.GetLength(0);
// m is number of columns
int m = grid.GetLength(1);
GFG l = new GFG();
// Call Quad method to create Linked List.
l.Quad(grid, n, m);
// Call Quad method to create Linked List.
l.printList(l.head);
}
}
// This code is contributed by lokesh (lokeshmvs21).
<script>
// Javascript program for above approach
// Node Class
class Node{
// Constructor to initialize the node object
constructor(val){
this.data = val
this.next = null
this.prev = null
this.top = null
this.bottom = null
}
}
// Linked List class
class LinkedList{
// Constructor to initialize head
constructor(){
this.head = null
}
// function to form square
// linked list of matrix.
Quad(grid, n, m){
// initialising A[0][0] as head.
this.head = new Node(grid[0][0])
// head is assigned to head.
let temp = this.head
// i is row index, j is column index
let i = 0
let j = 1
// loop till temp.top become equal to head.
while(temp.top != this.head){
// as we iterating over boundary
// of matrix so we will iterate
// over first(0) and last(n-1) row
// and first(0) and last(m-1) column.
// iterating over first i.e 0th row
// and connecting node.
if(j < m && i == 0){
temp.next = new Node(grid[i][j])
temp = temp.next
j += 1
}
// iterating over last i.e (m-1)th
// column and connecting Node.
else if (j == m && i < n - 1){
i = i + 1
temp.bottom = new Node(grid[i][j - 1])
temp = temp.bottom
}
// iterating over last i.e (n-1)th row
// and connecting Node.
else if (i == n - 1 && j <= m && j >= 1){
if (j == m) j = j - 1
j = j - 1
temp.prev = new Node(grid[i][j])
temp = temp.prev
}
// iterating over first i.e 0th column
// and connecting Node.
else if (i <= n - 1 && j == 0){
i = i - 1
temp.top = new Node(grid[i][j])
temp = temp.top
if(i == 1)
temp.top = this.head
}
}
}
// function to print Linked list.
printList(root){
let temp = root
// printing head of linked list
document.write(temp.data + " ")
// loop till temp.top
// become equal to head
while(temp.top != root){
// printing the node
if(temp.next){
document.write(temp.next.data + " ")
temp = temp.next
}
if(temp.prev){
document.write(temp.prev.data + " ")
temp = temp.prev
}
if(temp.bottom){
document.write(temp.bottom.data + " ")
temp = temp.bottom
}
if(temp.top){
document.write(temp.top.data + " ")
temp = temp.top
}
}
}
}
// Driver Code
let grid = [[13, 42, 93, 88],
[26, 38, 66, 42],
[75, 63, 78, 12]]
// n is number of rows
let n = grid.length
// m is number of column
let m = grid[0].length
// creation of object
let l = new LinkedList()
// Call Quad method to create Linked List.
l.Quad(grid, n, m)
// Call printList method to print list.
l.printList(l.head)
// This code is contributed by gfgking.
</script>
Output13 42 93 88 42 12 78 63 75 26
Time Complexity: O(N*M)
Auxiliary Space: O(N*M)
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