Java program to print spiral pattern

The spiral pattern (or matrix in spiral form) is frequently asked in Java interviews and academics. In this section, we will create a Java program to create a spiral pattern or spiral matrix.

What is Spiral Matrix or Spiral Pattern?

A spiral pattern is a number pattern that can be represented in matrix form. It is made up of a 2D array (m*n). Pattern may represent in two forms, either clockwise or anticlockwise.

In order to print a matrix in spiral pattern (clockwise) form, we need to follow the following traversal order:

Left to right (first row)
Top to bottom (Last column)
Right to left (last row)
Bottom to top (First column)

In order to print the spiral pattern (anticlockwise), reverse the above traversal order.

Let’s understand the pictorial representation of the spiral patterns, as shown in the following figure.

Java program to print the following spiral pattern on the console

For example, consider the following 5*5 matrix:

Java program to print the following spiral pattern on the console

The above matrix can be represented in spiral form as follows:

Java program to print the following spiral pattern on the console

We can follow any of the approaches to print the spiral pattern:

Iterative
Recursive

Throughout this section, we have used the iterative approach.

What should be the approach?

First, create a 2D array of size n.
Declare the variables to store the boundary of the array. Also declare a variable to store the size left to print the spiral.
For the movement over the array, use l, r, u, and d letters that represent the direction of movement left, right, up, and down, respectively.
Execute a loop from 1 to n*2 to fill the array in spiral form.

Spiral Pattern Java Program

SpiralPaternExample1.java

  public class SpiralPatternExample1  {  //defining method to print the spiral pattern or matrix  static void printSpiralPattern(int size)  {  //create two variables row and col to traverse rows and columns  int row = 0, col = 0;  int boundary = size – 1;  int sizeLeft = size – 1;  int flag = 1;  //variables r, l, u and d are used to determine the movement  // r = right, l = left, d = down, u = upper  char move = ‘r’;  //creating a 2D array for matrix  int[][] matrix =new int [size][size];  for (int i = 1; i < size * size + 1; i++)  {      //assigning values      matrix[row][col] = i;  //switch-case to determine the next index  switch (move)  {  //if right, go right  case ‘r’:      col += 1;      break;  //if left, go left  case ‘l’:      col -= 1;      break;  //if up, go up  case ‘u’:      row -= 1;      break;  //if down, go down  case ‘d’:      row += 1;      break;  }  //checks if the matrix has reached the array boundary  if (i == boundary)      {          //adds the left size for the next boundary          boundary = boundary + sizeLeft;          //decrease the size left by 1, if 2 rotations have been made          if (flag != 2)          {          flag = 2;          }          else          {          flag = 1;          sizeLeft -= 1;          }          //switch-case to rotate the movement          switch (move)          {          //if right, rotate to down          case ‘r’:              move = ‘d’;              break;          // if down, rotate to left          case ‘d’:              move = ‘l’;              break;          // if left, rotate to up          case ‘l’:              move = ‘u’;              break;          // if up, rotate to right          case ‘u’:              move = ‘r’;              break;          }      }      }  //printing the spiral matrix or pattern  //outer for loop for rows      for (row = 0; row < size; row++)      {          //inner for loop for columns          for (col = 0; col < size; col++)              {                  int n = matrix[row][col];                  if(n < 10)                      System.out.print(n +” “);                  else                      System.out.print(n +” “);              }          System.out.println();      }  }  //driver Code  public static void main(String args[])  {  //size of the array?s row and column  int size = 5;  System.out.println(“Spiral Matrix or Pattern is: n”);  //calling the method that prints the spiral pattern or matrix   printSpiralPattern(size);  }  }  

Output:

Java program to print the following spiral pattern on the console

Let’s see another spiral pattern.

In the following program, we have used the following approach:

Starting the indexing from i = 1 and j = 1. It can be observed that every value of the required matrix will be max(abs(i – n), abs(j – n)) + 1.

SpiralPatternExample2.java

  import java.util.Scanner;  import java.lang.Math;  public class SpiralPatternExample2  {  //function to print the spiral pattern  public static void printPattern(int n)  {  //detrmines the boundary size of the array  int size = 2 * n – 1;  //inner loop  for(int i = 1; i <= size; i++)  {  //outer loop      for(int j = 1; j <= size; j++)  {  //calculates and prints the values for pattern  System.out.print(Math.max(Math.abs(i – n), Math.abs(j – n)) + 1 + ” “);  }  System.out.println();  }  }  //driver code  public static void main(String args[])  {  Scanner sc = new Scanner(System.in);  System.out.print(“Enter the value of n: “);  int n = sc.nextInt();  System.out.println();  //function calling  printPattern(n);  }  }  

Output:

Java program to print the following spiral pattern on the console

SpiralPatternExample3.java

  public class SprialPatternExample3  {  public static void main(String args[])  {      int SIZE=10;      int i, j, N;      int[][] board = new int[SIZE][SIZE];      int left, top;      left = 0;      top  = SIZE – 1;      N    = 1;      for(i=1; i<=SIZE/2; i++, left++, top–)      {          //fill from left to right          for(j=left; j<=top; j++, N++)          {              board[left][j] = N;          }          //fill from top to down          for(j=left+1; j<=top; j++, N++)          {              board[j][top] = N;          }          //fill from right to left          for(j=top-1; j>=left; j–, N++)          {              board[top][j] = N;          }          //fill from down to top          for(j=top-1; j>=left+1; j–, N++)          {              board[j][left] = N;          }      }      //print the pattern      for(i=0; i<SIZE; i++)      {          for(j=0; j<SIZE; j++)          {              System.out.printf(“%-5d”, board[i][j]);          }      System.out.printf(“n”);      }  }  }  

Output:

Java program to print the following spiral pattern on the console

Print a Given Matrix in Spiral Form

There are many approaches that can be used to print a matrix in spiral form.

SpiralToMatrixExample.java

  import java.io.*;  public class SpiralToMatrixExample   {  //Function print matrix in spiral form  static void spiralPrint(int m, int n, int a[][])  {  int i, k = 0, l = 0;  /* k – starting row index  m – ending row index  l – starting column index  n – ending column index  i – iterator  */  while (k < m && l < n)   {  //prints the first row from the remaining rows  for (i = l; i < n; ++i)   {  System.out.print(a[k][i] + ” “);  }  k++;    //prints the last column from the remaining columns  for (i = k; i < m; ++i)   {  System.out.print(a[i][n – 1] + ” “);  }  n–;  //prints the last row from the remaining rows   if (k < m)   {  for (i = n – 1; i >= l; –i)   {  System.out.print(a[m – 1][i] + ” “);  }  m–;  }  //prints the first column from the remaining columns  if (l < n)   {  for (i = m – 1; i >= k; –i)   {  System.out.print(a[i][l] + ” “);  }  l++;  }  }  }  //driver Code  public static void main(String args[])  {  int row = 3;  int col = 6;  //array to print in spiral form  int arr[][] = { { 10, 20, 30, 40, 50, 60 }, { 70, 80, 90, 100, 110, 120 }, { 130, 140, 150, 160, 170, 180 } };  //function Calling  spiralPrint(row, col, arr);  }  }  

Output:

10 20 30 40 50 60 120 180 170 160 150 140 130 70 80 90 100 110

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Java program to print spiral pattern