Program to print Spiral Pattern
Given a number n as the size of the matrix, the task is to print a spiral pattern in 2D array of size n.
Examples:
Input: n = 5
Output: 1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
Preview of the Spiral Pattern:
Approach:
- Create a 2D array of size n
- Store the boundary of the array in boundary variable. Initially it will be n-1 and thereafter it changes after every rotation.
- Store the size left for the spiral printing in variable sizeLeft. Initially it will be n-1 and thereafter it will decrease by 1 after every 2 rotations.
- Create a flag to determine 2 rotations, as after every 2 rotations, the sizeLeft will decrease.
- Create a char variable move to store the current movement of the spiral pattern. It can have ‘r’ for right, ‘l’ for left, ‘d’ for down, and ‘u’ for up.
- Repeat the below steps till ‘i’ is in range [1, n^2]:
- Assign the value of i to the spiral pattern.
- Determine the next movement of the pattern.
- Check for pattern to reach boundary. If reached, modify the sizes and rotate the spiral pattern.
- Print the Spiral Pattern stored in the 2D array.
Below is Java implementation of the above approach:
Java
public class GFG { public static void printSpiral(int size) { // Create row and col // to traverse rows and columns int row = 0, col = 0; int boundary = size - 1; int sizeLeft = size - 1; int flag = 1; // Variable to determine the movement // r = right, l = left, d = down, u = upper char move = 'r'; // Array for matrix int matrix[][] = new int[size][size]; for (int i = 1; i < size * size + 1; i++) { // Assign the value 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; } // Check if the matrix // has reached array boundary if (i == boundary) { // Add the left size for the next boundary boundary += sizeLeft; // If 2 rotations has been made, // decrease the size left by 1 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; } } } // Print the matrix for (row = 0; row < size; row++) { for (col = 0; col < size; col++) { int n = matrix[row][col]; System.out.print((n < 10) ? (n + " ") : (n + " ")); } System.out.println(); } } // Driver Code public static void main(String[] args) { // Get the size of size int size = 5; // Print the Spiral Pattern printSpiral(size); } } |
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C#
// C# implementation of the approach using System; class GFG { public static void printSpiral(int size) { // Create row and col // to traverse rows and columns int row = 0, col = 0; int boundary = size - 1; int sizeLeft = size - 1; int flag = 1; // Variable to determine the movement // r = right, l = left, d = down, u = upper char move = 'r'; // Array for matrix int[, ] matrix = new int[size, size]; for (int i = 1; i < size * size + 1; i++) { // Assign the value 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; } // Check if the matrix // has reached array boundary if (i == boundary) { // Add the left size for the next boundary boundary += sizeLeft; // If 2 rotations has been made, // decrease the size left by 1 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; } } } // Print the matrix for (row = 0; row < size; row++) { for (col = 0; col < size; col++) { int n = matrix[row, col]; Console.Write((n < 10) ? (n + " ") : (n + " ")); } Console.WriteLine(); } } // Driver Code public static void Main(String[] args) { // Get the size of size int size = 5; // Print the Spiral Pattern printSpiral(size); } } /* This code contributed by PrinciRaj1992 */ |
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Output:
1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9
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