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Program to print Spiral Pattern

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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: 

  1. Create a 2D array of size n
  2. Store the boundary of the array in boundary variable. Initially it will be n-1 and thereafter it changes after every rotation.
  3. 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.
  4. Create a flag to determine 2 rotations, as after every 2 rotations, the sizeLeft will decrease.
  5. 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.
  6. 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.
  7. Print the Spiral Pattern stored in the 2D array.

Below is Java implementation of the above approach: 

C++




#include <iostream>
using namespace std;
 
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[size][size] = {0};
 
    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];
            if(n < 10)
                cout << n << " ";
            else
                cout << n << " ";
        }
 
        cout << endl;
    }
}
 
// Driver Code
int main()
{
 
    // Get the size of size
    int size = 5;
 
    // Print the Spiral Pattern
    printSpiral(size);
    return 0;
}
 
// This code is contributed by 29AjayKumar


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);
    }
}


Python3




def printSpiral(size):
    # Create row and col to traverse rows and columns
    row, col = 0, 0
 
    boundary = size - 1
    sizeLeft = size - 1
    flag = 1
 
    # Variable to determine the movement
    # r = right, l = left, d = down, u = upper
    move = 'r'
 
    # Array for matrix
    matrix = [[0 for j in range(size)] for i in range(size)]
 
    for i in range(1, size * size + 1):
        # Assign the value
        matrix[row][col] = i
 
        # switch-case to determine the next index
        if move == 'r':
            col += 1
        elif move == 'l':
            col -= 1
        elif move == 'u':
            row -= 1
        elif move == 'd':
            row += 1
 
        # Check if the matrix has reached array boundary
        if i == boundary:
            # Add the left size for the next boundary
            boundary += sizeLeft
 
            # If 2 rotations have been made,
            # decrease the size left by 1
            if flag != 2:
                flag = 2
            else:
                flag = 1
                sizeLeft -= 1
 
            # switch-case to rotate the movement
            if move == 'r':
                move = 'd'
            elif move == 'd':
                move = 'l'
            elif move == 'l':
                move = 'u'
            elif move == 'u':
                move = 'r'
 
    # Print the matrix
    for row in range(size):
        for col in range(size):
            n = matrix[row][col]
            print(n, end='  ' if n < 10 else ' ')
        print()
 
# Driver Code
size = 5
 
# Print the Spiral Pattern
printSpiral(size)


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 */


Javascript




<script>
 
function printSpiral(size)
{
     
    // Create row and col
    // to traverse rows and columns
    let row = 0, col = 0;
 
    let boundary = size - 1;
    let sizeLeft = size - 1;
    let flag = 1;
 
    // Variable to determine the movement
    // r = right, l = left, d = down, u = upper
    let move = 'r';
 
    // Array for matrix
    let matrix = new Array(size);
    for(let i = 0; i < size; i++)
    {
        matrix[i] = new Array(size).fill(0);
    }
 
    for(let 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++)
        {
            let n = matrix[row][col];
            if (n < 10)
                document.write(n + " ");
            else
                document.write(n + " ");
        }
        document.write("<br>");
    }
}
 
// Driver Code
 
// Get the size of size
let size = 5;
 
// Print the Spiral Pattern
printSpiral(size);
 
// This code is contributed by subham348
 
</script>


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 

Complexity Analysis:

  • Time complexity: O(m*n) where m is number of rows and n is number of columns of a given matrix
  • Auxiliary Space: O(1) because constant space has been used


Last Updated : 15 Feb, 2023
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