Print a given matrix in spiral form

Given a 2D array, print it in spiral form. See the following examples.

Examples:

Input:  1    2   3   4
        5    6   7   8
        9   10  11  12
        13  14  15  16
Output: 1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 
Explanation: The output is matrix in spiral format. 

Input:  1   2   3   4  5   6
        7   8   9  10  11  12
        13  14  15 16  17  18
Output: 1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11
Explanation :The output is matrix in spiral format.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1: This is a simple method to solve the following problem.

Approach: The problem can be solved by dividing the matrix into loops or squares or boundaries. It can be seen that the elements of the outer loop are printed first in a clockwise manner then the elements of the inner loop is printed. So printing the elements of a loop can be solved using four loops which prints all the elements. Every ‘for’ loop defines a single direction movement along with the matrix. The first for loop represents the movement from left to right, whereas the second crawl represents the movement from top to bottom, the third represents the movement from the right to left, and the fourth represents the movement from bottom to up.
Algorithm:
1. create and initilize variables k – starting row index, m – ending row index, l – starting column index, n – ending column index
2. Run a loop until all the squares of loops are printed.
3. In each outer loop traversal print the elements of a square in clockwise manner.
4. Print the top row, i.e. Print the elements of kth row from column index l to n, and increase the count of k.
5. Print the right column, i.e. Print the last column or n-1th column from row index k to m and decrease the count of n.
6. Print the bottom row, i.e. if k > m, then print the elements of m-1th row from column n-1 to l and decrease the count of m
7. Print the left column, i.e. if l < n, then print the elements of lth column from m-1th row to k and increase the count of l.

Implementation:

C++

#include <bits/stdc++.h> 
using namespace std; 
#define R 3 
#define C 6 
  
void spiralPrint(int m, int n, int a[R][C]) 
{ 
    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) { 
        /* Print the first row from 
               the remaining rows */
        for (i = l; i < n; ++i) { 
            cout << a[k][i] << " "; 
        } 
        k++; 
  
        /* Print the last column  
         from the remaining columns */
        for (i = k; i < m; ++i) { 
            cout << a[i][n - 1] << " "; 
        } 
        n--; 
  
        /* Print the last row from  
                the remaining rows */
        if (k < m) { 
            for (i = n - 1; i >= l; --i) { 
                cout << a[m - 1][i] << " "; 
            } 
            m--; 
        } 
  
        /* Print the first column from 
                   the remaining columns */
        if (l < n) { 
            for (i = m - 1; i >= k; --i) { 
                cout << a[i][l] << " "; 
            } 
            l++; 
        } 
    } 
} 
  
/* Driver program to test above functions */
int main() 
{ 
    int a[R][C] = { { 1, 2, 3, 4, 5, 6 }, 
                    { 7, 8, 9, 10, 11, 12 }, 
                    { 13, 14, 15, 16, 17, 18 } }; 
  
    spiralPrint(R, C, a); 
    return 0; 
} 
  
// This is code is contributed by rathbhupendra 

C

/* This code is adopted from the solution given  
   @ http:// effprog.blogspot.com/2011/01/ 
spiral-printing-of-two-dimensional.html */
  
#include <stdio.h> 
#define R 3 
#define C 6 
  
void spiralPrint(int m, int n, int a[R][C]) 
{ 
    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) { 
        /* Print the first row from the remaining rows */
        for (i = l; i < n; ++i) { 
            printf("%d ", a[k][i]); 
        } 
        k++; 
  
        /* Print the last column from the remaining columns */
        for (i = k; i < m; ++i) { 
            printf("%d ", a[i][n - 1]); 
        } 
        n--; 
  
        /* Print the last row from the remaining rows */
        if (k < m) { 
            for (i = n - 1; i >= l; --i) { 
                printf("%d ", a[m - 1][i]); 
            } 
            m--; 
        } 
  
        /* Print the first column from the remaining columns */
        if (l < n) { 
            for (i = m - 1; i >= k; --i) { 
                printf("%d ", a[i][l]); 
            } 
            l++; 
        } 
    } 
} 
  
/* Driver program to test above functions */
int main() 
{ 
    int a[R][C] = { { 1, 2, 3, 4, 5, 6 }, 
                    { 7, 8, 9, 10, 11, 12 }, 
                    { 13, 14, 15, 16, 17, 18 } }; 
  
    spiralPrint(R, C, a); 
    return 0; 
} 

Java

// Java program to print a given matrix in spiral form 
import java.io.*; 
  
class GFG { 
    // 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) { 
            // Print the first row from the remaining rows 
            for (i = l; i < n; ++i) { 
                System.out.print(a[k][i] + " "); 
            } 
            k++; 
  
            // Print the last column from the remaining columns 
            for (i = k; i < m; ++i) { 
                System.out.print(a[i][n - 1] + " "); 
            } 
            n--; 
  
            // Print 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--; 
            } 
  
            // Print 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 program 
    public static void main(String[] args) 
    { 
        int R = 3; 
        int C = 6; 
        int a[][] = { { 1, 2, 3, 4, 5, 6 }, 
                      { 7, 8, 9, 10, 11, 12 }, 
                      { 13, 14, 15, 16, 17, 18 } }; 
        spiralPrint(R, C, a); 
    } 
} 
  
// Contributed by Pramod Kumar 

Python3

# Python3 program to print  
# given matrix in spiral form 
def spiralPrint(m, n, a) : 
    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 and l < n) : 
          
        # Print the first row from 
        # the remaining rows  
        for i in range(l, n) : 
            print(a[k][i], end = " ") 
              
        k += 1
  
        # Print the last column from 
        # the remaining columns  
        for i in range(k, m) : 
            print(a[i][n - 1], end = " ") 
              
        n -= 1
  
        # Print the last row from 
        # the remaining rows  
        if ( k < m) : 
              
            for i in range(n - 1, (l - 1), -1) : 
                print(a[m - 1][i], end = " ") 
              
            m -= 1
          
        # Print the first column from 
        # the remaining columns  
        if (l < n) : 
            for i in range(m - 1, k - 1, -1) : 
                print(a[i][l], end = " ") 
              
            l += 1
  
# Driver Code 
a = [ [1, 2, 3, 4, 5, 6], 
      [7, 8, 9, 10, 11, 12], 
      [13, 14, 15, 16, 17, 18] ] 
        
R = 3; C = 6
spiralPrint(R, C, a) 
  
# This code is contributed by Nikita Tiwari. 

C#

// C# program to print a given 
// matrix in spiral form 
using System; 
  
class GFG { 
    // 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) { 
            // Print the first row  
            // from the remaining rows 
            for (i = l; i < n; ++i) { 
                Console.Write(a[k, i] + " "); 
            } 
            k++; 
  
            // Print the last column from the 
            // remaining columns 
            for (i = k; i < m; ++i) { 
                Console.Write(a[i, n - 1] + " "); 
            } 
            n--; 
  
            // Print the last row from  
            // the remaining rows  
            if (k < m) { 
                for (i = n - 1; i >= l; --i) { 
                    Console.Write(a[m - 1, i] + " "); 
                } 
                m--; 
            } 
  
            // Print the first column from  
            // the remaining columns 
            if (l < n) { 
                for (i = m - 1; i >= k; --i) { 
                    Console.Write(a[i, l] + " "); 
                } 
                l++; 
            } 
        } 
    } 
  
    // Driver program 
    public static void Main() 
    { 
        int R = 3; 
        int C = 6; 
        int[, ] a = { { 1, 2, 3, 4, 5, 6 }, 
                      { 7, 8, 9, 10, 11, 12 }, 
                      { 13, 14, 15, 16, 17, 18 } }; 
        spiralPrint(R, C, a); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php  
// PHP program to print a given 
// matrix in spiral form 
$R = 3; 
$C = 6; 
  
function spiralPrint($m, $n, &$a) 
{ 
    $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 < $n) 
    { 
        /* Print the first row from 
           the remaining rows */
        for ($i = $l; $i < $n; ++$i) 
        { 
            echo $a[$k][$i] . " "; 
        } 
        $k++; 
  
        /* Print the last column  
        from the remaining columns */
        for ($i = $k; $i < $m; ++$i) 
        { 
            echo $a[$i][$n - 1] . " "; 
        } 
        $n--; 
  
        /* Print the last row from 
           the remaining rows */
        if ($k < $m) 
        { 
            for ($i = $n - 1; $i >= $l; --$i) 
            { 
                echo $a[$m - 1][$i] . " "; 
            } 
            $m--; 
        } 
  
        /* Print the first column from 
           the remaining columns */
        if ($l < $n) 
        { 
            for ($i = $m - 1; $i >= $k; --$i) 
            { 
                echo $a[$i][$l] . " "; 
            } 
            $l++;  
        }      
    } 
} 
  
// Driver code 
$a = array(array(1, 2, 3, 4, 5, 6), 
           array(7, 8, 9, 10, 11, 12), 
           array(13, 14, 15, 16, 17, 18)); 
  
spiralPrint($R, $C, $a); 
  
// This code is contributed 
// by ChitraNayal 
?> 

Output:

1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11

Complexity Analysis:
- Time Complexity: O(m*n).
  To traverse the matrix O(m*n) time is required.
- Space Comepxlity:O(1).
  No extra space is required.

Method 2: This is a recursive approach.

Approach: The above problem can be solved by printing the boundary of the Matrix recursively. In each recursive call, we decrease the dimensions of the matrix. The idea of printing the boundary or loops is the same.
Algorithm:
1. create a recursive function that takes a matrix and some variables (k – starting row index, m – ending row index, l – starting column index, n – ending column index) as parameters
2. Check the base cases (stating index is less than or equal to ending index) and print the boundary elements in clockwise manner
3. Print the top row, i.e. Print the elements of kth row from column index l to n, and increase the count of k.
4. Print the right column, i.e. Print the last column or n-1th column from row index k to m and decrease the count of n.
5. Print the bottom row, i.e. if k > m, then print the elements of m-1th row from column n-1 to l and decrease the count of m
6. Print the left column, i.e. if l < n, then print the elements of lth column from m-1th row to k and increase the count of l.
7. Call the function recursively with the values of starting and ending indices of rows and columns.

Implementation:

C++

// C++. program for the above approach 
#include <iostream> 
using namespace std; 
  
#define R 4 
#define C 4 
  
// Function for printing matrix in spiral 
// form i, j: Start index of matrix, row  
// and column respectively m, n: End index 
// of matrix row and column respectively 
void print(int arr[R][C], int i,  
                     int j, int m, int n) 
{ 
    // If i or j lies outside the matrix 
    if (i >= m or j >= n) 
        return; 
  
    // Print First Row 
    for (int p = i; p < n; p++) 
        cout << arr[i][p] << " "; 
  
    // Print Last Column 
    for (int p = i + 1; p < m; p++) 
        cout << arr[p][n - 1] << " "; 
  
    // Print Last Row, if Last and 
    // First Row are not same 
    if ((m - 1) != i) 
        for (int p = n - 2; p >= j; p--) 
            cout << arr[m - 1][p] << " "; 
  
    // Print First Column,  if Last and 
    // First Column are not same 
    if ((n - 1) != j) 
        for (int p = m - 2; p > i; p--) 
            cout << arr[p][j] << " "; 
  
    print(arr, i + 1, j + 1, m - 1, n - 1); 
} 
  
// Driver Program 
int main() 
{ 
  
    int a[R][C] = { { 1, 2, 3, 4 }, 
                    { 5, 6, 7, 8 }, 
                    { 9, 10, 11, 12 }, 
                    { 13, 14, 15, 16 } }; 
  
    print(a, 0, 0, R, C); 
    return 0; 
} 
// This Code is contributed by Ankur Goel 

Java

// Java program for the above approach 
import java.util.*; 
  
class GFG 
{ 
    static int R = 4; 
    static int C = 4; 
  
    // Function for printing matrix in spiral 
    // form i, j: Start index of matrix, row  
    // and column respectively m, n: End index 
    // of matrix row and column respectively 
    static void print(int arr[][], int i, 
                      int j, int m, int n) 
    { 
        // If i or j lies outside the matrix 
        if (i >= m || j >= n) 
        { 
            return; 
        } 
  
        // Print First Row 
        for (int p = i; p < n; p++) 
        { 
            System.out.print(arr[i][p] + " "); 
        } 
  
        // Print Last Column 
        for (int p = i + 1; p < m; p++)  
        { 
            System.out.print(arr[p][n - 1] + " "); 
        } 
  
        // Print Last Row, if Last and 
        // First Row are not same 
        if ((m - 1) != i)  
        { 
            for (int p = n - 2; p >= j; p--)  
            { 
                System.out.print(arr[m - 1][p] + " "); 
            } 
        } 
  
        // Print First Column, if Last and 
        // First Column are not same 
        if ((n - 1) != j)  
        { 
            for (int p = m - 2; p > i; p--)  
            { 
                System.out.print(arr[p][j] + " "); 
            } 
        } 
        print(arr, i + 1, j + 1, m - 1, n - 1); 
    } 
  
    // Driver Code 
    public static void main(String[] args)  
    { 
        int a[][] = {{1, 2, 3, 4}, 
                     {5, 6, 7, 8}, 
                     {9, 10, 11, 12}, 
                     {13, 14, 15, 16}}; 
  
        print(a, 0, 0, R, C); 
    } 
} 
  
// This code is contributed by 29AjayKumar 

Python3

# Python3 program for the above approach 
  
# Function for printing matrix in spiral  
# form i, j: Start index of matrix, row   
# and column respectively m, n: End index  
# of matrix row and column respectively  
def printdata(arr, i, j, m, n): 
      
    # If i or j lies outside the matrix  
    if (i >= m or j >= n): 
        return; 
          
    # Print First Row  
    for p in range(i, n): 
        print(arr[i][p], end = " ") 
          
    # Print Last Column  
    for p in range(i + 1, m): 
        print(arr[p][n - 1], end = " ") 
          
    # Print Last Row, if Last and  
    # First Row are not same  
    if ((m - 1) != i):  
        for p in range(n - 2, j - 1, -1): 
            print(arr[m - 1][p], end = " ") 
              
    # Print First Column, if Last and  
    # First Column are not same  
    if ((n - 1) != j): 
        for p in range(m - 2, i, -1): 
            print(arr[p][j], end = " ") 
              
    printdata(arr, i + 1, j + 1, m - 1, n - 1)  
  
# Driver code 
R = 4
C = 4
arr = [ [ 1, 2, 3, 4 ], 
        [ 5, 6, 7, 8 ], 
        [ 9, 10, 11, 12 ], 
        [ 13, 14, 15, 16 ] ] 
          
printdata(arr, 0, 0, R, C) 
  
# This code is contributed by avsadityavardhan 

C#

// C# program for the above approach 
using System; 
      
class GFG 
{ 
    static int R = 4; 
    static int C = 4; 
  
    // Function for printing matrix in spiral 
    // form i, j: Start index of matrix, row  
    // and column respectively m, n: End index 
    // of matrix row and column respectively 
    static void print(int [,]arr, int i, 
                      int j, int m, int n) 
    { 
        // If i or j lies outside the matrix 
        if (i >= m || j >= n) 
        { 
            return; 
        } 
  
        // Print First Row 
        for (int p = i; p < n; p++) 
        { 
            Console.Write(arr[i, p] + " "); 
        } 
  
        // Print Last Column 
        for (int p = i + 1; p < m; p++)  
        { 
            Console.Write(arr[p, n - 1] + " "); 
        } 
  
        // Print Last Row, if Last and 
        // First Row are not same 
        if ((m - 1) != i)  
        { 
            for (int p = n - 2; p >= j; p--)  
            { 
                Console.Write(arr[m - 1, p] + " "); 
            } 
        } 
  
        // Print First Column, if Last and 
        // First Column are not same 
        if ((n - 1) != j)  
        { 
            for (int p = m - 2; p > i; p--)  
            { 
                Console.Write(arr[p, j] + " "); 
            } 
        } 
        print(arr, i + 1, j + 1, m - 1, n - 1); 
    } 
  
    // Driver Code 
    public static void Main(String[] args)  
    { 
        int [,]a = {{1, 2, 3, 4}, 
                    {5, 6, 7, 8}, 
                    {9, 10, 11, 12}, 
                    {13, 14, 15, 16}}; 
  
        print(a, 0, 0, R, C); 
    } 
}  
  
// This code is contributed by Princi Singh 

Output:

1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10

Complexity Analysis:
- Time Complexity: O(m*n).
  To traverse the matrix O(m*n) time is required.
- Space Comepxlity:O(1).
  No extra space is required.

Please write comments if you find the above code incorrect, or find other ways to solve the same problem.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

C++

Java

Python3

C#

Recommended Posts: