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 


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

spiral-matrix

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

Approach: The above problem can be solved using four for 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. The four variable along with their use is given below.

k - starting row index 
m - ending row index 
l - starting column index 
n - ending column index

Below is the implementation of the above approach:

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

Time Complexity: Time complexity of the above solution is O(mn).

Recursive 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.

Below is the Recursive implementation for the same:

C++

#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 to test 1/e law  
// for Secretary Problem : 
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 

C#

// C# Program to test 1/e law  
// for Secretary Problem : 
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

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

My Personal Notes

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

C++

C

Java

Python3

C#

PHP

C++

Java

C#

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