# Print a given matrix in spiral form

Given a 2D array, print it in spiral form.

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.

## Print a given matrix in spiral form using the simulation approach:

To solve the problem follow the below idea:

Draw the path that the spiral makes. We know that the path should turn clockwise whenever it would go out of bounds or into a cell that was previously visited

Follow the given steps to solve the problem:

• Let the array have R rows and C columns
• seen[r] denotes that the cell on the r-th row and c-th column was previously visited. Our current position is (r, c), facing direction di, and we want to visit R x C total cells.
• As we move through the matrix, our candidate’s next position is (cr, cc).
• If the candidate is in the bounds of the matrix and unseen, then it becomes our next position; otherwise, our next position is the one after performing a clockwise turn

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std; vector spiralOrder(vector >& matrix){    int m = matrix.size(), n = matrix[0].size();    vector ans;     if (m == 0)        return ans;     vector > seen(m, vector(n, false));    int dr[] = { 0, 1, 0, -1 };    int dc[] = { 1, 0, -1, 0 };     int x = 0, y = 0, di = 0;     // Iterate from 0 to m * n - 1    for (int i = 0; i < m * n; i++) {        ans.push_back(matrix[x][y]);        // on normal geeksforgeeks ui page it is showing        // 'ans.push_back(matrix[x])' which gets copied as        // this only and gives error on compilation,        seen[x][y] = true;        int newX = x + dr[di];        int newY = y + dc[di];         if (0 <= newX && newX < m && 0 <= newY && newY < n            && !seen[newX][newY]) {            x = newX;            y = newY;        }        else {            di = (di + 1) % 4;            x += dr[di];            y += dc[di];        }    }    return ans;} // Driver codeint main(){    vector > a{ { 1, 2, 3, 4 },                            { 5, 6, 7, 8 },                            { 9, 10, 11, 12 },                            { 13, 14, 15, 16 } };     // Function call    for (int x : spiralOrder(a)) {        cout << x << " ";    }    return 0;} // This code is contributed by Yashvendra Singh

## Java

 // Java program for the above approachimport java.util.*; class Solution {     // Function to print in spiral order    public static List spiralOrder(int[][] matrix)    {        List ans = new ArrayList();         if (matrix.length == 0)            return ans;         int m = matrix.length, n = matrix[0].length;        boolean[][] seen = new boolean[m][n];        int[] dr = { 0, 1, 0, -1 };        int[] dc = { 1, 0, -1, 0 };        int x = 0, y = 0, di = 0;         // Iterate from 0 to R * C - 1        for (int i = 0; i < m * n; i++) {            ans.add(matrix[x][y]);            seen[x][y] = true;            int cr = x + dr[di];            int cc = y + dc[di];             if (0 <= cr && cr < m && 0 <= cc && cc < n                && !seen[cr][cc]) {                x = cr;                y = cc;            }            else {                di = (di + 1) % 4;                x += dr[di];                y += dc[di];            }        }        return ans;    }     // 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 } };         // Function call        System.out.println(spiralOrder(a));    }}

## Python3

 # python3 program for the above approach  def spiralOrder(matrix):    ans = []     if (len(matrix) == 0):        return ans     m = len(matrix)    n = len(matrix[0])    seen = [[0 for i in range(n)] for j in range(m)]    dr = [0, 1, 0, -1]    dc = [1, 0, -1, 0]    x = 0    y = 0    di = 0     # Iterate from 0 to R * C - 1    for i in range(m * n):        ans.append(matrix[x][y])        seen[x][y] = True        cr = x + dr[di]        cc = y + dc[di]         if (0 <= cr and cr < m and 0 <= cc and cc < n and not(seen[cr][cc])):            x = cr            y = cc        else:            di = (di + 1) % 4            x += dr[di]            y += dc[di]    return ans  # Driver codeif __name__ == "__main__":    a = [[1, 2, 3, 4],         [5, 6, 7, 8],         [9, 10, 11, 12],         [13, 14, 15, 16]]     # Function call    for x in spiralOrder(a):        print(x, end=" ")    print()

## C#

 // C# program for the above approachusing System;using System.Collections.Generic; public class GFG {     // Function to print in spiral order    public static List spiralOrder(int[, ] matrix)    {        List ans = new List();         if (matrix.Length == 0)            return ans;         int R = matrix.GetLength(0), C                                     = matrix.GetLength(1);        bool[, ] seen = new bool[R, C];        int[] dr = { 0, 1, 0, -1 };        int[] dc = { 1, 0, -1, 0 };        int r = 0, c = 0, di = 0;         // Iterate from 0 to R * C - 1        for (int i = 0; i < R * C; i++) {            ans.Add(matrix[r, c]);            seen[r, c] = true;            int cr = r + dr[di];            int cc = c + dc[di];             if (0 <= cr && cr < R && 0 <= cc && cc < C                && !seen[cr, cc]) {                r = cr;                c = cc;            }            else {                di = (di + 1) % 4;                r += dr[di];                c += dc[di];            }        }        return ans;    }     // 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 } };         // Function call        spiralOrder(a).ForEach(i                               = > Console.Write(i + " "));    }} // This code is contributed by 29AjayKumar

## Javascript



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

Time Complexity: O(N), where N is the total number of elements in the input matrix. We add every element in the matrix to our final answer
Auxiliary Space: O(N), the information stored in seen and in ans.

## Print a given matrix in spiral form by dividing the matrix into cycles:

To solve the problem follow the below idea:

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 are printed. So printing the elements of a loop can be solved using four loops that print 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

Follow the given steps to solve the problem:

• Create and initialize variables k – starting row index, m – ending row index, l – starting column index, n – ending column index
• Run a loop until all the squares of loops are printed.
• In each outer loop traversal print the elements of a square in a clockwise manner.
• Print the top row, i.e. Print the elements of the kth row from column index l to n, and increase the count of k.
• 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.
• Print the bottom row, i.e. if k < m, then print the elements of the m-1th row from column n-1 to l and decrease the count of m
• 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.

Below is the implementation of the above approach:

## C++

 // C++ Program to print a matrix spirally #include using namespace std;#define R 4#define C 4 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 Code */int main(){    int a[R][C] = { { 1, 2, 3, 4 },                    { 5, 6, 7, 8 },                    { 9, 10, 11, 12 },                    { 13, 14, 15, 16 } };     // Function Call    spiralPrint(R, C, a);    return 0;} // This is code is contributed by rathbhupendra

## C

 // C program to print the array in a// spiral form #include #define R 4#define C 4 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 Code */int main(){    int a[R][C] = { { 1, 2, 3, 4 },                    { 5, 6, 7, 8 },                    { 9, 10, 11, 12 },                    { 13, 14, 15, 16 } };     // Function Call    spiralPrint(R, C, a);    return 0;}

## Java

 // Java program to print a given matrix in spiral formimport 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 Code    public static void main(String[] args)    {        int R = 4;        int C = 4;        int a[][] = { { 1, 2, 3, 4 },                      { 5, 6, 7, 8 },                      { 9, 10, 11, 12 },                      { 13, 14, 15, 16 } };         // Function Call        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 Codea = [[1, 2, 3, 4],     [5, 6, 7, 8],     [9, 10, 11, 12],     [13, 14, 15, 16]] R = 4C = 4 # Function CallspiralPrint(R, C, a) # This code is contributed by Nikita Tiwari.

## C#

 // C# program to print a given// matrix in spiral formusing 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 Code    public static void Main()    {        int R = 4;        int C = 4;        int[, ] a = { { 1, 2, 3, 4 },                      { 5, 6, 7, 8 },                      { 9, 10, 11, 12 },                      { 13, 14, 15, 16 } };         // Function Call        spiralPrint(R, C, a);    }} // This code is contributed by Sam007

## PHP

 = \$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),           array(5, 6, 7, 8),           array(9, 10, 11, 12),           array(13, 14, 15, 16)); // Function CallspiralPrint(\$R, \$C, \$a); // This code is contributed// by ChitraNayal?>

## Javascript



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

Time Complexity: O(M*N). To traverse the matrix O(M*M) time is required.
Auxiliary Space: O(1). No extra space is required.

## Print a given matrix in a spiral using recursion:

To solve the problem follow the below idea:

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

Follow the given steps to solve the problem:

• 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
• Check the base cases (starting index is less than or equal to the ending index) and print the boundary elements in a clockwise manner
• Print the top row, i.e. Print the elements of the kth row from column index l to n, and increase the count of k
• 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
• 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
• Print the left column, i.e. if l < n, then print the elements of the lth column from m-1th row to k and increase the count of l
• Call the function recursively with the values of starting and ending indices of rows and columns

Below is the implementation of the above approach:

## C++

 // C++. program for the above approach#include 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 respectivelyvoid 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 = j; 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 Codeint main(){    int a[R][C] = { { 1, 2, 3, 4 },                    { 5, 6, 7, 8 },                    { 9, 10, 11, 12 },                    { 13, 14, 15, 16 } };     // Function Call    print(a, 0, 0, R, C);    return 0;}// This Code is contributed by Ankur Goel

## Java

 // Java program for the above approachimport 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 } };         // Function Call        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 codeif __name__ == "__main__":    R = 4    C = 4    arr = [[1, 2, 3, 4],           [5, 6, 7, 8],           [9, 10, 11, 12],           [13, 14, 15, 16]]     # Function Call    printdata(arr, 0, 0, R, C) # This code is contributed by avsadityavardhan

## C#

 // C# program for the above approachusing 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 } };         // Function Call        print(a, 0, 0, R, C);    }} // This code is contributed by Princi Singh

## Javascript



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

Time Complexity: O(M*N). To traverse the matrix O(m*n) time is required.
Auxiliary Space: O(1). No extra space is required.

## Print a given matrix in spiral using DFS:

To solve the problem follow the below idea:

Another recursive approach is to consider DFS movement within the matrix (right->down->left->up->right->..->end). We do this by modifying the matrix itself such that when DFS algorithm visits each matrix cell it’s changed to a value which cannot be contained within the matrix. The DFS algorithm is terminated when it visits a cell such that all of its surrounding cells are already visited. The direction of the DFS search is controlled by a variable.

Follow the given steps to solve the problem:

• create a DFS function that takes matrix, cell indices, and direction
• checks are cell indices pointing to a valid cell (that is, not visited and in bounds)? if not, skip this cell
• print cell value
• mark matrix cell pointed by indicates as visited by changing it to a value not supported in the matrix
• check are surrounding cells valid? if not stop the algorithm, else continue
• if the direction is given right then check, if the cell to the right is valid. if so, DFS to the right cell given the steps above, else, change the direction to down and DFS downwards given the steps above
• else, if the direction given is down then check, if the cell to the down is valid. if so, DFS to the cell below given the steps above, else, change the direction to left and DFS leftwards given the steps above
• else, if the direction given is left then check, if the cell to the left is valid. if so, DFS to the left cell given the steps above, else, change the direction to up and DFS upwards given the steps above
• else, if the direction given is up then check, if the cell to the up is valid. if so, DFS to the upper cell given the steps above, else, change the direction to right and DFS rightwards given the steps above

Below is an implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std;#define R 4#define C 4 bool isInBounds(int i, int j){    if (i < 0 || i >= R || j < 0 || j >= C)        return false;    return true;} // check if the position is blockedbool isBlocked(int matrix[R][C], int i, int j){    if (!isInBounds(i, j))        return true;    if (matrix[i][j] == -1)        return true;    return false;} // DFS code to traverse spirallyvoid spirallyDFSTraverse(int matrix[R][C], int i, int j,                          int dir, vector& res){    if (isBlocked(matrix, i, j))        return;    bool allBlocked = true;    for (int k = -1; k <= 1; k += 2) {        allBlocked = allBlocked                     && isBlocked(matrix, k + i, j)                     && isBlocked(matrix, i, j + k);    }    res.push_back(matrix[i][j]);    matrix[i][j] = -1;    if (allBlocked) {        return;    }     // dir: 0 - right, 1 - down, 2 - left, 3 - up    int nxt_i = i;    int nxt_j = j;    int nxt_dir = dir;    if (dir == 0) {        if (!isBlocked(matrix, i, j + 1)) {            nxt_j++;        }        else {            nxt_dir = 1;            nxt_i++;        }    }    else if (dir == 1) {        if (!isBlocked(matrix, i + 1, j)) {            nxt_i++;        }        else {            nxt_dir = 2;            nxt_j--;        }    }    else if (dir == 2) {        if (!isBlocked(matrix, i, j - 1)) {            nxt_j--;        }        else {            nxt_dir = 3;            nxt_i--;        }    }    else if (dir == 3) {        if (!isBlocked(matrix, i - 1, j)) {            nxt_i--;        }        else {            nxt_dir = 0;            nxt_j++;        }    }    spirallyDFSTravserse(matrix, nxt_i, nxt_j, nxt_dir,                         res);} // To traverse spirallyvector spirallyTraverse(int matrix[R][C]){    vector res;    spirallyDFSTravserse(matrix, 0, 0, 0, res);    return res;} // Driver Codeint main(){    int a[R][C] = { { 1, 2, 3, 4 },                    { 5, 6, 7, 8 },                    { 9, 10, 11, 12 },                    { 13, 14, 15, 16 } };     // Function Call    vector res = spirallyTraverse(a);    int size = res.size();    for (int i = 0; i < size; ++i)        cout << res[i] << " ";    cout << endl;    return 0;} // code contributed by Ephi F

## Java

 // Java program for the above approach import java.io.*;import java.util.*; class GFG {    public static int R = 4, C = 4;    public static boolean isInBounds(int i, int j)    {        if (i < 0 || i >= R || j < 0 || j >= C)            return false;        return true;    }     // check if the position is blocked    public static boolean isBlocked(int[][] matrix, int i,                                    int j)    {        if (!isInBounds(i, j))            return true;        if (matrix[i][j] == -1)            return true;        return false;    }     // DFS code to traverse spirally    public static void    spirallyDFSTraverse(int[][] matrix, int i, int j,                         int dir, ArrayList res)    {        if (isBlocked(matrix, i, j))            return;        boolean allBlocked = true;        for (int k = -1; k <= 1; k += 2) {            allBlocked = allBlocked                         && isBlocked(matrix, k + i, j)                         && isBlocked(matrix, i, j + k);        }        res.add(matrix[i][j]);        matrix[i][j] = -1;        if (allBlocked) {            return;        }         // dir: 0 - right, 1 - down, 2 - left, 3 - up        int nxt_i = i;        int nxt_j = j;        int nxt_dir = dir;        if (dir == 0) {            if (!isBlocked(matrix, i, j + 1)) {                nxt_j++;            }            else {                nxt_dir = 1;                nxt_i++;            }        }        else if (dir == 1) {            if (!isBlocked(matrix, i + 1, j)) {                nxt_i++;            }            else {                nxt_dir = 2;                nxt_j--;            }        }        else if (dir == 2) {            if (!isBlocked(matrix, i, j - 1)) {                nxt_j--;            }            else {                nxt_dir = 3;                nxt_i--;            }        }        else if (dir == 3) {            if (!isBlocked(matrix, i - 1, j)) {                nxt_i--;            }            else {                nxt_dir = 0;                nxt_j++;            }        }        spirallyDFSTravserse(matrix, nxt_i, nxt_j, nxt_dir,                             res);    }     // to traverse spirally    public static ArrayList    spirallyTraverse(int[][] matrix)    {        ArrayList res = new ArrayList();        spirallyDFSTravserse(matrix, 0, 0, 0, res);        return res;    }     // 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 } };         // Function Call        ArrayList res = spirallyTraverse(a);        int size = res.size();        for (int i = 0; i < size; ++i)            System.out.print(res.get(i) + " ");        System.out.println();    }} // This code is contributed by Manu Pathria

## Python3

 # Python3 program for the above approach R = 4C = 4  def isInBounds(i, j):    global R    global C    if (i < 0 or i >= R or j < 0 or j >= C):        return False    return True # Check if the position is blocked  def isBlocked(matrix, i, j):    if (not isInBounds(i, j)):        return True    if (matrix[i][j] == -1):        return True     return False # DFS code to traverse spirally  def spirallyDFSTraverse(matrix, i, j, Dir, res):    if (isBlocked(matrix, i, j)):        return     allBlocked = True    for k in range(-1, 2, 2):        allBlocked = allBlocked and isBlocked(            matrix, k + i, j) and isBlocked(matrix, i, j + k)     res.append(matrix[i][j])    matrix[i][j] = -1    if (allBlocked):        return     # dir: 0 - right, 1 - down, 2 - left, 3 - up    nxt_i = i    nxt_j = j    nxt_dir = Dir    if (Dir == 0):        if (not isBlocked(matrix, i, j + 1)):            nxt_j += 1        else:            nxt_dir = 1            nxt_i += 1     elif(Dir == 1):        if (not isBlocked(matrix, i + 1, j)):            nxt_i += 1        else:            nxt_dir = 2            nxt_j -= 1     elif(Dir == 2):        if (not isBlocked(matrix, i, j - 1)):            nxt_j -= 1        else:            nxt_dir = 3            nxt_i -= 1     elif(Dir == 3):        if (not isBlocked(matrix, i - 1, j)):            nxt_i -= 1        else:            nxt_dir = 0            nxt_j += 1     spirallyDFSTravserse(matrix, nxt_i, nxt_j, nxt_dir, res) # To traverse spirally  def spirallyTraverse(matrix):    res = []    spirallyDFSTravserse(matrix, 0, 0, 0, res)    return res  # Driver codeif __name__ == "__main__":    a = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]     # Function Call    res = spirallyTraverse(a)    print(*res) # This code is contributed by rag2127

## C#

 // C# program for the above approach using System;using System.Collections.Generic; class GFG {     public static int R = 4, C = 4;    public static bool isInBounds(int i, int j)    {        if (i < 0 || i >= R || j < 0 || j >= C)            return false;         return true;    }     // Check if the position is blocked    public static bool isBlocked(int[, ] matrix, int i,                                 int j)    {        if (!isInBounds(i, j))            return true;        if (matrix[i, j] == -1)            return true;        return false;    }     // DFS code to traverse spirally    public static void spirallyDFSTraverse(int[, ] matrix,                                            int i, int j,                                            int dir,                                            List res)    {        if (isBlocked(matrix, i, j))            return;         bool allBlocked = true;         for (int k = -1; k <= 1; k += 2) {            allBlocked = allBlocked                         && isBlocked(matrix, k + i, j)                         && isBlocked(matrix, i, j + k);        }        res.Add(matrix[i, j]);        matrix[i, j] = -1;         if (allBlocked) {            return;        }         // dir: 0 - right, 1 - down, 2 - left, 3 - up        int nxt_i = i;        int nxt_j = j;        int nxt_dir = dir;        if (dir == 0) {            if (!isBlocked(matrix, i, j + 1)) {                nxt_j++;            }            else {                nxt_dir = 1;                nxt_i++;            }        }        else if (dir == 1) {            if (!isBlocked(matrix, i + 1, j)) {                nxt_i++;            }            else {                nxt_dir = 2;                nxt_j--;            }        }        else if (dir == 2) {            if (!isBlocked(matrix, i, j - 1)) {                nxt_j--;            }            else {                nxt_dir = 3;                nxt_i--;            }        }        else if (dir == 3) {            if (!isBlocked(matrix, i - 1, j)) {                nxt_i--;            }            else {                nxt_dir = 0;                nxt_j++;            }        }        spirallyDFSTravserse(matrix, nxt_i, nxt_j, nxt_dir,                             res);    }     // To traverse spirally    public static List spirallyTraverse(int[, ] matrix)    {        List res = new List();        spirallyDFSTravserse(matrix, 0, 0, 0, res);        return res;    }     // Driver code    static public void Main()    {        int[, ] a = { { 1, 2, 3, 4 },                      { 5, 6, 7, 8 },                      { 9, 10, 11, 12 },                      { 13, 14, 15, 16 } };         // Function Call        List res = spirallyTraverse(a);        int size = res.Count;         for (int i = 0; i < size; ++i)            Console.Write(res[i] + " ");         Console.WriteLine();    }} // This code is contributed by avanitrachhadiya2155

## Javascript



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

Time Complexity: O(M*N). To traverse the matrix O(M*N) time is required.
Auxiliary Space: O(1). No extra space is required (without consideration of the stack used by the recursion).

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

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