# 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 ` `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 ` `#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

 `= ``\$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 ` `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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Article Tags :

76

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.