# Spiral Pattern

Given a number N, the task is to print the following pattern:-

Examples:

```Input : N = 4
Output : 4 4 4 4 4 4 4
4 3 3 3 3 3 4
4 3 2 2 2 3 4
4 3 2 1 2 3 4
4 3 2 2 2 3 4
4 3 3 3 3 3 4
4 4 4 4 4 4 4

Input : N = 2
Output : 2 2 2
2 1 2
2 2 2
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach 1: The common observation is that the square thus formed will be of size (2*N-1)x(2*N-1). Fill the first row and column, last row and column with N, and then gradually decrease N and fill the remaining rows and columns similarly. Decrease N every time after filling 2 rows and 2 columns.

Below is the implementation of the above approach:

## C++

 `// C++ program to print the ` `// spiral pattern ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print the pattern ` `void` `pattern(``int` `value) ` `{ ` `    ``// Declare a square matrix ` `    ``int` `row = 2 * value - 1; ` `    ``int` `column = 2 * value - 1; ` `    ``int` `arr[row][column]; ` ` `  `    ``int` `i, j, k; ` ` `  `    ``for` `(k = 0; k < value; k++) { ` ` `  `        ``// store the first row ` `        ``// from 1st column to last column ` `        ``j = k; ` `        ``while` `(j < column - k) { ` `            ``arr[k][j] = value - k; ` `            ``j++; ` `        ``} ` ` `  `        ``// store the last column ` `        ``// from top to bottom ` `        ``i = k + 1; ` `        ``while` `(i < row - k) { ` `            ``arr[i][row - 1 - k] = value - k; ` `            ``i++; ` `        ``} ` ` `  `        ``// store the last row ` `        ``// from last column to 1st column ` `        ``j = column - k - 2; ` `        ``while` `(j >= k) { ` `            ``arr[column - k - 1][j] = value - k; ` `            ``j--; ` `        ``} ` ` `  `        ``// store the first column ` `        ``// from bottom to top ` `        ``i = row - k - 2; ` `        ``while` `(i > k) { ` `            ``arr[i][k] = value - k; ` `            ``i--; ` `        ``} ` `    ``} ` ` `  `    ``// print the pattern ` `    ``for` `(i = 0; i < row; i++) { ` `        ``for` `(j = 0; j < column; j++) { ` `            ``cout << arr[i][j] << ``" "``; ` `        ``} ` `        ``cout << endl; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 5; ` `    ``pattern(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to print ` `// the spiral pattern ` `class` `GFG { ` ` `  `    ``// Function to print the pattern ` `    ``static` `void` `pattern(``int` `value) ` `    ``{ ` `        ``// Declare a square matrix ` `        ``int` `row = ``2` `* value - ``1``; ` `        ``int` `column = ``2` `* value - ``1``; ` `        ``int``[][] arr = ``new` `int``[row][column]; ` ` `  `        ``int` `i, j, k; ` ` `  `        ``for` `(k = ``0``; k < value; k++) { ` ` `  `            ``// store the first row ` `            ``// from 1st column to last column ` `            ``j = k; ` `            ``while` `(j < column - k) { ` `                ``arr[k][j] = value - k; ` `                ``j++; ` `            ``} ` ` `  `            ``// store the last column ` `            ``// from top to bottom ` `            ``i = k + ``1``; ` `            ``while` `(i < row - k) { ` `                ``arr[i][row - ``1` `- k] = value - k; ` `                ``i++; ` `            ``} ` ` `  `            ``// store the last row ` `            ``// from last column ` `            ``// to 1st column ` `            ``j = column - k - ``2``; ` `            ``while` `(j >= k) { ` `                ``arr[column - k - ``1``][j] = value - k; ` `                ``j--; ` `            ``} ` ` `  `            ``// store the first column ` `            ``// from bottom to top ` `            ``i = row - k - ``2``; ` `            ``while` `(i > k) { ` `                ``arr[i][k] = value - k; ` `                ``i--; ` `            ``} ` `        ``} ` ` `  `        ``// print the pattern ` `        ``for` `(i = ``0``; i < row; i++) { ` `            ``for` `(j = ``0``; j < column; j++) { ` `                ``System.out.print(arr[i][j] + ``" "``); ` `            ``} ` `            ``System.out.println(); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``5``; ` `        ``pattern(n); ` `    ``} ` `} ` ` `  `// This code is contributed ` `// by ChitraNayal `

## Python 3

 `# Python 3 program to print  ` `# the spiral pattern ` ` `  `# Function to print the pattern ` `def` `pattern(value): ` `     `  `    ``# Declare a square matrix ` `    ``row ``=` `2` `*` `value ``-` `1` `    ``column ``=` `2` `*` `value ``-` `1` `    ``arr ``=` `[[``0` `for` `i ``in` `range``(row)]  ` `              ``for` `j ``in` `range` `(column)] ` ` `  `    ``for` `k ``in` `range``( value): ` ` `  `        ``# store the first row ` `        ``# from 1st column to  ` `        ``# last column ` `        ``j ``=` `k ` `        ``while` `(j < column ``-` `k): ` `            ``arr[k][j] ``=` `value ``-` `k ` `            ``j ``+``=` `1` ` `  `        ``# store the last column ` `        ``# from top to bottom ` `        ``i ``=` `k ``+` `1` `        ``while` `(i < row ``-` `k): ` `            ``arr[i][row ``-` `1` `-` `k] ``=` `value ``-` `k ` `            ``i ``+``=` `1` ` `  `        ``# store the last row ` `        ``# from last column  ` `        ``# to 1st column ` `        ``j ``=` `column ``-` `k ``-` `2` `        ``while` `j >``=` `k : ` `            ``arr[column ``-` `k ``-` `1``][j] ``=` `value ``-` `k ` `            ``j ``-``=` `1` ` `  `        ``# store the first column ` `        ``# from bottom to top ` `        ``i ``=` `row ``-` `k ``-` `2` `        ``while` `i > k : ` `            ``arr[i][k] ``=` `value ``-` `k ` `            ``i ``-``=` `1` ` `  `    ``# print the pattern ` `    ``for` `i ``in` `range``(row): ` `        ``for` `j ``in` `range``(column): ` `            ``print``(arr[i][j], end ``=` `" "``) ` `        ``print``() ` `     `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``n ``=` `5` `    ``pattern(n) ` ` `  `# This code is contributed ` `# by ChitraNayal `

## C#

 `// C# program to print ` `// the spiral pattern ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to print the pattern ` `    ``static` `void` `pattern(``int` `value) ` `    ``{ ` ` `  `        ``// Declare a square matrix ` `        ``int` `row = 2 * value - 1; ` `        ``int` `column = 2 * value - 1; ` `        ``int``[, ] arr = ``new` `int``[row, column]; ` ` `  `        ``int` `i, j, k; ` ` `  `        ``for` `(k = 0; k < value; k++) { ` ` `  `            ``// store the first row ` `            ``// from 1st column to ` `            ``// last column ` `            ``j = k; ` `            ``while` `(j < column - k) { ` `                ``arr[k, j] = value - k; ` `                ``j++; ` `            ``} ` ` `  `            ``// store the last column ` `            ``// from top to bottom ` `            ``i = k + 1; ` `            ``while` `(i < row - k) { ` `                ``arr[i, row - 1 - k] = value - k; ` `                ``i++; ` `            ``} ` ` `  `            ``// store the last row ` `            ``// from last column ` `            ``// to 1st column ` `            ``j = column - k - 2; ` `            ``while` `(j >= k) { ` `                ``arr[column - k - 1, j] = value - k; ` `                ``j--; ` `            ``} ` ` `  `            ``// store the first column ` `            ``// from bottom to top ` `            ``i = row - k - 2; ` `            ``while` `(i > k) { ` `                ``arr[i, k] = value - k; ` `                ``i--; ` `            ``} ` `        ``} ` ` `  `        ``// print the pattern ` `        ``for` `(i = 0; i < row; i++) { ` `            ``for` `(j = 0; j < column; j++) { ` `                ``Console.Write(arr[i, j] + ``" "``); ` `            ``} ` `            ``Console.Write(``"\n"``); ` `        ``} ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 5; ` `        ``pattern(n); ` `    ``} ` `} ` ` `  `// This code is contributed ` `// by ChitraNayal `

## PHP

 `= ``\$k``)  ` `        ``{ ` `            ``\$arr``[``\$column` `- ``\$k` `- 1][``\$j``] = ``\$value` `- ``\$k``; ` `            ``\$j``--; ` `        ``} ` ` `  `        ``// store the first column ` `        ``// from bottom to top ` `        ``\$i` `= ``\$row` `- ``\$k` `- 2; ` `        ``while` `(``\$i` `> ``\$k``)  ` `        ``{ ` `            ``\$arr``[``\$i``][``\$k``] = ``\$value` `- ``\$k``; ` `            ``\$i``--; ` `        ``} ` `    ``} ` ` `  `    ``// print the pattern ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$row``; ``\$i``++) ` `    ``{ ` `        ``for` `(``\$j` `= 0; ``\$j` `< ``\$column``; ``\$j``++) ` `        ``{ ` `            ``echo` `\$arr``[``\$i``][``\$j``] . ``" "``; ` `        ``} ` `        ``echo` `"\n"``; ` `    ``} ` `} ` ` `  `// Driver code ` `\$n` `= 5; ` `pattern(``\$n``); ` ` `  `// This code is contributed  ` `// by ChitraNayal ` `?> `

Output:

```5 5 5 5 5 5 5 5 5
5 4 4 4 4 4 4 4 5
5 4 3 3 3 3 3 4 5
5 4 3 2 2 2 3 4 5
5 4 3 2 1 2 3 4 5
5 4 3 2 2 2 3 4 5
5 4 3 3 3 3 3 4 5
5 4 4 4 4 4 4 4 5
5 5 5 5 5 5 5 5 5
```

Approach 2: Starting the indexing from i = 1 and j = 1, it can be observed that every value of the required matrix will be max(abs(i – n), abs(j – n)) + 1.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `// Function to print the required pattern ` `void` `pattern(``int` `n) ` `{ ` ` `  `    ``// Calculating boundary size ` `    ``int` `p = 2 * n - 1; ` ` `  `    ``for` `(``int` `i = 1; i <= p; i++) { ` `        ``for` `(``int` `j = 1; j <= p; j++) { ` ` `  `            ``// Printing the values ` `            ``cout << max(``abs``(i - n), ``abs``(j - n)) + 1 << ``" "``; ` `        ``} ` `        ``cout << endl; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 5; ` ` `  `    ``pattern(n); ` ` `  `    ``return` `0; ` `} ` `// This code is contributed by : Vivek kothari `

Output:

```5 5 5 5 5 5 5 5 5
5 4 4 4 4 4 4 4 5
5 4 3 3 3 3 3 4 5
5 4 3 2 2 2 3 4 5
5 4 3 2 1 2 3 4 5
5 4 3 2 2 2 3 4 5
5 4 3 3 3 3 3 4 5
5 4 4 4 4 4 4 4 5
5 5 5 5 5 5 5 5 5
```

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