# Find perimeter of shapes formed with 1s in binary matrix

Given a matrix of N rows and M columns, consist of 0’s and 1’s. The task is to find the perimeter of subfigure consisting only 1’s in the matrix. Perimeter of single 1 is 4 as it can be covered from all 4 side. Perimeter of double 11 is 6.

```

|  1  |           |  1    1  |

```

Examples:

```Input : mat[][] =
{
1, 0,
1, 1,
}
Output : 8
Cell (1,0) and (1,1) making a L shape whose perimeter is 8.

Input :  mat[][] =
{
0, 1, 0, 0, 0,
1, 1, 1, 0, 0,
1, 0, 0, 0, 0
}
Output : 12 ```

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

The idea is to traverse the matrix, find all ones and find their contribution in perimeter. The maximum contribution of a 1 is four if it is surrounded by all 0s. The contribution reduces by one with 1 around it.

Algorithm for solving this problem:

1. Traverse the whole matrix and find the cell having value equal to 1.
2. Calculate the number of closed side for that cell and add, 4 – number of closed side to the total perimeter.

Below is the implementation of this approach:

## C++

 `// C++ program to find perimeter of area coverede by ` `// 1 in 2D matrix consisits of 0's and  1's. ` `#include ` `using` `namespace` `std; ` `#define R 3 ` `#define C 5 ` ` `  `// Find the number of covered side for mat[i][j]. ` `int` `numofneighbour(``int` `mat[][C], ``int` `i, ``int` `j) ` `{ ` `    ``int` `count = 0; ` ` `  `    ``// UP ` `    ``if` `(i > 0 && mat[i - 1][j]) ` `        ``count++; ` ` `  `    ``// LEFT ` `    ``if` `(j > 0 && mat[i][j - 1]) ` `        ``count++; ` ` `  `    ``// DOWN ` `    ``if` `(i < R-1 && mat[i + 1][j]) ` `        ``count++; ` ` `  `    ``// RIGHT ` `    ``if` `(j < C-1 && mat[i][j + 1]) ` `        ``count++; ` ` `  `    ``return` `count; ` `} ` ` `  `// Returns sum of perimeter of shapes formed with 1s ` `int` `findperimeter(``int` `mat[R][C]) ` `{ ` `    ``int` `perimeter = 0; ` ` `  `    ``// Traversing the matrix and finding ones to ` `    ``// calculate their contribution. ` `    ``for` `(``int` `i = 0; i < R; i++) ` `        ``for` `(``int` `j = 0; j < C; j++) ` `            ``if` `(mat[i][j]) ` `                ``perimeter += (4 - numofneighbour(mat, i ,j)); ` ` `  `    ``return` `perimeter; ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `    ``int` `mat[R][C] = ` `    ``{ ` `        ``0, 1, 0, 0, 0, ` `        ``1, 1, 1, 0, 0, ` `        ``1, 0, 0, 0, 0, ` `    ``}; ` ` `  `    ``cout << findperimeter(mat) << endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find perimeter of area ` `// coverede by 1 in 2D matrix consisits  ` `// of 0's and 1's ` `class` `GFG { ` `     `  `    ``static` `final` `int` `R = ``3``; ` `    ``static` `final` `int` `C = ``5``; ` `     `  `    ``// Find the number of covered side  ` `    ``// for mat[i][j]. ` `    ``static` `int` `numofneighbour(``int` `mat[][],  ` `                            ``int` `i, ``int` `j)  ` `    ``{ ` `         `  `        ``int` `count = ``0``; ` `     `  `        ``// UP ` `        ``if` `(i > ``0` `&& mat[i - ``1``][j] == ``1``) ` `            ``count++; ` `     `  `        ``// LEFT ` `        ``if` `(j > ``0` `&& mat[i][j - ``1``] == ``1``) ` `            ``count++; ` `     `  `        ``// DOWN ` `        ``if` `(i < R - ``1` `&& mat[i + ``1``][j] == ``1``) ` `            ``count++; ` `     `  `        ``// RIGHT ` `        ``if` `(j < C - ``1` `&& mat[i][j + ``1``] == ``1``) ` `            ``count++; ` `     `  `        ``return` `count; ` `    ``} ` `     `  `    ``// Returns sum of perimeter of shapes ` `    ``// formed with 1s ` `    ``static` `int` `findperimeter(``int` `mat[][])  ` `    ``{ ` `         `  `        ``int` `perimeter = ``0``; ` `     `  `        ``// Traversing the matrix and  ` `        ``// finding ones to calculate  ` `        ``// their contribution. ` `        ``for` `(``int` `i = ``0``; i < R; i++) ` `            ``for` `(``int` `j = ``0``; j < C; j++) ` `                ``if` `(mat[i][j] == ``1``) ` `                    ``perimeter += (``4` `-  ` `                    ``numofneighbour(mat, i, j)); ` `     `  `        ``return` `perimeter; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `mat[][] = {{``0``, ``1``, ``0``, ``0``, ``0``},  ` `                       ``{``1``, ``1``, ``1``, ``0``, ``0``},  ` `                       ``{``1``, ``0``, ``0``, ``0``, ``0``}}; ` `                        `  `        ``System.out.println(findperimeter(mat)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python 3

 `# Python 3 program to find perimeter of area  ` `# covered by 1 in 2D matrix consisits of 0's and 1's. ` ` `  `R ``=` `3` `C ``=` `5` ` `  `# Find the number of covered side for mat[i][j]. ` `def` `numofneighbour(mat, i, j): ` ` `  `    ``count ``=` `0``; ` ` `  `    ``# UP ` `    ``if` `(i > ``0` `and` `mat[i ``-` `1``][j]): ` `        ``count``+``=` `1``; ` ` `  `    ``# LEFT ` `    ``if` `(j > ``0` `and` `mat[i][j ``-` `1``]): ` `        ``count``+``=` `1``; ` ` `  `    ``# DOWN ` `    ``if` `(i < R``-``1` `and` `mat[i ``+` `1``][j]): ` `        ``count``+``=` `1` ` `  `    ``# RIGHT ` `    ``if` `(j < C``-``1` `and` `mat[i][j ``+` `1``]): ` `        ``count``+``=` `1``; ` ` `  `    ``return` `count; ` ` `  `# Returns sum of perimeter of shapes formed with 1s ` `def` `findperimeter(mat): ` ` `  `    ``perimeter ``=` `0``; ` ` `  `    ``# Traversing the matrix and finding ones to ` `    ``# calculate their contribution. ` `    ``for` `i ``in` `range``(``0``, R): ` `        ``for` `j ``in` `range``(``0``, C): ` `            ``if` `(mat[i][j]): ` `                ``perimeter ``+``=` `(``4` `-` `numofneighbour(mat, i, j)); ` ` `  `    ``return` `perimeter; ` ` `  `# Driver Code ` `mat ``=` `[ [``0``, ``1``, ``0``, ``0``, ``0``], ` `        ``[``1``, ``1``, ``1``, ``0``, ``0``], ` `        ``[``1``, ``0``, ``0``, ``0``, ``0``] ] ` ` `  `print``(findperimeter(mat), end``=``"\n"``); ` ` `  `# This code is contributed by Akanksha Rai `

## C#

 `using` `System; ` ` `  `// C# program to find perimeter of area  ` `// coverede by 1 in 2D matrix consisits   ` `// of 0's and 1's  ` `public` `class` `GFG ` `{ ` ` `  `    ``public`  `const` `int` `R = 3; ` `    ``public` `const` `int` `C = 5; ` ` `  `    ``// Find the number of covered side   ` `    ``// for mat[i][j].  ` `    ``public` `static` `int` `numofneighbour(``int``[][] mat, ``int` `i, ``int` `j) ` `    ``{ ` ` `  `        ``int` `count = 0; ` ` `  `        ``// UP  ` `        ``if` `(i > 0 && mat[i - 1][j] == 1) ` `        ``{ ` `            ``count++; ` `        ``} ` ` `  `        ``// LEFT  ` `        ``if` `(j > 0 && mat[i][j - 1] == 1) ` `        ``{ ` `            ``count++; ` `        ``} ` ` `  `        ``// DOWN  ` `        ``if` `(i < R - 1 && mat[i + 1][j] == 1) ` `        ``{ ` `            ``count++; ` `        ``} ` ` `  `        ``// RIGHT  ` `        ``if` `(j < C - 1 && mat[i][j + 1] == 1) ` `        ``{ ` `            ``count++; ` `        ``} ` ` `  `        ``return` `count; ` `    ``} ` ` `  `    ``// Returns sum of perimeter of shapes  ` `    ``// formed with 1s  ` `    ``public` `static` `int` `findperimeter(``int``[][] mat) ` `    ``{ ` ` `  `        ``int` `perimeter = 0; ` ` `  `        ``// Traversing the matrix and   ` `        ``// finding ones to calculate   ` `        ``// their contribution.  ` `        ``for` `(``int` `i = 0; i < R; i++) ` `        ``{ ` `            ``for` `(``int` `j = 0; j < C; j++) ` `            ``{ ` `                ``if` `(mat[i][j] == 1) ` `                ``{ ` `                    ``perimeter += (4 - numofneighbour(mat, i, j)); ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``return` `perimeter; ` `    ``} ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``int``[][] mat = ``new` `int``[][] ` `        ``{ ` `            ``new` `int``[] {0, 1, 0, 0, 0}, ` `            ``new` `int``[] {1, 1, 1, 0, 0}, ` `            ``new` `int``[] {1, 0, 0, 0, 0} ` `        ``}; ` ` `  `        ``Console.WriteLine(findperimeter(mat)); ` `    ``} ` `} ` ` `  `// This code is contributed by Shrikant13 `

## PHP

 ` 0 && (``\$mat``[``\$i` `- 1][``\$j``]))  ` `        ``\$count``++;  ` ` `  `    ``// LEFT  ` `    ``if` `(``\$j` `> 0 && (``\$mat``[``\$i``][``\$j` `- 1]))  ` `        ``\$count``++;  ` ` `  `    ``// DOWN  ` `    ``if` `((``\$i` `< ``\$R``-1 )&& (``\$mat``[``\$i` `+ 1][``\$j``]))  ` `        ``\$count``++;  ` ` `  `    ``// RIGHT  ` `    ``if` `((``\$j` `< ``\$C``-1) && (``\$mat``[``\$i``][``\$j` `+ 1]))  ` `        ``\$count``++;  ` ` `  `    ``return` `\$count``;  ` `}  ` ` `  `// Returns sum of perimeter of shapes ` `// formed with 1s  ` `function` `findperimeter(``\$mat``)  ` `{  ` `    ``global` `\$R``;  ` `    ``global` `\$C``; ` `    ``\$perimeter` `= 0;  ` ` `  `    ``// Traversing the matrix and finding ones  ` `    ``// to calculate their contribution.  ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$R``; ``\$i``++)  ` `        ``for` `( ``\$j` `= 0; ``\$j` `< ``\$C``; ``\$j``++)  ` `            ``if` `(``\$mat``[``\$i``][``\$j``])  ` `                ``\$perimeter` `+= (4 -  ` `                ``numofneighbour(``\$mat``, ``\$i``, ``\$j``));  ` ` `  `    ``return` `\$perimeter``;  ` `}  ` ` `  `// Driver Code ` `\$mat` `= ``array``(``array``(0, 1, 0, 0, 0),  ` `             ``array``(1, 1, 1, 0, 0),  ` `             ``array``(1, 0, 0, 0, 0));  ` ` `  `echo` `findperimeter(``\$mat``), ``"\n"``;  ` ` `  `// This code is contributed by Sach_Code ` `?> `

Output:

```12
```

Time Complexity : O(RC).

This article is contributed by Anuj Chauhan(anuj0503). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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 arrow_drop_up

Article Tags :
Practice Tags :

1

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