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Boundary elements of a Matrix
• Difficulty Level : Basic
• Last Updated : 28 Apr, 2021

Printing Boundary Elements of a Matrix.

Given a matrix of size n x m. Print the boundary elements of the matrix. Boundary elements are those elements which are not surrounded by elements on all four directions, i.e. elements in first row, first column, last row and last column.

Examples:

```Input :
1 2 3 4
5 6 7 8
1 2 3 4
5 6 7 8
Output :
1 2 3 4
5     8
1     4
5 6 7 8
Explanation:The boundary elements of the
matrix is printed.

Input:
1 2 3
5 6 7
1 2 3
Output:
1 2 3
5   7
1 2 3
Explanation:The boundary elements of the
matrix is printed.```

Approach: The idea is simple. Traverse the matrix and check for every element if that element lies on the boundary or not, if yes then print the element else print space character.

• Algorithm :
1. Traverse the array from start to end.
2. Assign the outer loop to point the row and the inner row to traverse the elements of row.
3. If the element lies in the boundary of matrix, then print the element, i.e. if the element lies in 1st row, 1st column, last row, last column
4. If the element is not boundary element print a blank space.
• Implementation:

## C++

 `// C++ program to print boundary element of``// matrix.``#include ``using` `namespace` `std;` `const` `int` `MAX = 100;` `void` `printBoundary(``int` `a[][MAX], ``int` `m, ``int` `n)``{``    ``for` `(``int` `i = 0; i < m; i++) {``        ``for` `(``int` `j = 0; j < n; j++) {``            ``if` `(i == 0 || j == 0 || i == n - 1 || j == n - 1)``                ``cout << a[i][j] << ``" "``;``            ``else``                ``cout << ``" "``                     ``<< ``" "``;``        ``}``        ``cout << ``"\n"``;``    ``}``}` `// Driver code``int` `main()``{``    ``int` `a[MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };``    ``printBoundary(a, 4, 4);``    ``return` `0;``}`

## Java

 `// JAVA Code for Boundary elements of a Matrix``class` `GFG {` `    ``public` `static` `void` `printBoundary(``int` `a[][], ``int` `m,``                                     ``int` `n)``    ``{``        ``for` `(``int` `i = ``0``; i < m; i++) {``            ``for` `(``int` `j = ``0``; j < n; j++) {``                ``if` `(i == ``0``)``                    ``System.out.print(a[i][j] + ``" "``);``                ``else` `if` `(i == m - ``1``)``                    ``System.out.print(a[i][j] + ``" "``);``                ``else` `if` `(j == ``0``)``                    ``System.out.print(a[i][j] + ``" "``);``                ``else` `if` `(j == n - ``1``)``                    ``System.out.print(a[i][j] + ``" "``);``                ``else``                    ``System.out.print(``"  "``);``            ``}``            ``System.out.println(``""``);``        ``}``    ``}` `    ``/* Driver program to test above function */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[][] = { { ``1``, ``2``, ``3``, ``4` `}, { ``5``, ``6``, ``7``, ``8` `}, { ``1``, ``2``, ``3``, ``4` `}, { ``5``, ``6``, ``7``, ``8` `} };` `        ``printBoundary(a, ``4``, ``4``);``    ``}``}``// This code is contributed by Arnav Kr. Mandal.`

## Python

 `# Python program to print boundary element``# of the matrix.` `MAX` `=` `100`` ` `def` `printBoundary(a, m, n):``    ``for` `i ``in` `range``(m):``        ``for` `j ``in` `range``(n):``            ``if` `(i ``=``=` `0``):``                ``print` `a[i][j],``            ``elif` `(i ``=``=` `m``-``1``):``                ``print` `a[i][j],``            ``elif` `(j ``=``=` `0``):``                ``print` `a[i][j],``            ``elif` `(j ``=``=` `n``-``1``):``                ``print` `a[i][j],``            ``else``:``                ``print` `" "``,``        ``print``        ` `# Driver code``a ``=` `[ [ ``1``, ``2``, ``3``, ``4` `], [ ``5``, ``6``, ``7``, ``8` `],``    ``[ ``1``, ``2``, ``3``, ``4` `], [ ``5``, ``6``, ``7``, ``8` `] ]``printBoundary(a, ``4``, ``4``)` `# This code is contributed by Sachin Bisht`

## C#

 `// C# Code for Boundary``// elements of a Matrix``using` `System;` `class` `GFG {` `    ``public` `static` `void` `printBoundary(``int``[, ] a,``                                     ``int` `m,``                                     ``int` `n)``    ``{``        ``for` `(``int` `i = 0; i < m; i++) {``            ``for` `(``int` `j = 0; j < n; j++) {``                ``if` `(i == 0)``                    ``Console.Write(a[i, j] + ``" "``);``                ``else` `if` `(i == m - 1)``                    ``Console.Write(a[i, j] + ``" "``);``                ``else` `if` `(j == 0)``                    ``Console.Write(a[i, j] + ``" "``);``                ``else` `if` `(j == n - 1)``                    ``Console.Write(a[i, j] + ``" "``);``                ``else``                    ``Console.Write(``"  "``);``            ``}``            ``Console.WriteLine(``" "``);``        ``}``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{``        ``int``[, ] a = { { 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 },``                      ``{ 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 } };` `        ``printBoundary(a, 4, 4);``    ``}``}` `// This code is contributed by ajit`

## PHP

 ``

Output:

```1 2 3 4
5     8
1     4
5 6 7 8```
• Complexity Analysis:
• Time Complexity: O(n*n), where n is the size of array.
This is achieved by single traversal of the matrix.
• Space Complexity: O(1).
Since a constant space is needed.

Finding sum of boundary elements

Given an matrix of size n x m. Find the sum of boundary elements of the matrix. Boundary elements are those elements which is not surrounded by elements on all four directions, i.e. elements in first row, first column, last row and last column.

Examples:

```Input :
1 2 3 4
5 6 7 8
1 2 3 4
5 6 7 8
Output : 54
Explanation:The boundary elements of the matrix
1 2 3 4
5     8
1     4
5 6 7 8
Sum = 1+2+3+4+5+8+1+4+5+6+7+8 =54

Input :
1 2 3
5 6 7
1 2 3
Output : 24
Explanation:The boundary elements of the matrix
1 2 3
5   7
1 2 3
Sum = 1+2+3+5+7+1+2+3 = 24```

Approach: The idea is simple. Traverse the matrix and check for every element if that element lies on the boundary or not, if yes then add them to get the sum of all the boundary elements.

• Algorithm :
1. Create a variable to store the sum and Traverse the array from start to end.
2. Assign the outer loop to point the row and the inner row to traverse the elements of row.
3. If the element lies in the boundary of matrix then add th element to the sum, i.e. if the element lies in 1st row, 1st column, last row, last column
4. print the sum.
• Implementation:

## C++

 `// C++ program to find sum of boundary elements``// of matrix.``#include ``using` `namespace` `std;` `const` `int` `MAX = 100;` `int` `getBoundarySum(``int` `a[][MAX], ``int` `m, ``int` `n)``{``    ``long` `long` `int` `sum = 0;``    ``for` `(``int` `i = 0; i < m; i++) {``        ``for` `(``int` `j = 0; j < n; j++) {``            ``if` `(i == 0)``                ``sum += a[i][j];``            ``else` `if` `(i == m - 1)``                ``sum += a[i][j];``            ``else` `if` `(j == 0)``                ``sum += a[i][j];``            ``else` `if` `(j == n - 1)``                ``sum += a[i][j];``        ``}``    ``}``    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `a[][MAX] = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 1, 2, 3, 4 }, { 5, 6, 7, 8 } };``    ``long` `long` `int` `sum = getBoundarySum(a, 4, 4);``    ``cout << ``"Sum of boundary elements is "` `<< sum;``    ``return` `0;``}`

## Java

 `// JAVA Code for Finding sum of boundary elements``class` `GFG {` `    ``public` `static` `long` `getBoundarySum(``int` `a[][], ``int` `m,``                                      ``int` `n)``    ``{``        ``long` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < m; i++) {``            ``for` `(``int` `j = ``0``; j < n; j++) {``                ``if` `(i == ``0``)``                    ``sum += a[i][j];``                ``else` `if` `(i == m - ``1``)``                    ``sum += a[i][j];``                ``else` `if` `(j == ``0``)``                    ``sum += a[i][j];``                ``else` `if` `(j == n - ``1``)``                    ``sum += a[i][j];``            ``}``        ``}``        ``return` `sum;``    ``}` `    ``/* Driver program to test above function */``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[][] = { { ``1``, ``2``, ``3``, ``4` `}, { ``5``, ``6``, ``7``, ``8` `}, { ``1``, ``2``, ``3``, ``4` `}, { ``5``, ``6``, ``7``, ``8` `} };``        ``long` `sum = getBoundarySum(a, ``4``, ``4``);``        ``System.out.println(``"Sum of boundary elements"``                           ``+ ``" is "` `+ sum);``    ``}``}``// This code is contributed by Arnav Kr. Mandal.`

## Python

 `# Python program to print boundary element``# of the matrix.` `MAX` `=` `100`` ` `def` `printBoundary(a, m, n):``    ``sum` `=` `0``    ``for` `i ``in` `range``(m):``        ``for` `j ``in` `range``(n):``            ``if` `(i ``=``=` `0``):``                ``sum` `+``=` `a[i][j]``            ``elif` `(i ``=``=` `m``-``1``):``                ``sum` `+``=` `a[i][j]``            ``elif` `(j ``=``=` `0``):``                ``sum` `+``=` `a[i][j]``            ``elif` `(j ``=``=` `n``-``1``):``                ``sum` `+``=` `a[i][j]``    ``return` `sum``    ` `# Driver code``a ``=` `[ [ ``1``, ``2``, ``3``, ``4` `], [ ``5``, ``6``, ``7``, ``8` `],``    ``[ ``1``, ``2``, ``3``, ``4` `], [ ``5``, ``6``, ``7``, ``8` `] ]``sum` `=` `printBoundary(a, ``4``, ``4``)``print` `"Sum of boundary elements is"``, ``sum` `# This code is contributed by Sachin Bisht`

## C#

 `// C# Code for Finding sum``// of boundary elements``using` `System;` `class` `GFG {``    ``public` `static` `long` `getBoundarySum(``int``[, ] a,``                                      ``int` `m, ``int` `n)``    ``{``        ``long` `sum = 0;``        ``for` `(``int` `i = 0; i < m; i++) {``            ``for` `(``int` `j = 0; j < n; j++) {``                ``if` `(i == 0)``                    ``sum += a[i, j];``                ``else` `if` `(i == m - 1)``                    ``sum += a[i, j];``                ``else` `if` `(j == 0)``                    ``sum += a[i, j];``                ``else` `if` `(j == n - 1)``                    ``sum += a[i, j];``            ``}``        ``}``        ``return` `sum;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main()``    ``{``        ``int``[, ] a = { { 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 },``                      ``{ 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 } };``        ``long` `sum = getBoundarySum(a, 4, 4);``        ``Console.WriteLine(``"Sum of boundary"``                          ``+ ``" elements is "` `+ sum);``    ``}``}` `// This code is contributed by ajit`

## PHP

 ``

## Javascript

 ``

Output:

`Sum of boundary elements is 54`
• Complexity Analysis:
• Time Complexity: O(n*n), where n is the size of the array.
This is achieved by single traversal of the matrix.
• Space Complexity: O(1).
Since a constant space is needed.

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