Boundary elements of a Matrix

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1. 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 that are not surrounded by elements in all four directions, i.e. elements in the 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

Input:
1 2 3
5 6 7
1 2 3

Output:

1 2 3
5 7
1 2 3

Approach: To solve the problem follow the below idea:

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

Follow the given steps to solve the problem:

Traverse the array from start to end
Assign the outer loop to point to the row and the inner row to traverse the elements of row
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
If the element is not a boundary element print a blank space

Below is the implementation of the above approach:

C++

// C++ program to print boundary element of
// matrix.
#include <bits/stdc++.h>
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 == m - 1
                || j == n - 1)
                cout << a[i][j] << " ";
            else
                cout << " "
                     << " ";
        }
        cout << "\n";
    }
}
 
// Driver code
int main()
{
    int a[4][MAX] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 } };
 
    // Function call
    printBoundary(a, 4, 4);
    return 0;
}

Java

// JAVA Code for Boundary elements of a Matrix
import java.io.*;
public 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 code */
    public static void main(String[] args)
    {
        int a[][] = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 } };
 
        // Function call
        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
if __name__ == "__main__":
    a = [[1, 2, 3, 4], [5, 6, 7, 8],
         [1, 2, 3, 4], [5, 6, 7, 8]]
 
    # Function call
    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 } };
 
        // Function call
        printBoundary(a, 4, 4);
    }
}
 
// This code is contributed by ajit

PHP

<?php
// PHP program to print 
// boundary element of 
// matrix.
$MAX = 100;
 
function printBoundary($a, $m, $n)
{
    global $MAX;
    for ($i = 0; $i < $m; $i++) 
    {
        for ($j = 0; $j < $n; $j++) 
        {
            if ($i == 0)
                echo $a[$i][$j], " ";
            else if ($i == $m - 1)
                echo $a[$i][$j], " ";
            else if ($j == 0)
                echo $a[$i][$j], " ";
            else if ($j == $n - 1)
                echo $a[$i][$j], " ";
            else
                echo " ", " ";
        }
        echo "\n";
    }
}
 
// Driver code
$a = array(array( 1, 2, 3, 4 ), 
           array( 5, 6, 7, 8 ), 
           array( 1, 2, 3, 4 ), 
           array( 5, 6, 7, 8 ));
 
// Function call
printBoundary($a, 4, 4);
     
// This code is contributed 
// by akt_mit
?>

Javascript

<script>
   
// JavaScript Code for Boundary
// elements of a Matrix
function printBoundary(a, m, n)
{
    for (var i = 0; i < m; i++) {
        for (var j = 0; j < n; j++) {
            if (i == 0)
                document.write(a[i][j] + '\xa0');
            else if (i == m - 1)
                document.write(a[i][j] + '\xa0');
            else if (j == 0)
                document.write(a[i][j] + '\xa0');
            else if (j == n - 1)
                document.write(a[i][j] + '\xa0');
            else
                document.write("\xa0\xa0\xa0");
        }
        document.write("\xa0<br>");
    }
}
// Driver Code
var a = [ [ 1, 2, 3, 4 ],
              [ 5, 6, 7, 8 ],
              [ 1, 2, 3, 4 ],
              [ 5, 6, 7, 8 ]];
printBoundary(a, 4, 4);
 
 
</script>

Output

Time Complexity: O(N²), where N is the size of the array.
Auxiliary Space: O(1)

2. Finding sum of boundary elements:

Given a matrix of size n x m. Find the sum of boundary elements of the matrix. Boundary elements are those elements which are not surrounded by elements in all four directions, i.e. elements in the 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

To solve the problem follow the below idea:

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

Follow the given steps to solve the problem:

Create a variable to store the sum and Traverse the array from start to end
Assign the outer loop to point to the row and the inner row to traverse the elements of the row
If the element lies in the boundary of the matrix then add the element to the sum, i.e. if the element lies in the 1st row, 1st column, last row, and last column
Print the sum

Below is the implementation of the above approach:

C++

// C++ program to find sum of boundary elements
// of matrix.
#include <bits/stdc++.h>
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 } };
 
    // Function call
    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
import java.io.*;
public 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 */
    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);
 
        // Function call
        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
if __name__ == "__main__":
    a = [[1, 2, 3, 4], [5, 6, 7, 8],
         [1, 2, 3, 4], [5, 6, 7, 8]]
 
    # Function call
    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 } };
 
        // Function call
        long sum = getBoundarySum(a, 4, 4);
        Console.WriteLine("Sum of boundary"
                          + " elements is " + sum);
    }
}
 
// This code is contributed by ajit

PHP

<?php
// PHP program to find 
// sum of boundary 
// elements of matrix.
 
function getBoundarySum($a, 
                        $m, $n)
{
    $sum = 0;
    for ($i = 0; $i < $m; $i++)
    {
        for ( $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
$a = array(array(1, 2, 3, 4), 
           array(5, 6, 7, 8), 
           array(1, 2, 3, 4), 
           array(5, 6, 7, 8));
            
// Function call
$sum = getBoundarySum($a, 4, 4);
echo "Sum of boundary elements is ", $sum;
 
// This code is contributed by ajit
?>

Javascript

<script>
 
// Javascript code for finding sum
// of boundary elements
function getBoundarySum(a, m, n)
{
    let sum = 0;
    for(let i = 0; i < m; i++)
    {
        for(let 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
let a = [ [ 1, 2, 3, 4 ], 
          [ 5, 6, 7, 8 ], 
          [ 1, 2, 3, 4 ], 
          [ 5, 6, 7, 8 ] ];
let sum = getBoundarySum(a, 4, 4);
 
document.write("Sum of boundary elements" + 
               " is " + sum);
                
// This code is contributed by rameshtravel07  
 
</script>

Output

Sum of boundary elements is 54

Time Complexity: O(N²), where N is the size of the array
Auxiliary Space: O(1)

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Last Updated : 25 May, 2023