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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 <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 == n - 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 } };
    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



<?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
$a = array(array( 1, 2, 3, 4 ),
           array( 5, 6, 7, 8 ),
           array( 1, 2, 3, 4 ),
           array( 5, 6, 7, 8 ));
printBoundary($a, 4, 4);
     
// This code is contributed
// by akt_mit
?>
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 <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 } };
    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



<?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));
            
$sum = getBoundarySum($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
  • 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.
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