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++
#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";
}
}
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
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("");
}
}
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);
}
}
|
Python
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
if __name__ == "__main__":
a = [[1, 2, 3, 4], [5, 6, 7, 8],
[1, 2, 3, 4], [5, 6, 7, 8]]
printBoundary(a, 4, 4)
|
C#
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(" ");
}
}
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);
}
}
|
PHP
<?php
$MAX = 100;
function printBoundary($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";
}
}
$a = array(array( 1, 2, 3, 4 ),
array( 5, 6, 7, 8 ),
array( 1, 2, 3, 4 ),
array( 5, 6, 7, 8 ));
printBoundary($a, 4, 4);
?>
|
Javascript
<script>
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>");
}
}
var a = [ [ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ],
[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8 ]];
printBoundary(a, 4, 4);
</script>
|
Output1 2 3 4
5 8
1 4
5 6 7 8
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++
#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;
}
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
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;
}
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);
}
}
|
Python
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
if __name__ == "__main__":
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
|
C#
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;
}
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);
}
}
|
PHP
<?php
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;
}
$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;
?>
|
Javascript
<script>
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;
}
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);
</script>
|
OutputSum 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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