Interchange elements of first and last rows in matrix
Given a 4 x 4 matrix, we have to interchange the elements of the first and last row and show the resulting matrix
Examples:
Input: 3 4 5 0
2 6 1 2
2 7 1 2
2 1 1 2Output: 2 1 1 2
2 6 1 2
2 7 1 2
3 4 5 0Input: 9 7 5 1
2 3 4 1
5 6 6 5
1 2 3 1Output: 1 2 3 1
2 3 4 1
5 6 6 5
9 7 5 1
Interchange elements of first and last rows in the matrix using the swap function:
To solve the problem follow the below idea:
The approach is very simple, we can simply swap the elements of the first and last row of the matrix in order to get the desired matrix as output
Below is the implementation of the approach:
C++
// C++ code to swap the element of first // and last row and display the result #include <iostream> using namespace std; #define n 4 void interchangeFirstLast( int m[][n]) { int rows = n; // swapping of element between first // and last rows for ( int i = 0; i < n; i++) { int t = m[0][i]; m[0][i] = m[rows - 1][i]; m[rows - 1][i] = t; } } // Driver code int main() { // input in the array int m[n][n] = { { 8, 9, 7, 6 }, { 4, 7, 6, 5 }, { 3, 2, 1, 8 }, { 9, 9, 7, 7 } }; // Function call interchangeFirstLast(m); // printing the interchanged matrix for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) cout << m[i][j] << " " ; cout << endl; } } // This code is contributed by Anant Agarwal. |
Java
// Java code to swap the element of first // and last row and display the result import java.io.*; public class Interchange { static void interchangeFirstLast( int m[][]) { int rows = m.length; // swapping of element between first // and last rows for ( int i = 0 ; i < m[ 0 ].length; i++) { int t = m[ 0 ][i]; m[ 0 ][i] = m[rows - 1 ][i]; m[rows - 1 ][i] = t; } } // Driver code public static void main(String args[]) throws IOException { // input in the array int m[][] = { { 8 , 9 , 7 , 6 }, { 4 , 7 , 6 , 5 }, { 3 , 2 , 1 , 8 }, { 9 , 9 , 7 , 7 } }; // Function call interchangeFirstLast(m); // printing the interchanged matrix for ( int i = 0 ; i < m.length; i++) { for ( int j = 0 ; j < m[ 0 ].length; j++) System.out.print(m[i][j] + " " ); System.out.println(); } } } |
Python3
# Python3 code to swap the element # of first and last row and display # the result def interchangeFirstLast(mat, n, m): rows = n # swapping of element between # first and last rows for i in range (n): t = mat[ 0 ][i] mat[ 0 ][i] = mat[rows - 1 ][i] mat[rows - 1 ][i] = t # Driver code if __name__ = = "__main__" : mat = [[ 8 , 9 , 7 , 6 ], [ 4 , 7 , 6 , 5 ], [ 3 , 2 , 1 , 8 ], [ 9 , 9 , 7 , 7 ]] n = 4 m = 4 # Function call interchangeFirstLast(mat, n, m) # printing the interchanged matrix for i in range (n): for j in range (m): print (mat[i][j], end = " " ) print ( "\n" ) # This code is contributed by Shrikant13. |
C#
// C# code to swap the element of first // and last row and display the result using System; class GFG { public static void interchangeFirstLast( int [][] m) { int rows = m.Length; // swapping of element between first // and last rows for ( int i = 0; i < m[0].Length; i++) { int t = m[0][i]; m[0][i] = m[rows - 1][i]; m[rows - 1][i] = t; } } // Driver code public static void Main( string [] args) { // input in the array int [][] m = new int [][] { new int [] { 8, 9, 7, 6 }, new int [] { 4, 7, 6, 5 }, new int [] { 3, 2, 1, 8 }, new int [] { 9, 9, 7, 7 } }; // Function call interchangeFirstLast(m); // printing the interchanged matrix for ( int i = 0; i < m.Length; i++) { for ( int j = 0; j < m[0].Length; j++) { Console.Write(m[i][j] + " " ); } Console.WriteLine(); } } } // This code is contributed by Shrikant13 |
PHP
<?php // PHP code to swap the element of first // and last row and display the result $n = 4; function interchangeFirstLast(& $m ) { global $n ; $rows = $n ; // swapping of element between first // and last rows for ( $i = 0; $i < $n ; $i ++) { $t = $m [0][ $i ]; $m [0][ $i ] = $m [ $rows - 1][ $i ]; $m [ $rows - 1][ $i ] = $t ; } } // Driver code // input in the array $m = array ( array (8, 9, 7, 6), array (4, 7, 6, 5), array ( 3, 2, 1, 8), array (9, 9, 7, 7)); // Function call interchangeFirstLast( $m ); // printing the interchanged matrix for ( $i = 0; $i < $n ; $i ++) { for ( $j = 0; $j < $n ; $j ++) echo $m [ $i ][ $j ] . " " ; echo "\n" ; } ?> |
Javascript
<script> // Javascript code to swap the element of first // and last row and display the result function interchangeFirstLast(m) { let rows = m.length; // swapping of element between first // and last rows for (let i = 0; i < m[0].length; i++) { let t = m[0][i]; m[0][i] = m[rows-1][i]; m[rows-1][i] = t; } } // Driver code // input in the array let m = [[8, 9, 7, 6], [4, 7, 6, 5], [3, 2, 1, 8], [9, 9, 7, 7]] interchangeFirstLast(m); // printing the interchanged matrix for (let i = 0; i < m.length; i++) { for (let j = 0; j < m[0].length; j++) document.write(m[i][j] + " " ); document.write( "<br>" ); } // This code is contributed by avanitrachhadiya2155 </script> |
Output
9 9 7 7 4 7 6 5 3 2 1 8 8 9 7 6
Time Complexity: O(N)
Auxiliary Space: O(1)
Please Login to comment...