Given a matrix arr[][] of size M*N, where M is the number of rows and N is the number of columns. The task is to reverse the rows and columns of the matrix alternatively i.e start with reversing the 1st row, then the 2nd column, and so on.
Examples:
Input: arr[][] = {
{3, 4, 1, 8},
{11, 23, 43, 21},
{12, 17, 65, 91},
{71, 56, 34, 24}
}
Output: {
{8, 56, 4, 24},
{11, 17, 43, 12},
{91, 65, 23, 21},
{71, 1, 34, 3}
}
Explanation: Operations to be followed:
- Reverse the first row
- Reverse the second column
- Reverse the third row
- Reverse the fourth row
Input: { {11, 23, 43, 21}, {12, 17, 65, 91}, {71, 56, 34, 24} }
Output: { {21, 56, 23, 71}, {12, 17, 65, 91}, {24, 34, 43, 11} }
Approach: The task can be solved by simply running two while loops for traversing rows and columns alternatively. In the end, print the resultant matrix.
Below is the implementation of the above approach:
C++
// C++ program to find Reverse the// rows and columns of a matrix alternatively.#include <bits/stdc++.h>using namespace std;const int N = 4;const int M = 4;// Print matrix elementsvoid showArray(int arr[][N]){ for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) cout << arr[i][j] << " "; cout << endl; }}// Function to Reverse the rows and columns// of a matrix alternatively.void reverseAlternate(int arr[][N]){ int turn = 0; while (turn < M && turn < N) { if (turn % 2 == 0) { int start = 0, end = N - 1, temp; while (start < end) { temp = arr[turn][start]; arr[turn][start] = arr[turn][end]; arr[turn][end] = temp; start += 1; end -= 1; } turn += 1; } if (turn % 2 == 1) { int start = 0, end = M - 1, temp; while (start < end) { temp = arr[start][turn]; arr[start][turn] = arr[end][turn]; arr[end][turn] = temp; start += 1; end -= 1; } turn += 1; } }}// Driver codeint main(){ int matrix[][N] = { { 3, 4, 1, 8 }, { 11, 23, 43, 21 }, { 12, 17, 65, 91 }, { 71, 56, 34, 24 } }; reverseAlternate(matrix); showArray(matrix);} |
Java
// Java program for the above approachimport java.util.*;public class GFG{ static int N = 4; static int M = 4; // Print matrix elements static void showArray(int arr[][]) { for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) System.out.print(arr[i][j] + " "); System.out.println(); } } // Function to Reverse the rows and columns // of a matrix alternatively. static void reverseAlternate(int arr[][]) { int turn = 0; while (turn < M && turn < N) { if (turn % 2 == 0) { int start = 0, end = N - 1, temp; while (start < end) { temp = arr[turn][start]; arr[turn][start] = arr[turn][end]; arr[turn][end] = temp; start += 1; end -= 1; } turn += 1; } if (turn % 2 == 1) { int start = 0, end = M - 1, temp; while (start < end) { temp = arr[start][turn]; arr[start][turn] = arr[end][turn]; arr[end][turn] = temp; start += 1; end -= 1; } turn += 1; } } } // Driver Code public static void main(String args[]) { int matrix[][] = { { 3, 4, 1, 8 }, { 11, 23, 43, 21 }, { 12, 17, 65, 91 }, { 71, 56, 34, 24 } }; reverseAlternate(matrix); showArray(matrix); }}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python3 program to find Reverse the # rows and columns of a matrix alternatively.N = 4M = 4# Print matrix elementsdef showArray(arr): for i in range(M): for j in range(N): print(arr[i][j], end = " ") print()# Function to Reverse the rows and columns# of a matrix alternatively.def reverseAlternate(arr): turn = 0 while turn < M and turn < N: if (turn % 2 == 0): start = 0 end = N - 1 while (start < end): temp = arr[turn][start] arr[turn][start] = arr[turn][end] arr[turn][end] = temp start += 1 end -= 1 turn += 1 if (turn % 2 == 1): start = 0 end = M - 1 while (start < end): temp = arr[start][turn] arr[start][turn] = arr[end][turn] arr[end][turn] = temp start += 1 end -= 1 turn += 1# Driver codematrix = [ [ 3, 4, 1, 8 ], [ 11, 23, 43, 21 ], [ 12, 17, 65, 91 ], [ 71, 56, 34, 24 ] ] reverseAlternate(matrix)showArray(matrix)# This code is contributed by Potta Lokesh |
C#
// C# program to find Reverse the// rows and columns of a matrix alternatively.using System;class GFG { const int N = 4; const int M = 4; // Print matrix elements static void showArray(int[, ] arr) { for (int i = 0; i < M; i++) { for (int j = 0; j < N; j++) Console.Write(arr[i, j] + " "); Console.WriteLine(); } } // Function to Reverse the rows and columns // of a matrix alternatively. static void reverseAlternate(int[, ] arr) { int turn = 0; while (turn < M && turn < N) { if (turn % 2 == 0) { int start = 0, end = N - 1, temp; while (start < end) { temp = arr[turn, start]; arr[turn, start] = arr[turn, end]; arr[turn, end] = temp; start += 1; end -= 1; } turn += 1; } if (turn % 2 == 1) { int start = 0, end = M - 1, temp; while (start < end) { temp = arr[start, turn]; arr[start, turn] = arr[end, turn]; arr[end, turn] = temp; start += 1; end -= 1; } turn += 1; } } } // Driver code public static void Main() { int[, ] matrix = { { 3, 4, 1, 8 }, { 11, 23, 43, 21 }, { 12, 17, 65, 91 }, { 71, 56, 34, 24 } }; reverseAlternate(matrix); showArray(matrix); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript program to find Reverse the // rows and columns of a matrix alternatively. const N = 4; const M = 4; // Print matrix elements const showArray = (arr) => { for (let i = 0; i < M; i++) { for (let j = 0; j < N; j++) document.write(`${arr[i][j]} `); document.write("<br/>"); } } // Function to Reverse the rows and columns // of a matrix alternatively. const reverseAlternate = (arr) => { let turn = 0; while (turn < M && turn < N) { if (turn % 2 == 0) { let start = 0, end = N - 1, temp; while (start < end) { temp = arr[turn][start]; arr[turn][start] = arr[turn][end]; arr[turn][end] = temp; start += 1; end -= 1; } turn += 1; } if (turn % 2 == 1) { let start = 0, end = M - 1, temp; while (start < end) { temp = arr[start][turn]; arr[start][turn] = arr[end][turn]; arr[end][turn] = temp; start += 1; end -= 1; } turn += 1; } } } // Driver code let matrix = [[3, 4, 1, 8], [11, 23, 43, 21], [12, 17, 65, 91], [71, 56, 34, 24]]; reverseAlternate(matrix); showArray(matrix); // This code is contributed by rakeshsahni</script> |
Output
8 56 4 24 11 17 43 12 91 65 23 21 71 1 34 3
Time Complexity: O(M*N)
Space Complexity: O(1), no additional extra space is used.
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