In this HackerRank Matrix Layer Rotation problem, You are given a 2D matrix of dimension m x n and a positive integer r. You have to rotate the matrix r times and print the resultant matrix. Rotation should be in the anti-clockwise direction.
Problem solution in Python programming.
M, N, R = map(int, input().split())
mx, my = [], []
for i in range(M):
mx.append([int(x) for x in input().split()])
for k in range(min(M, N) // 2):
tmp = []
i = j = k
m, n = M - k, N - k
while i < m - 1:
tmp.append(mx[i][j])
i += 1
while j < n - 1:
tmp.append(mx[i][j])
j += 1
while i > k:
tmp.append(mx[i][j])
i -= 1
while j > k:
tmp.append(mx[i][j])
j -= 1
my.append(tmp[-(R % len(tmp)):] + tmp[:-(R % len(tmp))])
for k in range(len(my)):
tmp = []
c = 0
i = j = k
m, n = M - k, N - k
while i < m - 1:
mx[i][j] = my[k][c]
i += 1
c += 1
while j < n - 1:
mx[i][j] = my[k][c]
j += 1
c += 1
while i > k:
mx[i][j] = my[k][c]
i -= 1
c += 1
while j > k:
mx[i][j] = my[k][c]
j -= 1
c += 1
[print(' '.join([str(b) for b in a])) for a in mx]
Problem solution in Java Programming.
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int m = sc.nextInt();
int n = sc.nextInt();
int r = sc.nextInt();
int total_len = m * n;
int matrix[] = new int[total_len + n];
int nr_row = 0, nr_col, level = 0;
for (int i = 0; i < m; i++) {
nr_row = i * 2 < m ? nr_row + 1 : i * 2 == m ? i : nr_row - 1;
nr_col = 0;
for (int j = 0; j < n; j++) {
nr_col = j * 2 < n ? nr_col + 1 : j * 2 == n ? j : nr_col - 1;
int nr = sc.nextInt();
int new_row = i, new_col = j;
level = Math.min(nr_row, nr_col) - 1;
int rotations = r % ((2 * ((m + n) - 4 * level)) - (m > 2 && n > 2 ? 4 : 0));
for (int k = 0; k < rotations; k++) {
if (new_row == level) {
if (new_col > level ) {
new_col -= 1;
} else {
new_row += 1;
}
} else if (new_row == m - level - 1) {
if (new_col < n - level - 1) {
new_col += 1;
} else {
new_row -= 1;
}
} else if (new_col == level) {
if (new_row < m - 1) {
new_row += 1;
} else {
new_col += 1;
}
} else if (new_col == n - level - 1) {
if (new_row > 0) {
new_row -= 1;
} else {
new_col -= 1;
}
}
}
matrix[new_row * n + new_col] = nr;
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
System.out.print(matrix[i * n + j] + " ");
}
System.out.println();
}
}
}
Problem solution in C++ programming.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int W,H,R;
cin>>H>>W>>R;
int ncycles=min(W,H)/2;
int m[W*H];
int cycles[150][1196];
int i,x,y;
int cycle,cyclelen;
for(i=0;i<W*H;i++)cin>>m[i];
for(cycle=0;cycle<ncycles;cycle++){
i=0;
x=y=cycle;
for(;y<H-cycle-1;y++)cycles[cycle][i++]=m[W*y+x];
for(;x<W-cycle-1;x++)cycles[cycle][i++]=m[W*y+x];
for(;y>cycle;y--)cycles[cycle][i++]=m[W*y+x];
for(;x>cycle;x--)cycles[cycle][i++]=m[W*y+x];
}
for(cycle=0;cycle<ncycles;cycle++){
cyclelen=2*(W-2*cycle)+2*(H-2*cycle)-4;
i=-R%cyclelen+cyclelen;
x=y=cycle;
for(;y<H-cycle-1;y++)m[W*y+x]=cycles[cycle][i++%cyclelen];
for(;x<W-cycle-1;x++)m[W*y+x]=cycles[cycle][i++%cyclelen];
for(;y>cycle;y--)m[W*y+x]=cycles[cycle][i++%cyclelen];
for(;x>cycle;x--)m[W*y+x]=cycles[cycle][i++%cyclelen];
}
for(y=0;y<H;y++){
for(x=0;x<W;x++){
if(x)cout<<' ';
cout<<m[W*y+x];
}
cout<<endl;
}
return 0;
}
Problem solution in C programming.
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int** matrix;
void init(int rows, int cols) {
matrix = malloc( sizeof( int*) * rows);
for (int row = 0; row < rows; row++) {
*(matrix + row) = malloc( sizeof(int) * cols);
}
//Load Matrices
int scan_var = -1;
for (int row = 0; row < rows; row++) {
for (int col = 0; col< cols; col++) {
scanf("%d ", &scan_var);
matrix[row][col] = scan_var;
}
}//end loading
}
void print(int** matrix, int rows, int cols) {
for (int row = 0; row < rows; row++) {
for (int col = 0; col< cols; col++) {
printf("%d ", matrix[row][col]);
}
printf("n");
}
}
void rotate(int** matrix, int top, int bottom, int left, int right) {
int temp;
temp = matrix[bottom][right];
//Bottom Row of matrix
for (int i = right; i >= left + 1; i--)
matrix[bottom][i] = matrix[bottom][i-1];
//Left Column of matrix
for (int i = bottom; i >= top + 1; i--)
matrix[i][left] = matrix[i-1][left];
//Top Row of matrix
for (int i = left; i <= right - 1; i++)
matrix[top][i] = matrix[top][i+1];
//Right Column of matrix
for (int i = top; i <= bottom - 1; i++)
matrix[i][right] = matrix[i+1][right];
matrix[bottom - 1][right] = temp;
}
int main(int argc, char* argv[]) {
int rows, cols, r;
scanf("%d %d %dn", &rows, &cols, &r);
init(rows, cols);
/*int rotations = 0;
for (int rotations = 0; rotations < r; rotations++) {
//rotate the entire matrix one "square" at a time
//moving in from outermost square
int i = 0;
while( i < (rows - 1 - i) && i < (cols - i - 1)) {
rotate(matrix, 0 + i, rows - i - 1, 0 + i, cols - 1 - i);
i++;
}
}*/
int i = 0;
int items_per_row, items_per_col, total_items, actual_rotations;
while( i < (rows - i - 1) && i < (cols - i - 1)) {
items_per_row = cols - (2 * i);
items_per_col = rows - (2 * i) - 2; //subtract 2 for items in row
total_items = items_per_row * 2 + items_per_col * 2;
actual_rotations = r % total_items;
for (int rotations = 0; rotations < actual_rotations; rotations++) {
rotate(matrix, 0 + i, rows - i - 1, 0 + i, cols - 1 - i);
}
i++;
}
print(matrix, rows, cols);
return 0;
}
Problem solution in JavaScript programming.
function valueAt(matrix, x, y, rot) {
var level, limits = { x:0, y:0 }, levels = { x: 0, y: 0};
levels.x = (x < matrix[y].length/2) ? x : matrix[y].length - x - 1;
levels.y = (y < matrix.length/2) ? y : matrix.length - y - 1;
level = Math.min(levels.x, levels.y);
limits.x = matrix[y].length-level-1;
limits.y = matrix.length-level-1;
var max = (matrix.length-level*2)*2 + (matrix[y].length-level*2)*2 - 4;
//process.stdout.write("n! " + max + " : " + rot + " -x " + limits.x + " -y " + limits.y + "n");
rot = rot % max;
for (; rot > 0; rot--)
if (y == level && x < limits.x) x += 1;
else if (x == limits.x && y < limits.y) y += 1;
else if (y == limits.y && x > level) x -= 1;
else if (x == level && y > level) y-=1;
return matrix[y][x];
}
function processData(input) {
var numbers, lines = input.split("n");
var dims = lines[0].split(' ').map(function(v) { return parseInt(v); });
var rows = dims[0];
var cols = dims[1];
var rots = dims[2];
var matrix = [];
for (var y = 0; y < lines.length-1; y++) {
numbers = lines[y+1].split(' ');
matrix.push(numbers);
//for (var x = 0; x < numbers.length; )
}
for (var y = 0; y < matrix.length; y++) {
for (var x = 0; x < matrix[y].length; x++)
process.stdout.write(valueAt(matrix, x, y, rots) + " ");
process.stdout.write("n");
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});
process.stdin.on("end", function () {
processData(_input);
});