HackerRank Flipping the Matrix problem solution

In this HackerRank Flipping the Matrix problem solution, Sean invented a game involving a 2n x 2n matrix where each cell of the matrix contains an integer. He can reverse any of its rows or columns any number of times. The goal of the game is to maximize the sum of the elements in the n x n submatrix located in the upper-left quadrant of the matrix.

Given the initial configurations for q matrices, help Sean reverse the rows and columns of each matrix in the best possible way so that the sum of the elements in the matrix’s upper-left quadrant is maximal.

HackerRank Flipping the Matrix problem solution in Python.

import math
q = int(input().strip())

for a0 in range(q):
    n = int(input().strip())
    a =[[0 for j in range(2*n)] for i in range(2*n)]
    
    for x in range(2 * n):
        c = [int(c_temp) for c_temp in input().strip().split(' ')]        
        a[x] = c
     
    
    v = 0
    for i in range(n):
        for j in range(n):
            l = []
            l.append(a[i][j]) # current matrix
            l.append(a[2 * n - 1 - i][j])  # bottom left
            l.append(a[i][2 * n - 1- j]) # top right
            l.append(a[2* n - 1 - i][2 * n - 1- j]) # bottom right
    
            maxv = max(l)
            #print(l)
            #print(max(l))
            
            v += maxv

    print(v)

Flipping the Matrix problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner in = new Scanner(System.in);
        int queries = in.nextInt();
        int size = 0;
        int length;
        int total;
        for(int i = 0; i < queries; i++){
            total = 0;
            size = in.nextInt();
            length = size * 2;
            int[][] matrix = new int[length][length];
            for(int row = 0; row < length; row++){
                for(int column = 0; column < length; column++){
                    matrix[column][row] = in.nextInt();
                }
            }
            int max = 0;
            for(int row = 0; row < size; row++){
                for(int column = 0; column < size; column++){
                    max = Integer.MIN_VALUE;
                    if(matrix[row][column] > max)
                        max = matrix[row][column];
                    if(matrix[row][length - column - 1] > max)
                        max = matrix[row][length - column - 1];
                    if(matrix[length - row - 1][column] > max)
                        max = matrix[length - row - 1][column];
                    if(matrix[length - row - 1][length - column - 1] > max)
                        max = matrix[length - row - 1][length - column - 1];
                    total += max;
                }
            }
            System.out.println(total);
        }
    }
}

Problem solution in C++.

#include <cstdio>
#include <cassert>
#include <algorithm>
using namespace std;

const int MAXN = 300;

int matrix[MAXN][MAXN];

void Work() {
  int n;
  assert(scanf("%d", &n) == 1);
  int halfn = n;
  n *= 2;

  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      assert(scanf("%d", &matrix[i][j]) == 1);
    }
  }

  int ans = 0;
  for (int i = 0; i < halfn; ++i) {
    for (int j = 0; j < halfn; ++j) {
      int t_max = matrix[i][j];
      int ti = i;
      int tj = j;

      ti = n - 1 - ti;
      t_max = max(t_max, matrix[ti][tj]);

      tj = n - 1 - tj;
      t_max = max(t_max, matrix[ti][tj]);

      ti = n - 1 - ti;
      t_max = max(t_max, matrix[ti][tj]);

      ans += t_max;
    }
  }
  printf("%dn", ans);
}

int main() {
  int q;
  assert(scanf("%d", &q) == 1);
  while (q--) {
    Work();
  }
  return 0;
}

Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int maxm(int a,int b,int c,int d){
    int max;
    if(a>=b && a>=c && a>=d){
        max=a;
    }
    else if(a<=b && b>=c && b>=d)
        max=b;
    else if(c>=b && a<=c && c>=d)
        max=c;
   else
        max=d;
   return max;
        }
int main() {
    int q;
    scanf("%d",&q);
    while(q--){
        int n,i,j;
        scanf("%d",&n);
        int x[2*n][2*n];
        for(i=0;i<2*n;i++){
            for(j=0;j<2*n;j++)
                scanf("%d",&x[i][j]);
        }
        int m=2*n;
        int sum=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                sum+=maxm(x[i][j],x[i][m-1-j],x[m-1-i][j],x[m-1-i][m-1-j]);
            }
        }
        printf("%dn",sum);
    }

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}