# Flipping the Matrix

### Problem Statement :

```Sean invented a game involving a 2n * 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 * 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.

Example
matrix = [[1,2],[3,4]]
1 2
3 4
It is 2*2 and we want to maximize the top left quadrant, a 1*1 matrix. Reverse row 1:

1 2
4 3
And now reverse column 0:

4 2
1 3
The maximal sum is 4.

Function Description

Complete the flippingMatrix function in the editor below.

flippingMatrix has the following parameters:
- int matrix[2n][2n]: a 2-dimensional array of integers

Returns
- int: the maximum sum possible.

Input Format

The first line contains an integer q, the number of queries.

The next q sets of lines are in the following format:

The first line of each query contains an integer, n.
Each of the next 2n lines contains 2n space-separated integers matrix[i][j] in row i of the matrix.

Constraints
1 <= q <= 16
1 <= n <= 128
0 <= matrix[i][j] <= 4096, where 0 <= i,j < 2n.```

### Solution :

```                            ```Solution in C :

In C++ :

#include <bits/stdc++.h>

#define FI(i,a,b) for(int i=(a);i<=(b);i++)
#define FD(i,a,b) for(int i=(a);i>=(b);i--)

#define LL long long
#define Ldouble long double
#define PI 3.1415926535897932384626

#define PII pair<int,int>
#define PLL pair<LL,LL>
#define mp make_pair
#define fi first
#define se second

using namespace std;

int q, n, s[299][299];

int main(){
scanf("%d", &q);
while(q--){
scanf("%d", &n);
FI(i, 1, 2 * n) FI(j, 1, 2 * n) scanf("%d", &s[i][j]);

int ans = 0;
FI(i, 1, n) FI(j, 1, n){
int i2 = n + n + 1 - i;
int j2 = n + n + 1 - j;
int mx = max(max(max(s[i][j], s[i][j2]), s[i2][j]), s[i2][j2]);

ans += mx;
}
printf("%d\n", ans);
}
return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args)throws IOException {
for(int t1=0;t1<t;t1++){
String[][] s=new String[2*n][2*n];
for(int i=0;i<2*n;i++)
long sum=0;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
sum+=Math.max(Math.max(Integer.parseInt(s[i][j]),
Integer.parseInt(s[2*n-1-i][j])),
Math.max(Integer.parseInt(s[i][2*n-1-j]),
Integer.parseInt(s[2*n-1-i][2*n-1-j])));
}
}
System.out.println(sum);
}
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

int q;
scanf("%d",&q);
while(q--){
int n;
scanf("%d",&n);
int matrix[2*n][2*n];
int sol[2*n][2*n];
for(int i=0;i<2*n;i++){
for(int j=0;j<2*n;j++){
scanf("%d",&matrix[i][j]);
sol[i][j]=0;
}
}
int sum=0;
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
int a = matrix[i][j];
if(matrix[2*n-1-i][2*n-1-j]>a && sol[2*n-1-i][2*n-1-j]==0){
a=matrix[2*n-1-i][2*n-1-j];
sol[2*n-1-i][2*n-1-j]++;
}
if(matrix[i][2*n-1-j]>a && sol[i][2*n-1-j]==0){
a=matrix[i][2*n-1-j];
sol[i][2*n-1-j]++;
}
if(matrix[2*n-1-i][j]>a && sol[2*n-1-i][j]==0){
a=matrix[2*n-1-i][j];
sol[2*n-1-i][j]++;
}
sum+=a;
}
}
printf("%d\n",sum);
}
return 0;
}

In Python3 :

q = int(input())
for _ in range(q):
n = int(input())
a = []
for y in range(2*n):
a.append([int(x) for x in input().split()])
suma = 0
for i in range(n):
for j in range(n):
suma += max(max(a[i][j],a[2*n-i-1][j]),max(a[i][2*n-j-1],a[2*n-i-1][2*n-j-1]))
print(suma)```
```

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