Maximice la array binaria invirtiendo la subarray una vez

Dada una array binaria de R filas y C columnas. Se nos permite cambiar a cualquier tamaño de subarray. Voltear significa cambiar 1 a 0 y 0 a 1. La tarea es maximizar el número de 1 en la array. Salida el número máximo de 1s.

Ejemplos: 

Input : R = 3, C =3
mat[][] = { 0, 0, 1,
            0, 0, 1,
            1, 0, 1 }

Output : 8
Flip
0 0 1
0 0 1
1 0 1

to get

1 1 1
1 1 1
0 1 1

Input : R = 2, C = 3
mat[][] = { 0, 0, 0,
            0, 0, 0 }
Output : 6

Cree una array ones[][] de R filas y C columnas, que calcula previamente el número de unos en la subarray de (0, 0) a (i, j) por 

// Common elements in ones[i-1][j] and 
// ones[i][j-1] are ones[i-1][j-1]
ones[i][j] = ones[i-1][j] + ones[i][j-1] - 
             ones[i-1][j-1] + (mat[i][j] == 1)

Dado que se nos permite voltear la subarray solo una vez. Iteramos sobre todas las subarray posibles de todos los tamaños posibles para cada celda (i, j) a (i + k – 1, i + k – 1). Calculamos el número total de unos después de completar los dígitos en la subarray elegida.

El número total de unos en la array final después de cambiar la subarray (i, j) a (i + k – 1) será Unos en toda la array – Unos en la subarray elegida + Ceros en la subarray elegida. Eso resulta ser: – 

ones[R][C] - cal(i, j, i + k-1, j + k - 1) + k*k - cal(i, j, i + k - 1, j + k - 1) 
where cal(a, b, c, d) denotes the number of ones in square submatrix of length c - a.
Now cal(x1, y1, x2, y2) can be define by: 
ones[x2][y2] - ones[x2][y1 - 1] - ones[x1 - 1][y2] + ones[x1 - 1][y1 - 1].

A continuación se muestra la implementación de este enfoque: 

// C++ program to find maximum number of ones after
// one flipping in Binary Matrix
#include <bits/stdc++.h>
#define R 3
#define C 3
using namespace std;
 
// Return number of ones in square submatrix of size
// k x k starting from (x, y)
int cal(int ones[R + 1][C + 1], int x, int y, int k)
{
    return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1]
           - ones[x + k - 1][y - 1] + ones[x - 1][y - 1];
}
 
// Return maximum number of 1s after flipping a submatrix
int sol(int mat[R][C])
{
    int ans = 0;
 
    // Precomputing the number of 1s
    int ones[R + 1][C + 1] = {0};
    for (int i = 1; i <= R; i++)
        for (int j = 1; j <= C; j++)
            ones[i][j] = ones[i - 1][j] + ones[i][j - 1] -
                         ones[i - 1][j - 1] +
                         (mat[i - 1][j - 1] == 1);
 
    // Finding the maximum number of 1s after flipping
    for (int k = 1; k <= min(R, C); k++)
        for (int i = 1; i + k - 1 <= R; i++)
            for (int j = 1; j + k - 1 <= C; j++)
                ans = max(ans, (ones[R][C] + k * k -
                                2 * cal(ones, i, j, k)));
    return ans;
}
 
// Driver code
int main()
{
    int mat[R][C] = {{0, 0, 1},
        { 0, 0, 1},
        { 1, 0, 1 }
    };
 
    cout << sol(mat) << endl;
 
    return 0;
}

// Java program to find maximum number of ones after
// one flipping in Binary Matrix
class GFG {
 
static final int R = 3;
static final int C = 3 ;
 
 
// Return number of ones in square submatrix of size
// k x k starting from (x, y)
static int cal(int ones[][], int x, int y, int k)
{
    return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1]
        - ones[x + k - 1][y - 1] + ones[x - 1][y - 1];
}
 
// Return maximum number of 1s after flipping a submatrix
static int sol(int mat[][])
{
    int ans = 0;
        int val =0;
    // Precomputing the number of 1s
    int ones[][] = new int[R + 1][C + 1];
    for (int i = 1; i <= R; i++)
        for (int j = 1; j <= C; j++) {
                    if(mat[i - 1][j - 1] == 1)
                        val=1;
        ones[i][j] = ones[i - 1][j] + ones[i][j - 1] -
                        ones[i - 1][j - 1] +
                        (val);
                }
    // Finding the maximum number of 1s after flipping
    for (int k = 1; k <= Math.min(R, C); k++)
        for (int i = 1; i + k - 1 <= R; i++)
            for (int j = 1; j + k - 1 <= C; j++)
                ans = Math.max(ans, (ones[R][C] + k * k -
                                2 * cal(ones, i, j, k)));
    return ans;
}
 
// Driver code
    static public void main(String[] args) {
        int mat[][] = {{0, 0, 1},
        { 0, 0, 1},
        { 1, 0, 1 }
    };
 
       System.out.println(sol(mat));
    }
}
 
// This code is contributed by Rajput-Ji

# Python 3 program to find maximum number of
# ones after one flipping in Binary Matrix
R = 3
C = 3
 
# Return number of ones in square submatrix
# of size k x k starting from (x, y)
def cal(ones, x, y, k):
    return (ones[x + k - 1][y + k - 1] -
            ones[x - 1][y + k - 1] -
            ones[x + k - 1][y - 1] +
            ones[x - 1][y - 1])
 
# Return maximum number of 1s after
# flipping a submatrix
def sol(mat):
    ans = 0
 
    # Precomputing the number of 1s
    ones = [[0 for i in range(C + 1)]
               for i in range(R + 1)]
    for i in range(1, R + 1, 1):
        for j in range(1, C + 1, 1):
            ones[i][j] = (ones[i - 1][j] + ones[i][j - 1] -
                          ones[i - 1][j - 1] +
                         (mat[i - 1][j - 1] == 1))
 
    # Finding the maximum number of 1s
    # after flipping
    for k in range(1, min(R, C) + 1, 1):
        for i in range(1, R - k + 2, 1):
            for j in range(1, C - k + 2, 1):
                ans = max(ans, (ones[R][C] + k * k - 2 *
                            cal(ones, i, j, k)))
    return ans
 
# Driver code
if __name__ == '__main__':
    mat = [[0, 0, 1],
           [0, 0, 1],
           [1, 0, 1]]
 
    print(sol(mat))
 
# This code is contributed by
# Sahil_Shelangia

// C# program to find maximum number of ones after
// one flipping in Binary Matrix
using System;   
 
public class GFG {
 
    static readonly int R = 3;
    static readonly int C = 3 ;
 
 
    // Return number of ones in square submatrix of size
    // k x k starting from (x, y)
    static int cal(int [,]ones, int x, int y, int k)
    {
        return ones[x + k - 1,y + k - 1] - ones[x - 1,y + k - 1]
            - ones[x + k - 1,y - 1] + ones[x - 1,y - 1];
    }
 
    // Return maximum number of 1s after flipping a submatrix
    static int sol(int [,]mat)
    {
        int ans = 0;
            int val =0;
        // Precomputing the number of 1s
        int [,]ones = new int[R + 1,C + 1];
        for (int i = 1; i <= R; i++)
            for (int j = 1; j <= C; j++) {
                        if(mat[i - 1,j - 1] == 1)
                            val=1;
            ones[i,j] = ones[i - 1,j] + ones[i,j - 1] -
                            ones[i - 1,j - 1] +
                            (val);
                    }
        // Finding the maximum number of 1s after flipping
        for (int k = 1; k <= Math.Min(R, C); k++)
            for (int i = 1; i + k - 1 <= R; i++)
                for (int j = 1; j + k - 1 <= C; j++)
                    ans = Math.Max(ans, (ones[R,C] + k * k -
                                    2 * cal(ones, i, j, k)));
        return ans;
    }
 
    // Driver code
    static public void Main() {
        int [,]mat = {{0, 0, 1},
        { 0, 0, 1},
        { 1, 0, 1 }
    };
 
    Console.WriteLine(sol(mat));
    }
}
// This code is contributed by 29AjayKumar

<?php
// PHP program to find maximum number of ones after
// one flipping in Binary Matrix
 
$R = 3;
$C = 3;
 
// Return number of ones in square submatrix of size
// k x k starting from (x, y)
function cal($ones, $x, $y, $k)
{
    return $ones[$x + $k - 1][$y + $k - 1] -
            $ones[$x - 1][$y + $k - 1] -
            $ones[$x + $k - 1][$y - 1] +
            $ones[$x - 1][$y - 1];
}
 
// Return maximum number of 1s after flipping a submatrix
function sol($mat)
{
    global $C,$R;
    $ans = 0;
 
    // Precomputing the number of 1s
    $ones=array_fill(0,$R + 1,array_fill(0,$C + 1,0));
    for ($i = 1; $i <= $R; $i++)
        for ($j = 1; $j <= $C; $j++)
            $ones[$i][$j] = $ones[$i - 1][$j] + $ones[$i][$j - 1] -
                        $ones[$i - 1][$j - 1] +
                        (int)($mat[$i - 1][$j - 1] == 1);
 
    // Finding the maximum number of 1s after flipping
    for ($k = 1; $k <= min($R, $C); $k++)
        for ($i = 1; $i + $k - 1 <= $R; $i++)
            for ($j = 1; $j + $k - 1 <= $C; $j++)
                $ans = max($ans, ($ones[$R][$C] + $k * $k -
                                2 * cal($ones, $i, $j, $k)));
    return $ans;
}
 
    // Driver code
    $mat = array(array(0, 0, 1),
        array( 0, 0, 1),
        array( 1, 0, 1 ));
 
    echo sol($mat);
 
// This code is contributed by mits
?>

<script>
// Javascript program to find maximum number of ones after
// one flipping in Binary Matrix
 
let R = 3;
let C = 3 ;
   
   
// Return number of ones in square submatrix of size
// k x k starting from (x, y)
function cal(ones, x, y, k)
{
    return ones[x + k - 1][y + k - 1] - ones[x - 1][y + k - 1]
        - ones[x + k - 1][y - 1] + ones[x - 1][y - 1];
}
   
// Return maximum number of 1s after flipping a submatrix
function sol(mat)
{
    let ans = 0;
        let val =0;
    // Precomputing the number of 1s
    let ones = new Array(R + 1);
     
    // Loop to create 2D array using 1D array
    for (var i = 0; i < ones.length; i++) {
        ones[i] = new Array(2);
    }
     
    for (var i = 0; i < ones.length; i++) {
        for (var j = 0; j < ones.length; j++) {
        ones[i][j] = 0;
    }
    }
 
    for (let i = 1; i <= R; i++)
        for (let j = 1; j <= C; j++) {
                    if(mat[i - 1][j - 1] == 1)
                        val=1;
        ones[i][j] = ones[i - 1][j] + ones[i][j - 1] -
                        ones[i - 1][j - 1] +
                        (val);
                }
    // Finding the maximum number of 1s after flipping
    for (let k = 1; k <= Math.min(R, C); k++)
        for (let i = 1; i + k - 1 <= R; i++)
            for (let j = 1; j + k - 1 <= C; j++)
                ans = Math.max(ans, (ones[R][C] + k * k -
                                2 * cal(ones, i, j, k)));
    return ans;
}
     
// driver function
 
        let mat = [[0, 0, 1],
        [ 0, 0, 1],
        [ 1, 0, 1 ]
    ];
   
       document.write(sol(mat));
 
// This code is contributed by susmitakundugoaldanga.
</script>   

8

Complejidad de tiempo: O(R*C*min(R, C))

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