Dada una array cuadrada, encuentre el producto máximo de cuatro elementos adyacentes de la array. Los elementos adyacentes de la array pueden ser arriba, abajo, izquierda, derecha, diagonal o antidiagonal. Los cuatro o más números deben estar adyacentes entre sí.
Nota: n debe ser mayor o igual a 4, es decir, n >= 4
C++
// C++ program to find out the maximum product
// in the matrix which four elements are
// adjacent to each other in one direction
#include <bits/stdc++.h>
using namespace std;
const int n = 5;
// function to find max product
int FindMaxProduct(int arr[][n], int n)
{
int max = 0, result;
// iterate the rows.
for (int i = 0; i < n; i++)
{
// iterate the columns.
for (int j = 0; j < n; j++)
{
// check the maximum product
// in horizontal row.
if ((j - 3) >= 0)
{
result = arr[i][j] * arr[i][j - 1] *
arr[i][j - 2] * arr[i][j - 3];
if (max < result)
max = result;
}
// check the maximum product
// in vertical row.
if ((i - 3) >= 0)
{
result = arr[i][j] * arr[i - 1][j] *
arr[i - 2][j] * arr[i - 3][j];
if (max < result)
max = result;
}
// check the maximum product in
// diagonal (going through down - right)
if ((i - 3) >= 0 && (j - 3) >= 0)
{
result = arr[i][j] * arr[i - 1][j - 1] *
arr[i - 2][j - 2] * arr[i - 3][j - 3];
if (max < result)
max = result;
}
// check the maximum product in
// diagonal (going through up - right)
if ((i - 3) >= 0 && (j - 1) <= 0)
{
result = arr[i][j] * arr[i - 1][j + 1] *
arr[i - 2][j + 2] * arr[i - 3][j + 3];
if (max < result)
max = result;
}
}
}
return max;
}
// Driver code
int main()
{
/* int arr[][4] = {{6, 2, 3, 4},
{5, 4, 3, 1},
{7, 4, 5, 6},
{8, 3, 1, 0}};*/
/* int arr[][5] = {{1, 2, 1, 3, 4},
{5, 6, 3, 9, 2},
{7, 8, 8, 1, 2},
{1, 0, 7, 9, 3},
{3, 0, 8, 4, 9}};*/
int arr[][5] = {{1, 2, 3, 4, 5},
{6, 7, 8, 9, 1},
{2, 3, 4, 5, 6},
{7, 8, 9, 1, 0},
{9, 6, 4, 2, 3}};
cout << FindMaxProduct(arr, n);
return 0;
}
Java
// Java program to find out the
// maximum product in the matrix
// which four elements are adjacent
// to each other in one direction
class GFG {
static final int n = 5;
// function to find max product
static int FindMaxProduct(int arr[][], int n)
{
int max = 0, result;
// iterate the rows.
for (int i = 0; i < n; i++)
{
// iterate the columns.
for (int j = 0; j < n; j++)
{
// check the maximum product
// in horizontal row.
if ((j - 3) >= 0)
{
result = arr[i][j] * arr[i][j - 1]
* arr[i][j - 2]
* arr[i][j - 3];
if (max < result)
max = result;
}
// check the maximum product
// in vertical row.
if ((i - 3) >= 0)
{
result = arr[i][j] * arr[i - 1][j]
* arr[i - 2][j]
* arr[i - 3][j];
if (max < result)
max = result;
}
// check the maximum product in
// diagonal (going through down - right)
if ((i - 3) >= 0 && (j - 3) >= 0)
{
result = arr[i][j] * arr[i - 1][j - 1]
* arr[i - 2][j - 2]
* arr[i - 3][j - 3];
if (max < result)
max = result;
}
// check the maximum product in
// diagonal (going through up - right)
if ((i - 3) >= 0 && (j - 1) <= 0)
{
result = arr[i][j] * arr[i - 1][j + 1]
* arr[i - 2][j + 2]
* arr[i - 3][j + 3];
if (max < result)
max = result;
}
}
}
return max;
}
// Driver code
public static void main(String[] args)
{
/* int arr[][4] = {{6, 2, 3, 4},
{5, 4, 3, 1},
{7, 4, 5, 6},
{8, 3, 1, 0}};*/
/* int arr[][5] = {{1, 2, 1, 3, 4},
{5, 6, 3, 9, 2},
{7, 8, 8, 1, 2},
{1, 0, 7, 9, 3},
{3, 0, 8, 4, 9}};*/
int arr[][] = { { 1, 2, 3, 4, 5 },
{ 6, 7, 8, 9, 1 },
{ 2, 3, 4, 5, 6 },
{ 7, 8, 9, 1, 0 },
{ 9, 6, 4, 2, 3 } };
System.out.print(FindMaxProduct(arr, n));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 program to find out the maximum
# product in the matrix which four elements
# are adjacent to each other in one direction
n = 5
# function to find max product
def FindMaxProduct(arr, n):
max = 0
# iterate the rows.
for i in range(n):
# iterate the columns.
for j in range( n):
# check the maximum product
# in horizontal row.
if ((j - 3) >= 0):
result = (arr[i][j] * arr[i][j - 1] *
arr[i][j - 2] * arr[i][j - 3])
if (max < result):
max = result
# check the maximum product
# in vertical row.
if ((i - 3) >= 0) :
result = (arr[i][j] * arr[i - 1][j] *
arr[i - 2][j] * arr[i - 3][j])
if (max < result):
max = result
# check the maximum product in
# diagonal going through down - right
if ((i - 3) >= 0 and (j - 3) >= 0):
result = (arr[i][j] * arr[i - 1][j - 1] *
arr[i - 2][j - 2] * arr[i - 3][j - 3])
if (max < result):
max = result
# check the maximum product in
# diagonal going through up - right
if ((i - 3) >= 0 and (j - 1) <= 0):
result = (arr[i][j] * arr[i - 1][j + 1] *
arr[i - 2][j + 2] * arr[i - 3][j + 3])
if (max < result):
max = result
return max
# Driver code
if __name__ == "__main__":
# int arr[][4] = {{6, 2, 3, 4},
# {5, 4, 3, 1},
# {7, 4, 5, 6},
# {8, 3, 1, 0}};
# int arr[][5] = {{1, 2, 1, 3, 4},
# {5, 6, 3, 9, 2},
# {7, 8, 8, 1, 2},
# {1, 0, 7, 9, 3},
# {3, 0, 8, 4, 9}};
arr = [[1, 2, 3, 4, 5],
[6, 7, 8, 9, 1],
[2, 3, 4, 5, 6],
[7, 8, 9, 1, 0],
[9, 6, 4, 2, 3]]
print(FindMaxProduct(arr, n))
# This code is contributed by ita_c
C#
// C# program to find out the
// maximum product in the matrix
// which four elements are adjacent
// to each other in one direction
using System;
public class GFG {
static int n = 5;
// Function to find max product
static int FindMaxProduct(int[, ] arr, int n)
{
int max = 0, result;
// iterate the rows
for (int i = 0; i < n; i++) {
// iterate the columns
for (int j = 0; j < n; j++) {
// check the maximum product
// in horizontal row.
if ((j - 3) >= 0) {
result = arr[i, j] * arr[i, j - 1]
* arr[i, j - 2]
* arr[i, j - 3];
if (max < result)
max = result;
}
// check the maximum product
// in vertical row.
if ((i - 3) >= 0) {
result = arr[i, j] * arr[i - 1, j]
* arr[i - 2, j]
* arr[i - 3, j];
if (max < result)
max = result;
}
// check the maximum product in
// diagonal going through down - right
if ((i - 3) >= 0 && (j - 3) >= 0) {
result = arr[i, j] * arr[i - 1, j - 1]
* arr[i - 2, j - 2]
* arr[i - 3, j - 3];
if (max < result)
max = result;
}
// check the maximum product in
// diagonal going through up - right
if ((i - 3) >= 0 && (j - 1) <= 0) {
result = arr[i, j] * arr[i - 1, j + 1]
* arr[i - 2, j + 2]
* arr[i - 3, j + 3];
if (max < result)
max = result;
}
}
}
return max;
}
// Driver Code
static public void Main()
{
int[, ] arr = { { 1, 2, 3, 4, 5 },
{ 6, 7, 8, 9, 1 },
{ 2, 3, 4, 5, 6 },
{ 7, 8, 9, 1, 0 },
{ 9, 6, 4, 2, 3 } };
Console.Write(FindMaxProduct(arr, n));
}
}
// This code is contributed by Shrikant13
PHP
<?php
// PHP program to find out the maximum product
// in the matrix which four elements are
// adjacent to each other in one direction
$n = 5;
// function to find max product
function FindMaxProduct( $arr, $n)
{
$max = 0; $result;
// iterate the rows.
for ( $i = 0; $i < $n; $i++)
{
// iterate the columns.
for ( $j = 0; $j < $n; $j++)
{
// check the maximum product
// in horizontal row.
if (($j - 3) >= 0)
{
$result = $arr[$i][$j] *
$arr[$i][$j - 1] *
$arr[$i][$j - 2] *
$arr[$i][$j - 3];
if ($max < $result)
$max = $result;
}
// check the maximum product
// in vertical row.
if (($i - 3) >= 0)
{
$result = $arr[$i][$j] *
$arr[$i - 1][$j] *
$arr[$i - 2][$j] *
$arr[$i - 3][$j];
if ($max < $result)
$max = $result;
}
// check the maximum product in
// diagonal going through down - right
if (($i - 3) >= 0 and ($j - 3) >= 0)
{
$result = $arr[$i][$j] *
$arr[$i - 1][$j - 1] *
$arr[$i - 2][$j - 2] *
$arr[$i - 3][$j - 3];
if ($max < $result)
$max = $result;
}
// check the maximum product in
// diagonal going through up - right
if (($i - 3) >= 0 and ($j - 1) <= 0)
{
$result = $arr[$i][$j] *
$arr[$i - 1][$j + 1] *
$arr[$i - 2][$j + 2] *
$arr[$i - 3][$j + 3];
if ($max < $result)
$max = $result;
}
}
}
return $max;
}
// Driver Code
$arr = array(array(1, 2, 3, 4, 5),
array(6, 7, 8, 9, 1),
array(2, 3, 4, 5, 6),
array(7, 8, 9, 1, 0),
array(9, 6, 4, 2, 3));
echo FindMaxProduct($arr, $n);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// Javascript program to find out the
// maximum product in the matrix
// which four elements are adjacent
// to each other in one direction
let n = 5;
// function to find max product
function FindMaxProduct(arr,n)
{
let max = 0, result;
// iterate the rows.
for (let i = 0; i < n; i++)
{
// iterate the columns.
for (let j = 0; j < n; j++)
{
// check the maximum product
// in horizontal row.
if ((j - 3) >= 0)
{
result = arr[i][j] * arr[i][j - 1]
* arr[i][j - 2]
* arr[i][j - 3];
if (max < result)
max = result;
}
// check the maximum product
// in vertical row.
if ((i - 3) >= 0)
{
result = arr[i][j] * arr[i - 1][j]
* arr[i - 2][j]
* arr[i - 3][j];
if (max < result)
max = result;
}
// check the maximum product in
// diagonal (going through down - right)
if ((i - 3) >= 0 && (j - 3) >= 0)
{
result = arr[i][j] * arr[i - 1][j - 1]
* arr[i - 2][j - 2]
* arr[i - 3][j - 3];
if (max < result)
max = result;
}
// check the maximum product in
// diagonal (going through up - right)
if ((i - 3) >= 0 && (j - 1) <= 0)
{
result = arr[i][j] * arr[i - 1][j + 1]
* arr[i - 2][j + 2]
* arr[i - 3][j + 3];
if (max < result)
max = result;
}
}
}
return max;
}
// Driver code
/* int arr[][4] = {{6, 2, 3, 4},
{5, 4, 3, 1},
{7, 4, 5, 6},
{8, 3, 1, 0}};*/
/* int arr[][5] = {{1, 2, 1, 3, 4},
{5, 6, 3, 9, 2},
{7, 8, 8, 1, 2},
{1, 0, 7, 9, 3},
{3, 0, 8, 4, 9}};*/
let arr = [[ 1, 2, 3, 4, 5 ],
[ 6, 7, 8, 9, 1 ],
[ 2, 3, 4, 5, 6 ],
[ 7, 8, 9, 1, 0 ],
[ 9, 6, 4, 2, 3 ]];
document.write(FindMaxProduct(arr, n));
// This code is contributed by sravan kumar
</script>
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
int maxPro(int a[6][5], int n, int m, int k)
{
int maxi(1), mp(1);
for (int i = 0; i < n; ++i)
{
// Window Product for each row.
int wp(1);
for (int l = 0; l < k; ++l)
{
wp *= a[i][l];
}
// Maximum window product for each row
mp = wp;
for (int j = k; j < m; ++j)
{
wp = wp * a[i][j] / a[i][j - k];
// Global maximum window product
maxi = max(maxi,max(mp,wp));
}
}
return maxi;
}
// Driver Code
int main()
{
int n = 6, m = 5, k = 4;
int a[6][5] = { { 1, 2, 3, 4, 5 },
{ 6, 7, 8, 9, 1 },
{ 2, 3, 4, 5, 6 },
{ 7, 8, 9, 1, 0 },
{ 9, 6, 4, 2, 3 },
{ 1, 1, 2, 1, 1 } };
cout << maxPro(a, n, m, k);
return 0;
}
Java
// Java implementation of the above approach
import java.io.*;
class GFG {
public static int maxPro(int[][] a,
int n, int m,
int k)
{
int maxi = 1, mp = 1;
for (int i = 0; i < n; ++i)
{
// Window Product for each row.
int wp = 1;
for (int l = 0; l < k; ++l)
{
wp *= a[i][l];
}
// Maximum window product for each row
mp = wp;
for (int j = k; j < m; ++j)
{
wp = wp * a[i][j] / a[i][j - k];
// Global maximum
// window product
maxi = Math.max(
maxi,
Math.max(mp, wp));
}
}
return maxi;
}
// Driver Code
public static void main(String[] args)
{
int n = 6, m = 5, k = 4;
int[][] a = new int[][] {
{ 1, 2, 3, 4, 5 }, { 6, 7, 8, 9, 1 },
{ 2, 3, 4, 5, 6 }, { 7, 8, 9, 1, 0 },
{ 9, 6, 4, 2, 3 }, { 1, 1, 2, 1, 1 }
};
// Function call
int maxpro = maxPro(a, n, m, k);
System.out.println(maxpro);
}
}
Python3
# Python implementation of the above approach def maxPro(a,n,m,k): maxi = 1 mp = 1 for i in range(n): # Window Product for each row. wp = 1 for l in range(k): wp *= a[i][l] # Maximum window product for each row mp = wp for j in range(k,m): wp = wp * a[i][j] / a[i][j - k] # Global maximum # window product maxi = max( maxi, max(mp, wp)) return maxi # Driver Code n = 6 m = 5 k = 4 a=[[1, 2, 3, 4, 5 ], [ 6, 7, 8, 9, 1 ], [ 2, 3, 4, 5, 6 ], [ 7, 8, 9, 1, 0 ], [ 9, 6, 4, 2, 3 ], [ 1, 1, 2, 1, 1 ]] # Function call maxpro = maxPro(a, n, m, k) print(maxpro) # This code is contributed by ab2127
C#
// C# implementation of the above approach
using System;
class GFG{
public static int maxPro(int[,] a, int n,
int m, int k)
{
int maxi = 1, mp = 1;
for(int i = 0; i < n; ++i)
{
// Window Product for each row.
int wp = 1;
for(int l = 0; l < k; ++l)
{
wp *= a[i, l];
}
// Maximum window product for each row
mp = wp;
for(int j = k; j < m; ++j)
{
wp = wp * a[i, j] / a[i, j - k];
// Global maximum
// window product
maxi = Math.Max(maxi,
Math.Max(mp, wp));
}
}
return maxi;
}
// Driver Code
static public void Main()
{
int n = 6, m = 5, k = 4;
int[,] a = {{ 1, 2, 3, 4, 5 }, { 6, 7, 8, 9, 1 },
{ 2, 3, 4, 5, 6 }, { 7, 8, 9, 1, 0 },
{ 9, 6, 4, 2, 3 }, { 1, 1, 2, 1, 1 }};
// Function call
int maxpro = maxPro(a, n, m, k);
Console.WriteLine(maxpro);
}
}
// This code is contributed by avanitrachhadiya2155
Javascript
<script>
// Javascript implementation of the above approach
function maxPro(a,n,m,k)
{
let maxi = 1, mp = 1;
for (let i = 0; i < n; ++i)
{
// Window Product for each row.
let wp = 1;
for (let l = 0; l < k; ++l)
{
wp *= a[i][l];
}
// Maximum window product for each row
mp = wp;
for (let j = k; j < m; ++j)
{
wp = wp * a[i][j] / a[i][j - k];
// Global maximum
// window product
maxi = Math.max(
maxi,
Math.max(mp, wp));
}
}
return maxi;
}
// Driver Code
let n = 6, m = 5, k = 4;
let a=[[1, 2, 3, 4, 5 ], [ 6, 7, 8, 9, 1 ],
[ 2, 3, 4, 5, 6 ], [ 7, 8, 9, 1, 0 ],
[ 9, 6, 4, 2, 3 ], [ 1, 1, 2, 1, 1 ]]
// Function call
let maxpro = maxPro(a, n, m, k);
document.write(maxpro);
// This code is contributed by rag2127
</script>
Publicación traducida automáticamente
Artículo escrito por shrikanth13 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA