Producto de todos los pares de la array dada

Posted on by Rudeus Greyrat

Dada una array arr[] de N enteros, la tarea es encontrar el producto de todos los pares posibles de la array dada, como: 

  • (arr[i], arr[i]) también se considera un par válido.
  • (arr[i], arr[j]) y (arr[j], arr[i]) se consideran dos pares diferentes.

Imprime el módulo de respuesta resultante 10^9+7.

Ejemplos:  

Entrada: arr[] = {1, 2} 
Salida: 16 
Explicación: 
Todos los pares válidos son (1, 1), (1, 2), (2, 1) y (2, 2). 
Por lo tanto, 1 * 1 * 1 * 2 * 2 * 1 * 2 * 2 = 16

Entrada: arr[] = {1, 2, 3} 
Salida: 46656 
Explicación: 
Todos los pares válidos son (1, 1), (1, 2), (1, 3), (2, 1), (2, 2) ), (2, 3), (3, 1), (3, 2) y (3, 3). 
Por lo tanto, el producto es 1*1*1**2*1*3*2*1*2*2*2*3*3*1*3*2*3*3 = 46656 

Enfoque ingenuo: para resolver el problema mencionado anteriormente, el método ingenuo es encontrar todos los pares posibles y calcular el producto de los elementos de cada par.

A continuación se muestra la implementación del enfoque anterior:  

C++

// C++ implementation to find the
// product of all the pairs from
// the given array
 
#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
 
// Function to return the product of
// the elements of all possible pairs
// from the array
int productPairs(int arr[], int n)
{
 
    // To store the required product
    int product = 1;
 
    // Nested loop to calculate all
    // possible pairs
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
 
            // Multiply the product of
            // the elements of the
            // current pair
            product *= (arr[i] % mod
                        * arr[j] % mod)
                       % mod;
            product = product % mod;
        }
    }
 
    // Return the final result
    return product % mod;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << productPairs(arr, n);
 
    return 0;
}

Java

// Java implementation to find the
// product of all the pairs from
// the given array
import java.util.*;
 
class GFG{
     
static final int mod = 1000000007;
 
// Function to return the product of
// the elements of all possible pairs
// from the array
static int productPairs(int arr[], int n)
{
 
    // To store the required product
    int product = 1;
 
    // Nested loop to calculate all
    // possible pairs
    for(int i = 0; i < n; i++)
    {
       for(int j = 0; j < n; j++)
       {
           
          // Multiply the product
          // of the elements of the
          // current pair
          product *= (arr[i] % mod *
                      arr[j] % mod) % mod;
          product = product % mod;
       }
    }
 
    // Return the final result
    return product % mod;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
 
    System.out.print(productPairs(arr, n));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 implementation to find the
# product of all the pairs from
# the given array
mod = 1000000007;
 
# Function to return the product of
# the elements of all possible pairs
# from the array
def productPairs(arr, n):
   
    # To store the required product
    product = 1;
 
    # Nested loop to calculate all
    # possible pairs
    for i in range(n):
        for j in range(n):
           
            # Multiply the product
            # of the elements of the
            # current pair
            product *= (arr[i] % mod *
                        arr[j] % mod) % mod;
            product = product % mod;
 
    # Return the final result
    return product % mod;
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3];
    n = len(arr);
 
    print(productPairs(arr, n));
 
# This code is contributed by 29AjayKumar

C#

// C# implementation to find the
// product of all the pairs from
// the given array
using System;
class GFG{
     
static readonly int mod = 1000000007;
 
// Function to return the product of
// the elements of all possible pairs
// from the array
static int productPairs(int []arr, int n)
{
 
    // To store the required product
    int product = 1;
 
    // Nested loop to calculate all
    // possible pairs
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
                 
            // Multiply the product
            // of the elements of the
            // current pair
            product *= (arr[i] % mod *
                        arr[j] % mod) % mod;
            product = product % mod;
        }
    }
 
    // Return the readonly result
    return product % mod;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
 
    Console.Write(productPairs(arr, n));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

<script>
 
//Javascript implementation to find the
// product of all the pairs from
// the given array
 
mod = 1000000007
 
// Function to return the product of
// the elements of all possible pairs
// from the array
function productPairs(arr, n)
{
 
    // To store the required product
    let product = 1;
 
    // Nested loop to calculate all
    // possible pairs
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
 
            // Multiply the product of
            // the elements of the
            // current pair
            product *= (arr[i] % mod
                        * arr[j] % mod)
                    % mod;
            product = product % mod;
        }
    }
 
    // Return the final result
    return product % mod;
}
 
// Driver code
    let arr = [ 1, 2, 3 ];
 
    let n = arr.length;
 
    document.write(productPairs(arr, n));
 
// This code is contributed by Mayank Tyagi
 
</script>
Salida: 
46656
 

Complejidad del tiempo: O(N 2 )

Enfoque eficiente: podemos observar que cada elemento aparece exactamente (2 * N) veces como uno de los elementos de un par (X, Y) . Exactamente N veces como X y exactamente N veces como Y .

A continuación se muestra la implementación del enfoque anterior:  

C++

// C++ implementation to Find the product
// of all the pairs from the given array
#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007
#define ll long long int
 
// Function to calculate
// (x^y)%1000000007
int power(int x, unsigned int y)
{
    int p = 1000000007;
 
    // Initialize result
    int res = 1;
 
    // Update x if it is more than
    // or equal to p
    x = x % p;
 
    while (y > 0) {
        // If y is odd, multiply x
        // with result
        if (y & 1)
            res = (res * x) % p;
 
        y = y >> 1;
        x = (x * x) % p;
    }
 
    // Return the final result
    return res;
}
 
// Function to return the product
// of the elements of all possible
// pairs from the array
ll productPairs(ll arr[], ll n)
{
 
    // To store the required product
    ll product = 1;
 
    // Iterate for every element
    // of the array
    for (int i = 0; i < n; i++) {
 
        // Each element appears (2 * n) times
        product
            = (product
               % mod
               * (int)power(
                     arr[i], (2 * n))
               % mod)
              % mod;
    }
 
    return product % mod;
}
 
// Driver code
int main()
{
    ll arr[] = { 1, 2, 3 };
    ll n = sizeof(arr) / sizeof(arr[0]);
 
    cout << productPairs(arr, n);
 
    return 0;
}

Java

// Java implementation to Find the product
// of all the pairs from the given array
import java.util.*;
 
class GFG{
static final int mod = 1000000007;
 
// Function to calculate
// (x^y)%1000000007
static int power(int x, int y)
{
    int p = 1000000007;
 
    // Initialize result
    int res = 1;
 
    // Update x if it is more than
    // or equal to p
    x = x % p;
 
    while (y > 0)
    {
         
        // If y is odd, multiply x
        // with result
        if (y % 2 == 1)
            res = (res * x) % p;
 
        y = y >> 1;
        x = (x * x) % p;
    }
 
    // Return the final result
    return res;
}
 
// Function to return the product
// of the elements of all possible
// pairs from the array
static int productPairs(int arr[], int n)
{
 
    // To store the required product
    int product = 1;
 
    // Iterate for every element
    // of the array
    for (int i = 0; i < n; i++)
    {
 
        // Each element appears (2 * n) times
        product = (product % mod *
                  (int)power(arr[i],
                            (2 * n)) % mod) % mod;
    }
 
    return product % mod;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
 
    System.out.print(productPairs(arr, n));
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to Find the product
# of all the pairs from the given array
mod = 1000000007
 
# Function to calculate
# (x^y)%1000000007
def power(x, y):
 
    p = 1000000007
 
    # Initialize result
    res = 1
 
    # Update x if it is more than
    # or equal to p
    x = x % p
 
    while (y > 0):
         
        # If y is odd, multiply x
        # with result
        if ((y & 1) != 0):
            res = (res * x) % p
 
        y = y >> 1
        x = (x * x) % p
 
    # Return the final result
    return res
 
# Function to return the product
# of the elements of all possible
# pairs from the array
def productPairs(arr, n):
 
    # To store the required product
    product = 1
 
    # Iterate for every element
    # of the array
    for i in range(n):
 
        # Each element appears (2 * n) times
        product = (product % mod *
          (int)(power(arr[i], (2 * n))) %
                            mod) % mod
 
    return (product % mod)
     
# Driver code
arr = [ 1, 2, 3 ]
n = len(arr)
 
print(productPairs(arr, n))
 
# This code is contributed by divyeshrabadiya07

C#

// C# implementation to Find the product
// of all the pairs from the given array
using System;
class GFG{
const int mod = 1000000007;
 
// Function to calculate
// (x^y)%1000000007
static int power(int x, int y)
{
    int p = 1000000007;
 
    // Initialize result
    int res = 1;
 
    // Update x if it is more than
    // or equal to p
    x = x % p;
 
    while (y > 0)
    {
         
        // If y is odd, multiply x
        // with result
        if (y % 2 == 1)
            res = (res * x) % p;
 
        y = y >> 1;
        x = (x * x) % p;
    }
 
    // Return the final result
    return res;
}
 
// Function to return the product
// of the elements of all possible
// pairs from the array
static int productPairs(int []arr, int n)
{
 
    // To store the required product
    int product = 1;
 
    // Iterate for every element
    // of the array
    for (int i = 0; i < n; i++)
    {
 
        // Each element appears (2 * n) times
        product = (product % mod *
                  (int)power(arr[i],
                            (2 * n)) % mod) % mod;
    }
 
    return product % mod;
}
 
// Driver code
public static void Main()
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
 
    Console.Write(productPairs(arr, n));
}
}
 
// This code is contributed by Code_Mech

Javascript

<script>
 
// Javascript implementation to Find the product
// of all the pairs from the given array
  
let mod = 1000000007;
 
// Function to calculate
// (x^y)%1000000007
function power(x, y)
{
    let p = 1000000007;
  
    // Initialize result
    let res = 1;
  
    // Update x if it is more than
    // or equal to p
    x = x % p;
  
    while (y > 0)
    {
          
        // If y is odd, multiply x
        // with result
        if (y % 2 == 1)
            res = (res * x) % p;
  
        y = y >> 1;
        x = (x * x) % p;
    }
  
    // Return the final result
    return res;
}
  
// Function to return the product
// of the elements of all possible
// pairs from the array
function productPairs(arr, n)
{
  
    // To store the required product
    let product = 1;
  
    // Iterate for every element
    // of the array
    for (let i = 0; i < n; i++)
    {
  
        // Each element appears (2 * n) times
        product = (product % mod *
                  power(arr[i],
                            (2 * n)) % mod) % mod;
    }
  
    return product % mod;
}
  
  // Driver Code
     
    let arr = [ 1, 2, 3 ];
    let n = arr.length;
  
    document.write(productPairs(arr, n));
     
</script>
Salida: 
46656
 

Complejidad de tiempo: O(N)
 

Publicación traducida automáticamente

Artículo escrito por spp____ y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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