Product of all the pairs from the given array

Given an array arr[] of N integers, the task is to find the product of all the pairs possible from the given array such as: 
 

  • (arr[i], arr[i]) is also considered as a valid pair.
  • (arr[i], arr[j]) and (arr[j], arr[i]) are considered as two different pairs.

Print the resultant answer modulus 10^9+7.
Examples: 
 

Input: arr[] = {1, 2} 
Output: 16 
Explanation: 
All valid pairs are (1, 1), (1, 2), (2, 1) and (2, 2). 
Hence, 1 * 1 * 1 * 2 * 2 * 1 * 2 * 2 = 16
Input: arr[] = {1, 2, 3} 
Output: 46656 
Explanation: 
All valid pairs are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2) and (3, 3). 
Hence the product is 1*1*1**2*1*3*2*1*2*2*2*3*3*1*3*2*3*3 = 46656 
 

 

Naive Approach: To solve the problem mentioned above the naive method is to find all the possible pairs and calculate the product of the elements of each pair.
Below is the implementation of the above approach: 
 



C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 
46656

 

Time Complexity: O(N2)
Efficient approach: We can observe that each element appears exactly (2 * N) times as one of the elements of a pair (X, Y). Exactly N times as X and exactly N times as Y.
Below is the implementation of the above approach: 
 

C++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Output: 
46656

 

Time Complexity: O(N)
 

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