Mean of array generated by products of all pairs of the given array

Given an array arr[] consisting of N integers, the task is to find the mean of the array formed by the products of unordered pairs of the given array.

Examples:

Input: arr[] = {2, 5, 7}
Output: 19.67
Explanation:
Product of unordered pairs of array arr[] are 2 * 5 = 10, 2 * 7 = 14 and 5 * 7 = 35.
Therefore, the resultant array of product of pairs is {10, 14, 35}.
Mean of the array of product of pairs is 59/3 = 19.67

Input: arr[] = {1, 2, 4, 8}
Output: 11.67
Explanation:
Product of unordered pairs of array arr[] are 1 * 2 = 2, 1 * 4 = 4, 1 * 8 = 8, 2 * 4 = 8, 2 * 8 = 16, 4 * 8 = 32.
Therefore, the resultant array of product of pairs is {2, 4, 8, 8, 16, 32}.
Mean of the array of product of pairs is 70/6 i.e., 11.67

Naive Approach: The simplest approach to solve the problem is to generate all possible pairs of array pairProductArray[] i.e., array formed by the product of unordered pairs of the array arr[]. Then, find the mean of the pairProductArray[]. Follow the steps below to solve the problem:

Generate all possible pairs of the array arr[] and store their products in pairProductArray[].
Initialize a variable sum to store the sum of the elements of pairProductArray[].
Divide the variable sum with the size of pairProductArray[] to get the required mean.
Finally, print the value of the sum as the resultant mean.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>

using namespace std;
 
// Function to find the mean of pair
// product array of arr[]

float pairProductMean(int arr[], int N)
{

    // Store product of pairs

    vector<int> pairArray;
 
    // Generate all unordered pairs

    for (int i = 0; i < N; i++) {

        for (int j = i + 1; j < N; j++) {
 
            int pairProduct

                = arr[i] * arr[j];
 
            // Store product of pairs

            pairArray.push_back(pairProduct);

        }

    }
 
    // Size of pairArray

    int length = pairArray.size();
 
    // Store sum of pairArray

    float sum = 0;

    for (int i = 0; i < length; i++)

        sum += pairArray[i];
 
    // Stores the mean of pairArray[]

    float mean;
 
    // Find mean of pairArray[]

    if (length != 0)

        mean = sum / length;

    else

        mean = 0;
 
    // Return the resultant mean

    return mean;
}
 
// Driver Code

int main()
{

    // Given array arr[]

    int arr[] = { 1, 2, 4, 8 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call

    cout << fixed << setprecision(2)

         << pairProductMean(arr, N);
 
    return 0;
}

Java

// Java program for the 
// above approach

import java.util.*;

class GFG{
 
// Function to find the mean 
// of pair product array of arr[]

static double  pairProductMean(int arr[], 

                               int N)
{

  // Store product of pairs

  Vector<Integer> pairArray = 

         new Vector<>();
 
  // Generate all unordered pairs

  for (int i = 0; i < N; i++) 

  {

    for (int j = i + 1; j < N; j++)

    {

      int pairProduct = arr[i] * 

                        arr[j];
 
      // Store product of pairs

      pairArray.add(pairProduct);

    }

  }
 
  // Size of pairArray

  int length = pairArray.size();
 
  // Store sum of pairArray

  float sum = 0;

  for (int i = 0; i < length; i++)

    sum += pairArray.get(i);
 
  // Stores the mean of 

  // pairArray[]

  float mean;
 
  // Find mean of pairArray[]

  if (length != 0)

    mean = sum / length;

  else

    mean = 0;
 
  // Return the resultant mean

  return mean;
}
 
// Driver Code

public static void main(String[] args)
{

  // Given array arr[]

  int arr[] = {1, 2, 4, 8};
 
  int N = arr.length;
 
  // Function Call
 
  System.out.format("%.2f", 

                    pairProductMean(arr, N));
}
}
 
// This code is contributed by shikhasingrajput

Python3

# Python3 program for the
# above approach
 
# Function to find the mean
# of pair product array of arr

def pairProductMean(arr, N):

    # Store product of pairs

    pairArray = [];
 
    # Generate all unordered

    # pairs

    for i in range(N):

        for j in range(i + 1, N):

            pairProduct = arr[i] * arr[j];
 
            # Store product of pairs

            pairArray.append(pairProduct);
 
    # Size of pairArray

    length = len(pairArray);
 
    # Store sum of pairArray

    sum = 0;

    for i in range(length):

        sum += pairArray[i];
 
    # Stores the mean of

    # pairArray

    mean = 0;
 
    # Find mean of pairArray

    if (length != 0):

        mean = sum / length;

    else:

        mean = 0;
 
    # Return the resultant 

    # mean

    return mean;
 
# Driver Code

if __name__ == '__main__':

    # Given array arr

    arr = [1, 2, 4, 8];
 
    N = len(arr);
 
    # Function Call

    print("{0:.2f}".format(

            pairProductMean(arr, N)))
 
# This code is contributed by Rajput-Ji

// C# program for the 
// above approach

using System;

using System.Collections.Generic;

class GFG{
 
// Function to find the mean 
// of pair product array of []arr

static double  pairProductMean(int []arr, 

                               int N)
{

  // Store product of pairs

  List<int> pairArray = 

         new List<int>();
 
  // Generate all unordered pairs

  for (int i = 0; i < N; i++) 

  {

    for (int j = i + 1; j < N; j++)

    {

      int pairProduct = arr[i] * 

                        arr[j];
 
      // Store product of pairs

      pairArray.Add(pairProduct);

    }

  }
 
  // Size of pairArray

  int length = pairArray.Count;
 
  // Store sum of pairArray

  float sum = 0;

  for (int i = 0; i < length; i++)

    sum += pairArray[i];
 
  // Stores the mean of 

  // pairArray[]

  float mean;
 
  // Find mean of pairArray[]

  if (length != 0)

    mean = sum / length;

  else

    mean = 0;
 
  // Return the resultant mean

  return mean;
}
 
// Driver Code

public static void Main(String[] args)
{

  // Given array []arr

  int []arr = {1, 2, 4, 8};
 
  int N = arr.Length;
 
  // Function Call

  Console.WriteLine("{0:F2}", 

                    pairProductMean(arr,

                                    N));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript program for the 
// above approach
 
    // Function to find the mean

    // of pair product array of arr

    function pairProductMean(arr , N) 

    {

        // Store product of pairs

        var pairArray = [];
 
        // Generate all unordered pairs

        for (i = 0; i < N; i++) {

            for (j = i + 1; j < N; j++) {

                var pairProduct = arr[i] * arr[j];
 
                // Store product of pairs

                pairArray.push(pairProduct);

            }

        }
 
        // Size of pairArray

        var length = pairArray.length;
 
        // Store sum of pairArray

        var sum = 0;

        for (i = 0; i < length; i++)

            sum += pairArray[i];
 
        // Stores the mean of

        // pairArray

        var mean;
 
        // Find mean of pairArray

        if (length != 0)

            mean = sum / length;

        else

            mean = 0;
 
        // Return the resultant mean

        return mean;

    }
 
    // Driver Code

        // Given array arr

        var arr = [ 1, 2, 4, 8 ];
 
        var N = arr.length;
 
        // Function Call
 
        document.write(pairProductMean(arr, N).toFixed(2));
 
// This code contributed by gauravrajput1
 
</script>

Output

11.67

Time Complexity: O(N²)
Auxiliary Space: O(N²)

Efficient Approach: The idea is to use the fact that every element arr[i] is multiplied with every element arr[j] which is on the right side of the element arr[i], more formally element at index i is multiplied to all the elements positioned at index j such that j > i. Follow the steps below to solve the problem:

Create a suffix sum array suffixSumArray[] for the given array arr[].
Initialize variable res to store the sum of product pairs of array arr[].
Iterate array arr[] and for each position i incrementing res with arr[i]*suffixSumArray[i+1].
Divide the variable res with N*(N – 1)/ 2 which is the number of possible products.
Finally, print the value of res as the resultant mean.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>

using namespace std;
 
// Function to find the mean of pair
// product array of arr[]

float pairProductMean(int arr[], int N)
{

    // Initializing suffix sum array

    int suffixSumArray[N];

    suffixSumArray[N - 1] = arr[N - 1];
 
    // Build suffix sum array

    for (int i = N - 2; i >= 0; i--) {

        suffixSumArray[i]

            = suffixSumArray[i + 1]

              + arr[i];

    }
 
    // Size of pairProductArray

    int length = (N * (N - 1)) / 2;
 
    // Stores sum of pairProductArray

    float res = 0;
 
    for (int i = 0; i < N - 1; i++) {

        res += arr[i]

               * suffixSumArray[i + 1];

    }
 
    // Store the mean

    float mean;
 
    // Find mean of pairProductArray

    if (length != 0)

        mean = res / length;

    else

        mean = 0;
 
    // Return the resultant mean

    return mean;
}
 
// Driver Code

int main()
{

    // Given array arr[]

    int arr[] = { 1, 2, 4, 8 };

    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call

    cout << fixed << setprecision(2)

         << pairProductMean(arr, N);
 
    return 0;
}

Java

// Java program for the above approach

import java.io.*;

import java.util.*;
 
class GFG{

// Function to find the mean of pair
// product array of arr[]

static float pairProductMean(int arr[], int N)
{

    // Initializing suffix sum array

    int suffixSumArray[] = new int[N];

    suffixSumArray[N - 1] = arr[N - 1];

    // Build suffix sum array

    for(int i = N - 2; i >= 0; i--) 

    {

        suffixSumArray[i] = suffixSumArray[i + 1] +

                                       arr[i];

    }

    // Size of pairProductArray

    int length = (N * (N - 1)) / 2;

    // Stores sum of pairProductArray

    float res = 0;

    for(int i = 0; i < N - 1; i++)

    {

        res += arr[i] * 

               suffixSumArray[i + 1];

    }

    // Store the mean

    float mean;

    // Find mean of pairProductArray

    if (length != 0)

        mean = res / length;

    else

        mean = 0;

    // Return the resultant mean

    return mean;
}
 
// Driver Code

public static void main (String[] args) 
{

    // Given array arr[]

    int arr[] = { 1, 2, 4, 8 };

    int N = arr.length;

    // Function call

    System.out.format("%.2f", 

                      pairProductMean(arr, N));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to find the mean of pair
# product array of arr[]

def pairProductMean(arr, N):

    # Initializing suffix sum array

    suffixSumArray = [0] * N

    suffixSumArray[N - 1] = arr[N - 1]
 
    # Build suffix sum array

    for i in range(N - 2, -1, -1):

        suffixSumArray[i] = suffixSumArray[i + 1] + arr[i]
 
    # Size of pairProductArray

    length = (N * (N - 1)) // 2
 
    # Stores sum of pairProductArray

    res = 0
 
    for i in range(N - 1):

        res += arr[i] * suffixSumArray[i + 1]
 
    # Store the mean

    mean = 0
 
    # Find mean of pairProductArray

    if (length != 0):

        mean = res / length

    else:

        mean = 0
 
    # Return the resultant mean

    return mean
 
# Driver Code

if __name__ == '__main__':

    # Given array arr[]

    arr = [ 1, 2, 4, 8 ]

    N = len(arr)
 
    # Function Call

    print(round(pairProductMean(arr, N), 2))
 
# This code is contributed by mohit kumar 29

// C# program for the above approach

using System;

class GFG{

// Function to find the mean of pair
// product array of arr[]

static double pairProductMean(int[] arr, int N)
{

    // Initializing suffix sum array

    int[] suffixSumArray = new int[N];

    suffixSumArray[N - 1] = arr[N - 1];

    // Build suffix sum array

    for(int i = N - 2; i >= 0; i--) 

    {

        suffixSumArray[i] = suffixSumArray[i + 1] +

                                       arr[i];

    }

    // Size of pairProductArray

    int length = (N * (N - 1)) / 2;

    // Stores sum of pairProductArray

    double res = 0;

    for(int i = 0; i < N - 1; i++)

    {

        res += arr[i] * 

               suffixSumArray[i + 1];

    }

    // Store the mean

    double mean;

    // Find mean of pairProductArray

    if (length != 0)

        mean = res / length;

    else

        mean = 0;

    // Return the resultant mean

    return mean;
}

// Driver code

public static void Main()
{

    // Given array arr[]

    int[] arr = { 1, 2, 4, 8 };

    int N = arr.Length;

    // Function call

    Console.WriteLine(string.Format("{0:0.00}",

                      pairProductMean(arr, N)));
}
}

// This code is contributed by code_hunt

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to find the mean of pair
// product array of arr[]

function pairProductMean(arr, N)
{

    // Initializing suffix sum array

    var suffixSumArray = Array(N);

    suffixSumArray[N - 1] = arr[N - 1];
 
    // Build suffix sum array

    for (var i = N - 2; i >= 0; i--) {

        suffixSumArray[i]

            = suffixSumArray[i + 1]

              + arr[i];

    }
 
    // Size of pairProductArray

    var length = (N * (N - 1)) / 2;
 
    // Stores sum of pairProductArray

    var res = 0;
 
    for (var i = 0; i < N - 1; i++) {

        res += arr[i]

               * suffixSumArray[i + 1];

    }
 
    // Store the mean

    var mean;
 
    // Find mean of pairProductArray

    if (length != 0)

        mean = res / length;

    else

        mean = 0;
 
    // Return the resultant mean

    return mean;
}
 
// Driver Code
// Given array arr[]

var arr = [ 1, 2, 4, 8 ];

var N = arr.length;
// Function Call
document.write( pairProductMean(arr, N).toFixed(2));
 
</script>

Output

11.67

Complexity analysis:

The time complexity of the given program is O(N), where N is the size of the input array. This is because the program iterates over the input array twice, once to build the suffix sum array, and once to calculate the sum of the pair product array.

The space complexity of the program is O(N), as it creates an array of size N to store the suffix sum array.

Article Tags :

Arrays