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#

// 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#

// 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.