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Product of the maximums of all subsets of an array

  • Difficulty Level : Hard
  • Last Updated : 22 Jul, 2021

Given an array arr[] consisting of N positive integers, the task is to find the product of the maximum of all possible subsets of the given array. Since the product can be very large, print it to modulo (109 + 7).

Examples:

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Input: arr[] = {1, 2, 3}
Output:
Explanation:
All possible subsets of the given array with their respective maximum elements are:



  1. {1}, the maximum element is 1.
  2. {2}, the maximum element is 2.
  3. {3}, the maximum element is 3.
  4. {1, 2}, the maximum element is 2.
  5. {1, 3}, the maximum element is 3.
  6. {2, 3}, the maximum element is 3.
  7. {1, 2, 3}, the maximum element is 3.

The product of all the above maximum element is 1*2*3*2*3*3*3 = 324.

Input: arr[] = {1, 1, 1, 1}
Output: 1

Naive Approach: The simplest approach to solve the given problem is to generate all possible subsets of the given array and find the product of the maximum of all the generated subsets modulo (109 + 7) as the resultant product.

Time Complexity: O(N*2N)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized based on the following observations:

  • The idea is to count the number of times each array element occurs as the maximum element among all possible subsets formed.
  • An array element arr[i] is a maximum if and only if all the elements except arr[i] are smaller than or equal to it.
  • Therefore, the number of subsets formed by all elements smaller than or equal to each array element arr[i] contributes to the count of subsets having arr[i] as the maximum element.

Follow the steps below to solve the problem:

  • Initialize a variable, say maximumProduct as 1 that stores the resultant product of maximum elements of all subsets.
  • Sort the given array arr[] in the increasing order.
  • Traverse the array from the end using the variable i and perform the following steps:
    • Find the number of subsets that are smaller than the current element arr[i] as (2i – 1) and store it in a variable say P.
    • Since the array element arr[i] contributes P number of times, therefore multiply the value arr[i], P times to the variable maximumProduct.
  • Find the product of the array element with maximumProduct for including all the subsets of size 1.
  • After completing the above steps, print the value of maximumProduct as the resultant maximum product.

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 product of the
// maximum of all possible subsets
long maximumProduct(int arr[], int N)
{
    long mod = 1000000007;
 
    // Sort the given array arr[]
    sort(arr, arr + N);
 
    // Stores the power of 2
    long power[N + 1];
    power[0] = 1;
 
    // Calculate the power of 2
    for (int i = 1; i <= N; i++) {
        power[i] = 2 * power[i - 1];
        power[i] %= mod;
    }
 
    // Stores the resultant product
    long result = 1;
 
    // Traverse the array from the back
    for (int i = N - 1; i > 0; i--) {
 
        // Find the value of 2^i - 1
        long value = (power[i] - 1);
 
        // Iterate value number of times
        for (int j = 0; j < value; j++) {
 
            // Multiply value with
            // the result
            result *= 1LL * arr[i];
            result %= mod;
        }
    }
 
    // Calculate the product of array
    // elements with result to consider
    // the subset of size 1
    for (int i = 0; i < N; i++) {
        result *= 1LL * arr[i];
        result %= mod;
    }
 
    // Return the resultant product
    return result;
}
 
// Driver Code
int main()
{
 
    int arr[] = { 1, 2, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << maximumProduct(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.Arrays;
 
class GFG{
 
// Function to find the product of the
// maximum of all possible subsets
static long maximumProduct(int arr[], int N)
{
    long mod = 1000000007;
 
    // Sort the given array arr[]
    Arrays.sort(arr);
 
    // Stores the power of 2
    long power[] = new long[N + 1];
    power[0] = 1;
 
    // Calculate the power of 2
    for(int i = 1; i <= N; i++)
    {
        power[i] = 2 * power[i - 1];
        power[i] %= mod;
    }
 
    // Stores the resultant product
    long result = 1;
 
    // Traverse the array from the back
    for(int i = N - 1; i > 0; i--)
    {
         
        // Find the value of 2^i - 1
        long value = (power[i] - 1);
 
        // Iterate value number of times
        for(int j = 0; j < value; j++)
        {
             
            // Multiply value with
            // the result
            result *= arr[i];
            result %= mod;
        }
    }
 
    // Calculate the product of array
    // elements with result to consider
    // the subset of size 1
    for(int i = 0; i < N; i++)
    {
        result *= arr[i];
        result %= mod;
    }
 
    // Return the resultant product
    return result;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3 };
    int N = arr.length;
     
    System.out.println(maximumProduct(arr, N));
}
}
 
// This code is contributed by rishavmahato348

Python3




# Python3 program for the above approach
 
# Function to find the product of the
# maximum of all possible subsets
def maximumProduct(arr, N):
     
    mod = 1000000007
 
    # Sort the given array arr[]
    arr = sorted(arr)
 
    # Stores the power of 2
    power = [0] * (N + 1)
    power[0] = 1
 
    # Calculate the power of 2
    for i in range(1, N + 1):
        power[i] = 2 * power[i - 1]
        power[i] %= mod
 
    # Stores the resultant product
    result = 1
 
    # Traverse the array from the back
    for i in range(N - 1, -1, -1):
         
        # Find the value of 2^i - 1
        value = (power[i] - 1)
 
        # Iterate value number of times
        for j in range(value):
             
            # Multiply value with
            # the result
            result *= arr[i]
            result %= mod
 
    # Calculate the product of array
    # elements with result to consider
    # the subset of size 1
    for i in range(N):
        result *= arr[i]
        result %= mod
         
    # Return the resultant product
    return result
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 2, 3 ]
    N = len(arr)
     
    print(maximumProduct(arr, N))
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the product of the
// maximum of all possible subsets
static long maximumProduct(int []arr, int N)
{
    long mod = 1000000007;
 
    // Sort the given array arr[]
    Array.Sort(arr);
 
    // Stores the power of 2
    long []power = new long[N + 1];
    power[0] = 1;
 
    // Calculate the power of 2
    for (int i = 1; i <= N; i++) {
        power[i] = 2 * power[i - 1];
        power[i] %= mod;
    }
 
    // Stores the resultant product
    long result = 1;
 
    // Traverse the array from the back
    for (int i = N - 1; i > 0; i--) {
 
        // Find the value of 2^i - 1
        long value = (power[i] - 1);
 
        // Iterate value number of times
        for (int j = 0; j < value; j++) {
 
            // Multiply value with
            // the result
            result *= 1 * arr[i];
            result %= mod;
        }
    }
 
    // Calculate the product of array
    // elements with result to consider
    // the subset of size 1
    for (int i = 0; i < N; i++) {
        result *= 1 * arr[i];
        result %= mod;
    }
 
    // Return the resultant product
    return result;
}
 
// Driver Code
public static void Main()
{
 
    int []arr = {1, 2, 3};
    int N = arr.Length;
    Console.Write(maximumProduct(arr, N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript




<script>
 
// JavaScript program for the above approach
 
 
// Function to find the product of the
// maximum of all possible subsets
function maximumProduct(arr, N)
{
    let mod = 1000000007;
 
    // Sort the given array arr[]
    arr.sort((a, b) =>  a - b);
 
    // Stores the power of 2
    let power = new Array(N + 1);
    power[0] = 1;
 
    // Calculate the power of 2
    for (let i = 1; i <= N; i++) {
        power[i] = 2 * power[i - 1];
        power[i] %= mod;
    }
 
    // Stores the resultant product
    let result = 1;
 
    // Traverse the array from the back
    for (let i = N - 1; i > 0; i--) {
 
        // Find the value of 2^i - 1
        let value = (power[i] - 1);
 
        // Iterate value number of times
        for (let j = 0; j < value; j++) {
 
            // Multiply value with
            // the result
            result *= 1 * arr[i];
            result %= mod;
        }
    }
 
    // Calculate the product of array
    // elements with result to consider
    // the subset of size 1
    for (let i = 0; i < N; i++) {
        result *= 1 * arr[i];
        result %= mod;
    }
 
    // Return the resultant product
    return result;
}
 
// Driver Code
 
let arr = [1, 2, 3 ];
let N = arr.length;
document.write(maximumProduct(arr, N));
 
</script>
Output: 
324

 

Time Complexity: O(N*log N)
Auxiliary Space: O(N)




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