Data Modelling & AI Data Structure & Algorithm

Product of the maximums of all subsets of an array

21 June 2025

0

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:

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

{1}, the maximum element is 1.

{2}, the maximum element is 2.

{3}, the maximum element is 3.

{1, 2}, the maximum element is 2.

{1, 3}, the maximum element is 3.

{2, 3}, the maximum element is 3.

{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.

C++

// C++ code for the approach
 
#include <bits/stdc++.h>
using namespace std;
 
#define mod 1000000007
 
// Function to find the product of the maximum of all
// possible subsets
long long int maxProduct(int arr[], int n) {
    long long int ans = 1;
 
    // Generate all possible subsets of the array
    for (int i = 0; i < (1 << n); i++) {
        long long int maxVal = INT_MIN;
 
        // Find the maximum value of each subset
        for (int j = 0; j < n; j++) {
            if ((i & (1 << j)) != 0) {
                maxVal = max(maxVal, (long long int)arr[j]);
            }
        }
 
        // Multiply the maximum value of each subset to the
        // answer
        if (maxVal != INT_MIN)
            ans = (ans * maxVal) % mod;
    }
 
    return ans;
}
 
// Function to print the product of the maximum of all
// possible subsets of the array
void printMaxProduct(int arr[], int n) {
    long long int ans = maxProduct(arr, n);
    cout << ans << endl;
}
 
// Driver code
int main() {
    // Input array
    int arr[] = { 1, 2, 3 };
    // Size of the array
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Print the product of the maximum of all
    // possible subsets of the array
    printMaxProduct(arr, n);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
public class Main {
    static final int mod = 1000000007;
 
    // Function to find the product of
    // the maximum of all possible subsets
    static long maxProduct(int[] arr, int n)
    {
        long ans = 1;
 
        // Generate all possible subsets
        // of the array
        for (int i = 0; i < (1 << n); i++) {
            long maxVal = Integer.MIN_VALUE;
 
            // Find the maximum value of
            // each subset
            for (int j = 0; j < n; j++) {
                if ((i & (1 << j)) != 0) {
                    maxVal = Math.max(maxVal, arr[j]);
                }
            }
 
            // Multiply the maximum value of
            // each subset to the answer
            if (maxVal != Integer.MIN_VALUE) {
                ans = (ans * maxVal) % mod;
            }
        }
 
        return ans;
    }
 
    // Function to print the product of the
    // maximum of all possible subsets
    // of the array
    static void printMaxProduct(int[] arr, int n)
    {
        long ans = maxProduct(arr, n);
        System.out.println(ans);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        // Input array
        int[] arr = { 1, 2, 3 };
        // Size of the array
        int n = arr.length;
 
        // Print the product of the maximum of all
        // possible subsets of the array
        printMaxProduct(arr, n);
    }
}

Python3

# Python3 code for the approach
MOD = int(1e9 + 7)
 
# Function to find the product of the maximum of all
# possible subsets
def maxProduct(arr, n):
  ans = 1
   
  # Generate all possible subsets of the array
  for i in range(1 << n):
    maxVal = float('-inf')
     
    # Find the maximum value of each subset
    for j in range(n):
      if i & (1 << j):
        maxVal = max(maxVal, arr[j])
        
    # Multiply the maximum value of each subset to the
    # answer
    if maxVal != float('-inf'):
      ans = (ans * maxVal) % MOD
           
           
  return ans
 
# Function to print the product of the maximum of all
# possible subsets of the array
def printMaxProduct(arr, n):
    ans = maxProduct(arr, n)
    print(ans)
   
# Driver code
if __name__ == '__main__':
    # Input array
    arr = [1, 2, 3]
    # Size of the array
    n = len(arr)
     
    # Print the product of the maximum of all
    # possible subsets of the array
    printMaxProduct(arr, n)

C#

using System;
 
public class GFG {
  static readonly int mod = 1000000007;
 
  // Function to find the product of
  // the maximum of all possible subsets
  static long MaxProduct(int[] arr, int n)
  {
    long ans = 1;
 
    // Generate all possible subsets
    // of the array
    for (int i = 0; i < (1 << n); i++) {
      long maxVal = int.MinValue;
 
      // Find the maximum value of
      // each subset
      for (int j = 0; j < n; j++) {
        if ((i & (1 << j)) != 0) {
          maxVal = Math.Max(maxVal, arr[j]);
        }
      }
 
      // Multiply the maximum value of
      // each subset to the answer
      if (maxVal != int.MinValue) {
        ans = (ans * maxVal) % mod;
      }
    }
 
    return ans;
  }
 
  // Function to print the product of the
  // maximum of all possible subsets
  // of the array
  static void PrintMaxProduct(int[] arr, int n)
  {
    long ans = MaxProduct(arr, n);
    Console.WriteLine(ans);
  }
 
  // Driver code
  public static void Main()
  {
    // Input array
    int[] arr = { 1, 2, 3 };
    // Size of the array
    int n = arr.Length;
 
    // Print the product of the maximum of all
    // possible subsets of the array
    PrintMaxProduct(arr, n);
  }
}
 
//This code is contributed by aeroabrar_31

Javascript

// JavaScript code to find the product of the maximum of all possible subsets
 
const mod = 1000000007;
 
function maxProduct(arr, n) {
    let ans = 1;
 
    // Generate all possible subsets of the array
    for (let i = 0; i < (1 << n); i++) {
        let maxVal = Number.MIN_SAFE_INTEGER;
 
        // Find the maximum value of each subset
        for (let j = 0; j < n; j++) {
            if ((i & (1 << j)) !== 0) {
                maxVal = Math.max(maxVal, arr[j]);
            }
        }
 
        // Multiply the maximum value of each subset to the answer
        if (maxVal !== Number.MIN_SAFE_INTEGER) {
            ans = (ans * maxVal) % mod;
        }
    }
 
    return ans;
}
 
function printMaxProduct(arr, n) {
    let ans = maxProduct(arr, n);
    console.log(ans);
}
 
// Driver code
let arr = [1, 2, 3];
let n = arr.length;
 
// Print the product of the maximum of all possible subsets of the array
printMaxProduct(arr, n);

Output

Time Complexity: O(N*2^N)
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:

Time Complexity: O(N* $2^N$ )
Auxiliary Space: O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

2 COMMENTS

Homepage 3 February 2026 At 10:23 am

… [Trackback]

[…] Find More Information here to that Topic: geeksforgeeks.org/product-of-the-maximums-of-all-subsets-of-an-array/ […]

Log in to leave a comment
https://vavadakazinoskachat.top 20 February 2026 At 9:40 pm

… [Trackback]

[…] Find More Information here on that Topic: geeksforgeeks.org/product-of-the-maximums-of-all-subsets-of-an-array/ […]

Log in to leave a comment

Product of the maximums of all subsets of an array

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

Adding Persistent Memory to Claude Code with the Lightweight memsearch Plugin

GLM-5 vs. MiniMax M2.5 vs. Gemini 3 Deep Think: Which Model Fits Your AI Agent Stack?

We Extracted OpenClaw’s Memory System and Open-Sourced It (memsearch)

2 COMMENTS

LEAVE A REPLY Cancel reply

Most Popular

The wait is over: Google just released Snapseed 4.0 for Android with a new pro camera

What to watch this weekend: Sally Field bonds with a talking octopus and British civil servants become ‘Legends’

Samsung Galaxy Buds 3 Pro receive stability update

Google might finally fix a Pixel problem that users have complained about for a decade

EDITOR PICKS

The wait is over: Google just released Snapseed 4.0 for Android with a new pro camera

What to watch this weekend: Sally Field bonds with a talking octopus and British civil servants become ‘Legends’

Samsung Galaxy Buds 3 Pro receive stability update

POPULAR POSTS

The wait is over: Google just released Snapseed 4.0 for Android with a new pro camera

What to watch this weekend: Sally Field bonds with a talking octopus and British civil servants become ‘Legends’

Samsung Galaxy Buds 3 Pro receive stability update

POPULAR CATEGORY

ABOUT US

FOLLOW US