GeeksforGeeks App
Open App
Browser
Continue

# Product of the maximums of all subsets of an array

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. {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.

## C++

 `// C++ code for the approach` `#include ``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

`324`

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 ``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

 ``

Output:

`324`

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

My Personal Notes arrow_drop_up