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 subsetslong 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 arrayvoid printMaxProduct(int arr[], int n) { long long int ans = maxProduct(arr, n); cout << ans << endl;}// Driver codeint 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 approachimport 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 approachMOD = int(1e9 + 7)# Function to find the product of the maximum of all# possible subsetsdef 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 arraydef printMaxProduct(arr, n): ans = maxProduct(arr, n) print(ans) # Driver codeif __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 subsetsconst 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 codelet arr = [1, 2, 3];let n = arr.length;// Print the product of the maximum of all possible subsets of the arrayprintMaxProduct(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 <bits/stdc++.h>using namespace std;// Function to find the product of the// maximum of all possible subsetslong 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 Codeint 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 approachimport java.util.Arrays;class GFG{// Function to find the product of the// maximum of all possible subsetsstatic 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 Codepublic 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 subsetsdef 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 Codeif __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 approachusing System;using System.Collections.Generic;class GFG{// Function to find the product of the// maximum of all possible subsetsstatic 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 Codepublic 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 subsetsfunction 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 Codelet arr = [1, 2, 3 ];let N = arr.length;document.write(maximumProduct(arr, N));</script> |
Output:
324
Time Complexity: O(N*
)
Auxiliary Space: O(N)
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