Count ways to make product of array elements even by replacements

• Difficulty Level : Hard
• Last Updated : 27 Apr, 2021

Given an array arr[] consisting of N integers, the task is to count the number of ways to make the product of array elements even by replacing array elements any number of times. Since the count can be very large, print the count modulo 109+7.

Examples:

Input: arr[] = {1, 3}
Output: 3
Explanation:
Operation 1: Replacing arr by 2. Therefore, arr[] modifies to {2, 3}. Product = 6.
Operation 2: Replacing arr by 10. Therefore, arr[] modifies to {1, 10}, Product= 10.
Operation 3: Replacing arr and arr by 2. Therefore, arr[] modifies to {2, 2}, Product = 4.
Hence, 3 possible ways exists.

Input: arr[] = {3}
Output:

Approach: The idea is to use Greedy Approach to solve this problem.
Follow the steps below to solve the problem:

• In order to make the product of the array even, at least one even array element must exist.
• Traverse the array. For every array element, the following two situations arise:
• If the array consists of a single element only, then only a single way exists to make the product of the array even.
• Otherwise, 2N – 1 ways.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach#include using namespace std; // Function to count ways to make// product of given array even.void makeProductEven(int arr[], int N){    int m = 1000000007, ans = 1;     // Calculate 2 ^ N    for (int i = 0; i < N; i++)    {        ans = (ans * 2) % m;    }     // Print the answer    cout << ans - 1;} // Driver Codeint main(){       // Given array    int arr[] = { 1, 3 };     // Size of the array    int N = sizeof(arr) / sizeof(arr);    makeProductEven(arr, N);    return 0;}

Java

 // Java program for above approach/*package whatever //do not write package name here */import java.io.*;class GFG{   // Method to count ways to make  // product of given array even.  static void makeProductEven(int arr[], int N)  {    int m = 1000000007, ans = 1;     // Calculate 2 ^ N    for (int i = 0; i < N; i++)    {      ans = (ans * 2) % m;    }     // Print the answer    System.out.println(ans - 1);  }  public static void main(String[] args)  {     // Given array    int arr[] = { 1, 3 };     // Size of the array    int N = arr.length;    makeProductEven(arr, N);  }} // This code is contributed by shubham agrawal

Python3

 # Python3 program for the above approach # Function to count ways to make# product of given array even.def makeProductEven(arr, N) :    m = 1000000007; ans = 1;     # Calculate 2 ^ N    for i in range(N) :        ans = (ans * 2) % m;     # Print the answer    print(ans - 1); # Driver Codeif __name__ == "__main__" :     # Given array    arr = [ 1, 3 ];     # Size of the array    N = len(arr);     makeProductEven(arr, N);     # This code is contributed by AnkThon

C#

 // C# program for above approach/*package whatever //do not write package name here */using System;public class GFG{   // Method to count ways to make  // product of given array even.  static void makeProductEven(int []arr, int N)  {    int m = 1000000007, ans = 1;     // Calculate 2 ^ N    for (int i = 0; i < N; i++)    {      ans = (ans * 2) % m;    }     // Print the answer    Console.WriteLine(ans - 1);  }     // Driver code  public static void Main(String[] args)  {     // Given array    int []arr = { 1, 3 };     // Size of the array    int N = arr.Length;    makeProductEven(arr, N);  }} // This code is contributed by shikhasingrajput

Javascript


Output
3

Time Complexity: O(N)

Auxiliary Space: O(1)

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