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Count distinct ways to replace array elements such that product of the array becomes even

  • Last Updated : 26 Mar, 2021

Given an array arr[] consisting of N odd integers, the task is to count the different number of ways to make the product of all array elements even, by repeatedly changing any set of elements to any values. Since the count can be very large, print it to modulo 109 + 7.

Examples:

Input: arr[] = {1, 3}
Output: 3
Explanation: All possible ways to make the product of array elements odd are as follows: 
Replace arr[0] by any even integer. The array arr[] modifies to {even, 3}. Therefore, the product of the array = even * 3 = even. 
Replace arr[1] by any even integer. The array arr[] modifies to {1, even}. Therefore, the product of the array = 1 * even = even. 
Replace arr[0] and arr[1] by even integers. Since both array elements become even, the product of the array becomes even. Therefore, the total number of distinct ways to make the array even is 3.

Input: arr[] = {1, 2, 3, 4, 5}
Output: 31

Approach: The idea to solve the given problem is based on the observation that the product of an array is even only when at least one even element is present in the array. Therefore, the total number of distinct ways can be calculated by the number of distinct subsets of the given array.



Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
#define M 1000000007
using namespace std;
 
// Function to find the value of (x^y)
long long power(long long x, long long y,
                long long p)
{
    // Stores the result
    long long res = 1;
 
    while (y > 0) {
 
        // If y is odd, then
        // multiply x with res
        if (y & 1)
            res = (res * x) % p;
 
        // y must be even now
        y = y >> 1;
 
        // Update x
        x = (x * x) % p;
    }
    return res;
}
 
// Function to count the number of ways
// to make the product of an array even
// by replacing array elements
int totalOperations(int arr[], int N)
{
    // Find the value ( 2 ^ N ) % M
    long long res = power(2, N, M);
 
    // Exclude empty subset
    res--;
 
    // Print the answer
    cout << res;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    totalOperations(arr, N);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
 
class GFG{
     
static long M = 1000000007;
 
// Function to find the value of (x^y)
static long power(long x, long y, long p)
{
     
    // Stores the result
    long res = 1;
  
    while (y > 0)
    {
         
        // If y is odd, then
        // multiply x with res
        if ((y & 1) > 0)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1;
  
        // Update x
        x = (x * x) % p;
    }
    return res;
}
  
// Function to count the number of ways
// to make the product of an array even
// by replacing array elements
static int totalOperations(int arr[], int N)
{
     
    // Find the value ( 2 ^ N ) % M
    long res = power(2, N, M);
  
    // Exclude empty subset
    res--;
  
    // Print the answer
    System.out.print(res);
    return 0;
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int N = arr.length;
     
    totalOperations(arr, N);
}
}
 
// This code is contributed by rag2127

Python3




# Python3 program for the above approach
M = 1000000007
 
# Function to find the value of (x^y)
def power(x, y, p):
     
    global M
     
    # Stores the result
    res = 1
 
    while (y > 0):
 
        # If y is odd, then
        # multiply x with res
        if (y & 1):
            res = (res * x) % p;
 
        # y must be even now
        y = y >> 1
 
        # Update x
        x = (x * x) % p
 
    return res
 
# Function to count the number of ways
# to make the product of an array even
# by replacing array elements
def totalOperations(arr, N):
     
    # Find the value ( 2 ^ N ) % M
    res = power(2, N, M)
 
    # Exclude empty subset
    res-=1
 
    # Print the answer
    print (res)
 
# Driver Code
if __name__ == '__main__':
     
    arr = [1, 2, 3, 4, 5]
    N = len(arr)
 
    totalOperations(arr, N)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the above approach
using System;
class GFG {
     
static long M = 1000000007;
 
// Function to find the value of (x^y)
static long power(long x, long y, long p)
{
     
    // Stores the result
    long res = 1;
  
    while (y > 0)
    {
         
        // If y is odd, then
        // multiply x with res
        if ((y & 1) > 0)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1;
  
        // Update x
        x = (x * x) % p;
    }
    return res;
}
  
// Function to count the number of ways
// to make the product of an array even
// by replacing array elements
static int totalOperations(int[] arr, int N)
{
     
    // Find the value ( 2 ^ N ) % M
    long res = power(2, N, M);
  
    // Exclude empty subset
    res--;
  
    // Print the answer
    Console.Write(res);
    return 0;
}
 
// Calculating gcd
static int gcd(int a, int b)
{
    if (b == 0)
        return a;
         
    return gcd(b, a % b);
}
 
// Driver code
static void Main()
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int N = arr.Length;
     
    totalOperations(arr, N);
}
}
 
// This code is contributed by sanjoy_62.

Javascript




<script>
 
    let M = 1000000007;
    // Function to find the value of (x^y)
    function power(x,y,p)
    {
        // Stores the result
        let res = 1;
        while (y > 0)
        {
              
            // If y is odd, then
            // multiply x with res
            if ((y & 1) > 0)
                res = (res * x) % p;
       
            // y must be even now
            y = y >> 1;
       
            // Update x
            x = (x * x) % p;
        }
        return res;
    }
     
    // Function to count the number of ways
    // to make the product of an array even
    // by replacing array elements
    function totalOperations(arr,N)
    {
        // Find the value ( 2 ^ N ) % M
        let res = power(2, N, M);
        // Exclude empty subset
        res--;
        // Print the answer
        document.write(res);
    }
     
    // Driver Code
    let arr=[ 1, 2, 3, 4, 5];
    let N = arr.length;
    totalOperations(arr, N);
     
    // This code is contributed by avanitrachhadiya2155
     
</script>
Output: 
31

 

Time Complexity: O(log N)
Auxiliary Space: O(1)

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