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 1000000007using 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 elementsint 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 Codeint 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 approachimport 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 elementsstatic 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 Codepublic 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 approachM = 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 elementsdef 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 Codeif __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 approachusing 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 elementsstatic 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 gcdstatic int gcd(int a, int b){ if (b == 0) return a; return gcd(b, a % b);}// Driver codestatic 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> |
31
Time Complexity: O(log N)
Auxiliary Space: O(1)
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