# Minimum divide by 2 operations required to make GCD odd for given Array

• Last Updated : 22 Oct, 2021

Given an array arr[] of N positive integers, the task is to find the minimum number of operations required to make the GCD of array element odd such that in each operation an array element can be divided by 2.

Examples:

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Input: arr[] = {4, 6}
Output: 1
Explanation:
Below are the operations performed:
Operation 1: Divide the array element arr[0](= 4) by 2 modifies the array to {2, 6}.
Operation 2: Divide the array element arr[0](= 2) by 2 modifies the array to {1, 6}.
After the above operations, the GCD of the array elements is 1 which is odd. Therefore, the minimum number of operations required is 2.

Input: arr[] = {2, 4, 1}
Output: 0

Approach: The given problem can be solved based on the observation by finding the count of powers of 2 for each array element and the minimum power of 2(say C) will give the minimum operations because after dividing that element by 2C the element becomes odd and that results in the GCD of the array as odd.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; // Function to find the minimum number// of operations to make the GCD of// the array oddint minimumOperations(int arr[], int N){    // Stores the minimum operations    // required    int mini = INT_MAX;     for (int i = 0; i < N; i++) {         // Stores the powers of two for        // the current array element        int count = 0;         // Dividing by 2        while (arr[i] % 2 == 0) {            arr[i] = arr[i] / 2;             // Increment the count            count++;        }         // Update the minimum operation        // required        if (mini > count) {            mini = count;        }    }     // Return the result required    return mini;} // Driver Codeint main(){    int arr[] = { 4, 6 };    int N = sizeof(arr) / sizeof(arr[0]);     cout << minimumOperations(arr, N);     return 0;}

## Java

 // Java program for the above approachclass GFG{ // Function to find the minimum number// of operations to make the GCD of// the array oddpublic static int minimumOperations(int arr[], int N){       // Stores the minimum operations    // required    int mini = Integer.MAX_VALUE;     for (int i = 0; i < N; i++) {         // Stores the powers of two for        // the current array element        int count = 0;         // Dividing by 2        while (arr[i] % 2 == 0) {            arr[i] = arr[i] / 2;             // Increment the count            count++;        }         // Update the minimum operation        // required        if (mini > count) {            mini = count;        }    }     // Return the result required    return mini;} // Driver Codepublic static void  main(String args[]){    int arr[] = { 4, 6 };    int N = arr.length;     System.out.println(minimumOperations(arr, N)); }} // This code is contributed by saurabh_jaiswal.

## Python3

 # python program for the above approacINT_MAX = 2147483647 # Function to find the minimum number# of operations to make the GCD of# the array odddef minimumOperations(arr, N):     # Stores the minimum operations    # required    mini = INT_MAX     for i in range(0, N):         # Stores the powers of two for        # the current array element        count = 0         # Dividing by 2        while (arr[i] % 2 == 0):            arr[i] = arr[i] // 2             # Increment the count            count += 1         # Update the minimum operation        # required        if (mini > count):            mini = count     # Return the result required    return mini # Driver Codeif __name__ == "__main__":     arr = [4, 6]    N = len(arr)     print(minimumOperations(arr, N)) # This code is contributed by rakeshsahni

## C#

 // C# program for the above approachusing System;class GFG {     // Function to find the minimum number    // of operations to make the GCD of    // the array odd    public static int minimumOperations(int[] arr, int N)    {         // Stores the minimum operations        // required        int mini = Int32.MaxValue;         for (int i = 0; i < N; i++) {             // Stores the powers of two for            // the current array element            int count = 0;             // Dividing by 2            while (arr[i] % 2 == 0) {                arr[i] = arr[i] / 2;                 // Increment the count                count++;            }             // Update the minimum operation            // required            if (mini > count) {                mini = count;            }        }         // Return the result required        return mini;    }     // Driver Code    public static void Main(string[] args)    {        int[] arr = { 4, 6 };        int N = arr.Length;         Console.WriteLine(minimumOperations(arr, N));    }} // This code is contributed by ukasp.

## Javascript



Output:
1

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

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