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:
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++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the minimum number // of operations to make the GCD of // the array odd int 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 Code int main()
{ int arr[] = { 4, 6 };
int N = sizeof (arr) / sizeof (arr[0]);
cout << minimumOperations(arr, N);
return 0;
} |
// Java program for the above approach 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 = 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 Code public 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. |
# python program for the above approach INT_MAX = 2147483647
# Function to find the minimum number # of operations to make the GCD of # the array odd def 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 Code if __name__ = = "__main__" :
arr = [ 4 , 6 ]
N = len (arr)
print (minimumOperations(arr, N))
# This code is contributed by rakeshsahni |
// C# program for the above approach using 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. |
<script> // Javascript program for the above approach // Function to find the minimum number // of operations to make the GCD of // the array odd function minimumOperations(arr, N)
{ // Stores the minimum operations
// required
let mini = Number.MAX_SAFE_INTEGER;
for (let i = 0; i < N; i++)
{
// Stores the powers of two for
// the current array element
let count = 0;
// Dividing by 2
while (arr[i] % 2 == 0) {
arr[i] = Math.floor(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 let arr = [4, 6]; let N = arr.length; document.write(minimumOperations(arr, N)); // This code is contributed by saurabh_jaiswal. </script> |
1
Time Complexity: O(N*log N)
Auxiliary Space: O(1)