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Make all the array elements odd with minimum operations of given type

  • Difficulty Level : Medium
  • Last Updated : 20 Feb, 2022

Given an array arr[] consisting of even integers. At each move, you can select any even number X from the array and divide all the occurrences of X by 2. The task is to find the minimum number of moves needed so that all the elements in the array become odd.

Examples:  

Input: arr[] = {40, 6, 40, 20} 
Output:
Move 1: Select 40 and divide all the occurrences 
of 40 by 2 to get {20, 6, 20, 20} 
Move 2: Select 20 and divide all the occurrences 
of 20 by 2 to get {10, 6, 10, 10} 
Move 3: Select 10 and divide all the occurrences 
of 10 by 2 to get {5, 6, 5, 5}. 
Move 4: Select 6 and divide it by 2 to get {5, 3, 5, 5}.

Input: arr[] = {2, 4, 16, 8} 
Output:
 

Approach: This problem can be solved using greedy approach. At every move, take the largest remaining even number in the array and divide it by 2. The largest is taken because there is a chance that it can become equal to some other element in the array after it is divided by 2 which minimizes the total operations.

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of
// minimum operations required
int minOperations(int arr[], int n)
{
 
    // Insert all the elements in a set
    set<int> s;
    for (int i = 0; i < n; i++) {
        s.insert(arr[i]);
    }
 
    // To store the number of moves
    int moves = 0;
 
    // While the set is not empty
    while (s.empty() == 0) {
 
        // The last element of the set
        int z = *(s.rbegin());
 
        // If the number is even
        if (z % 2 == 0) {
            moves++;
            s.insert(z / 2);
        }
 
        // Remove the element from the set
        s.erase(z);
    }
 
    return moves;
}
 
// Driver code
int main()
{
    int arr[] = { 40, 6, 40, 20 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << minOperations(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.io.*;
import java.util.*;
 
class GFG
{
    // Function to return the count of
    // minimum operations required
    static int minOperations(int arr[], int n)
    {
 
        // Insert all the elements in a set
        TreeSet<Integer> s = new TreeSet<Integer>();
        for (int i = 0; i < n; i++)
        {
            s.add(arr[i]);
        }
         
        // To store the number of moves
        int moves = 0;
 
        // While the set is not empty
        while (s.size() != 0)
        {
 
            // The last element of the set
            Integer z = s.last();
 
            // If the number is even
            if (z % 2 == 0)
            {
                moves++;
                s.add(z / 2);
            }
 
            // Remove the element from the set
            s.remove(z);
        }
 
        return moves;
    }
 
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 40, 6, 40, 20 };
        int n = arr.length;
 
        System.out.println(minOperations(arr, n));
 
    }
}
 
// This code is contributed by ApurvaRaj

Python3




# Python3 implementation of the approach
from collections import OrderedDict as mpp
 
# Function to return the count of
# minimum operations required
def minOperations(arr, n):
 
    # Insert all the elements in a set
    s = mpp()
    for i in range(n):
        s[arr[i]] = 1
 
    # To store the number of moves
    moves = 0
 
    # While the set is not empty
    while (len(s) > 0):
 
        # The last element of the set
        z = sorted(list(s.keys()))[-1]
 
        # If the number is even
        if (z % 2 == 0):
            moves += 1
            s[z / 2] = 1
 
        # Remove the element from the set
        del s[z]
 
    return moves
 
# Driver code
 
arr = [40, 6, 40, 20]
n = len(arr)
 
print(minOperations(arr, n))
 
# This code is contributed by mohit kumar 29

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
    // Function to return the count of
    // minimum operations required
    static int minOperations(int []arr, int n)
    {
 
        // Insert all the elements in a set
        SortedSet<int> s = new SortedSet<int>();
        for (int i = 0; i < n; i++)
        {
            s.Add(arr[i]);
        }
         
        // To store the number of moves
        int moves = 0;
 
        // While the set is not empty
        while (s.Count != 0)
        {
 
            // The last element of the set
            int z = s.Max;
 
            // If the number is even
            if (z % 2 == 0)
            {
                moves++;
                s.Add(z / 2);
            }
 
            // Remove the element from the set
            s.Remove(z);
        }
 
        return moves;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 40, 6, 40, 20 };
        int n = arr.Length;
 
        Console.WriteLine(minOperations(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to find all the distinct
// remainders when n is divided by
// all the elements from
// the range [1, n + 1]
function findRemainders(n)
{
     
    // Set will be used to store
    // the remainders in order
    // to eliminate duplicates
    var vc = new Set();
 
    // Find the remainders
    for(var i = 1; i <= Math.ceil(Math.sqrt(n)); i++)
        vc.add(parseInt(n / i));
    for(var i = parseInt(n / Math.ceil(Math.sqrt(n))) - 1;
            i >= 0; i--)
        vc.add(i);
 
    // Print the contents of the set
    [...vc].sort((a, b) => a - b).forEach(it => {
        document.write(it + " ");
    });
}
 
// Driver code
var n = 5;
 
findRemainders(n);
 
// This code is contributed by famously
 
</script>
Output: 
4

 

Time Complexity: O(n)

Auxiliary Space: O(n)


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