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Maximum possible elements which are divisible by 2
  • Difficulty Level : Hard
  • Last Updated : 23 Jul, 2019

Given an integer array arr of size N. The task is to find the maximum possible elements in the array which are divisible by 2 after modifying the array. One can perform below operation an arbitrary number of times(possibly zero times).

Replace any two elements in the array with their sum.

Examples:

Input : arr = [1, 2, 3, 1, 3]
Output : 3
After adding elements at index 0 and 2, and index 3 and 4, array becomes arr=[4, 2, 4].

Input : arr = [1, 2, 3, 4, 5]
Output : 3
After adding 1 and 3, array becomes arr=[4, 2, 4, 5].



Approach :
First, observation is that we don’t need to modify elements which are divisible by 2(i.e., even numbers). Then we left with odd numbers. Addition of two numbers will give an even number which is divisible by 2.

So finally, the result will be:

count_even + count_odd/2.

Below is the implementation of the above approach:

CPP




// CPP program to find maximum possible
// elements which divisible by 2
#include <bits/stdc++.h>
using namespace std;
  
// Function to find maximum possible
// elements which divisible by 2
int Divisible(int arr[], int n)
{
    // To store count of even numbers
    int count_even = 0;
  
    for (int i = 0; i < n; i++)
        if (arr[i] % 2 == 0)
            count_even++;
  
    // All even numbers and half of odd numbers
    return count_even + (n - count_even) / 2;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
  
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function call
    cout << Divisible(arr, n);
  
    return 0;
}

Java




// Java program to find maximum possible 
// elements which divisible by 2 
class GFG 
{
  
    // Function to find maximum possible 
    // elements which divisible by 2 
    static int Divisible(int arr[], int n) 
    
        // To store count of even numbers 
        int count_even = 0
      
        for (int i = 0; i < n; i++) 
            if (arr[i] % 2 == 0
                count_even++; 
      
        // All even numbers and half of odd numbers 
        return count_even + (n - count_even) / 2
    
      
    // Driver code 
    public static void main (String[] args) 
    {
        int arr[] = { 1, 2, 3, 4, 5 }; 
      
        int n = arr.length; 
      
        // Function call 
        System.out.println(Divisible(arr, n)); 
    }
}
  
// This code is contributed by AnkitRai01

Python3




# Python3 program to find maximum possible
# elements which divisible by 2
  
# Function to find maximum possible
# elements which divisible by 2
def Divisible(arr, n):
    # To store count of even numbers
    count_even = 0
  
    for i in range(n):
        if (arr[i] % 2 == 0):
            count_even+=1
  
    # All even numbers and half of odd numbers
    return count_even + (n - count_even) // 2
  
# Driver code
  
arr=[1, 2, 3, 4, 5]
  
n = len(arr)
  
# Function call
print(Divisible(arr, n))
  
# This code is contribute by mohit kumar 29

C#




// C# program to find maximum possible 
// elements which divisible by 2 
using System;
  
class GFG
{
      
    // Function to find maximum possible 
    // elements which divisible by 2 
    static int Divisible(int []arr, int n) 
    
        // To store count of even numbers 
        int count_even = 0; 
      
        for (int i = 0; i < n; i++) 
            if (arr[i] % 2 == 0) 
                count_even++; 
      
        // All even numbers and half of odd numbers 
        return count_even + (n - count_even) / 2; 
    
      
    // Driver code 
    static public void Main ()
    {
          
        int []arr = { 1, 2, 3, 4, 5 }; 
        int n = arr.Length; 
          
        // Function call 
        Console.Write(Divisible(arr, n)); 
    }
}
  
// This code is contributed by ajit.
Output:
3

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