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 2int 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 codeint 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 2def 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 callprint(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. |
3
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