# Maximize count of equal numbers in Array of numbers upto N by replacing pairs with their sum

Given an array arr[] containing natural numbers from 1 to N, the task is to find the maximum number of elements that can be made equal after the below operations:

1. Remove any pair of elements from the array and insert their sum to an array.
2. Repeat the above operation any numbers of times to maximize the count of equal elements.

Examples:

Input: arr[] = {1, 2, 3, 4}
Output:
Explanation:
We can perform following operations:
{1, 2, 3, 4} -> {3, 3, 4} -> 2 elements are equal

Input: arr[] = {1 2 3 4 5 6}
Output:
Explanation:
{1, 2, 3, 4, 5, 6} -> {7, 2, 3, 4, 5} -> {7, 7, 3, 4} -> {7, 7, 37} -> 3 elements are equal

Approach: The key observation in the problem is that:

• If N is even, we can make a maximum count of equal elements by • If N is odd, we can make the maximum count of equal elements by Therefore, the answer will always be Below is the implementation of the above approach:

## C++

 // C++ implementation of // the above approach   #include  using namespace std;   // Function to count maximum number // of array elements equal int countEqual(int n) {     return (n + 1) / 2; }   // Driver Code int main() {     int arr[] = { 1, 2, 3, 4, 5, 6 };       int n = sizeof(arr) / sizeof(arr);       // Function Call     cout << countEqual(n);     return 0; }

## Java

 // Java implementation of // the above approach import java.io.*;   class GFG{   // Function to count maximum number // of array elements equal static int countEqual(int n) {     return (n + 1) / 2; }   // Driver Code public static void main (String[] args) {     int arr[] = { 1, 2, 3, 4, 5, 6 };       int n = arr.length;       // Function call     System.out.println(countEqual(n)); } }   // This code is contributed by AnkitRai01

## Python3

 # Python3 implementation of  # the above approach    # Function to count maximum number  # of array elements equal  def countEqual(n):       return (n + 1) // 2   # Driver Code  lst = [ 1, 2, 3, 4, 5, 6 ] n = len(lst)    # Function call print(countEqual(n))   # This code is contributed by vishu2908

## C#

 // C# implementation of // the above approach using System; class GFG{   // Function to count maximum number // of array elements equal static int countEqual(int n) {     return (n + 1) / 2; }   // Driver Code public static void Main(String[] args) {     int []arr = {1, 2, 3, 4, 5, 6};     int n = arr.Length;       // Function call     Console.WriteLine(countEqual(n)); } }   // This code is contributed by Rajput-Ji

Output:

3



Performance Analysis:

• Time Complexity: O(1)
• Auxiliary Space: O(1)

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