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

• Last Updated : 31 May, 2022

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 equalint countEqual(int n){    return (n + 1) / 2;} // Driver Codeint main(){    int arr[] = { 1, 2, 3, 4, 5, 6 };     int n = sizeof(arr) / sizeof(arr[0]);     // Function Call    cout << countEqual(n);    return 0;}

## Java

 // Java implementation of// the above approachimport java.io.*; class GFG{ // Function to count maximum number// of array elements equalstatic int countEqual(int n){    return (n + 1) / 2;} // Driver Codepublic 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 equaldef countEqual(n):     return (n + 1) // 2 # Driver Codelst = [ 1, 2, 3, 4, 5, 6 ]n = len(lst) # Function callprint(countEqual(n)) # This code is contributed by vishu2908

## C#

 // C# implementation of// the above approachusing System;class GFG{ // Function to count maximum number// of array elements equalstatic int countEqual(int n){    return (n + 1) / 2;} // Driver Codepublic 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

## Javascript

 

Output:

3

Performance Analysis:

Time Complexity: O(1), as we are not using any loops or recursion.

Auxiliary Space: O(1), as we are not using any extra space.

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