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

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  • Last Updated : 31 May, 2022
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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

1+N=2+(N-1)=3+(N-2)=...

  • If N is odd, we can make the maximum count of equal elements by

N=1+(N-1)=2+(N-2)=...
 

Therefore, the answer will always be 

\lceil \frac{N}{2} \rceil
 

Below is the implementation of the above approach:

 

C++




// C++ implementation of
// the above approach
 
#include <bits/stdc++.h>
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[0]);
 
    // 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

Javascript




<script>
 
// Javascript implementation of
// the above approach
 
// Function to count maximum number
// of array elements equal
function countEqual(n)
{
    return parseInt((n + 1) / 2);
}
 
// Driver Code
var arr = [ 1, 2, 3, 4, 5, 6 ];
var n = arr.length;
 
// Function Call
document.write( countEqual(n));
 
// This code is contributed by rrrtnx.
</script>

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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