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Maximum median possible by generating an Array of size N with sum X

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  • Last Updated : 24 Dec, 2021

Given two positive integers N and X. The task is to print the maximum median possible by generating an Array of size N with sum X

Examples: 

Input: N = 1, X = 7
Output: 7
Explanation: Array can be: [7], median is the 1st element, i.e., 7.

Input: N = 7, X = 18
Output: 4
Explanation: One of the possible arrays can be: [0, 1, 2, 3, 4, 4, 4]. The median = ceil(n/2)th element  = ceil(7/2) = 5th element, i.e., 4.

 

Approach:  Consider that the median needs to be maximized so the greedy approach can be to make all the elements before the position of the median element as zero and equally divide the sum X among the rest of the elements.
Follow the below steps to solve the problem:

  • If n = 1, print X.
  • For n >= 2.
  • Create a variable median_pos = ceil((double)(n)/2.0).
  • Decrement median_pos, as to represent the index value.
  • Create a variable median = X/(n-median_pos).
  • Print median.

Below is the implementation for the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum median possible
int maximizeMedian(int n, int X)
{
    // If only 1 element present
    if (n == 1) {
        return X;
    }
    else {
        // Position of median
        int median_pos = ceil((double)(n) / (2.0));
        median_pos--;
        int median = X / (n - median_pos);
        return median;
    }
    return 0;
}
 
// Driver Code
int main()
{
    int n = 1, X = 7;
    cout << maximizeMedian(n, X);
}


Java




// Java program for the above approach
import java.util.*;
public class GFG
{
   
    // Function to find the maximum median possible
    static int maximizeMedian(int n, int X)
    {
       
        // If only 1 element present
        if (n == 1) {
            return X;
        }
        else {
            // Position of median
            int median_pos
                = (int)Math.ceil((double)(n) / (2.0));
            median_pos--;
            int median = X / (n - median_pos);
            return median;
        }
    }
 
    // Driver Code
    public static void main(String args[])
    {
        int n = 1, X = 7;
        System.out.println(maximizeMedian(n, X));
    }
}
 
// This code is contributed by Samim Hossain Mondal.


Python3




# Python code for the above approach
 
# Function to find the maximum median possible
def maximizeMedian(n, X):
 
    # If only 1 element present
    if (n == 1):
        return X
    else:
        # Position of median
        median_pos = (n) // (2.0)
        median_pos -= 1
        median = X // (n - median_pos)
        return median
    return 0
 
# Driver Code
 
n = 1
X = 7
print(maximizeMedian(n, X))
 
# This code is contributed by gfgking


C#




// C# program for the above approach
using System;
class GFG
{
   
    // Function to find the maximum median possible
    static int maximizeMedian(int n, int X)
    {
       
        // If only 1 element present
        if (n == 1) {
            return X;
        }
        else {
            // Position of median
            int median_pos
                = (int)Math.Ceiling((double)(n) / (2.0));
            median_pos--;
            int median = X / (n - median_pos);
            return median;
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 1, X = 7;
        Console.WriteLine(maximizeMedian(n, X));
    }
}
 
// This code is contributed by ukasp.


Javascript




<script>
       // JavaScript code for the above approach
 
       // Function to find the maximum median possible
       function maximizeMedian(n, X)
       {
        
           // If only 1 element present
           if (n == 1) {
               return X;
           }
           else
           {
            
               // Position of median
               let median_pos = Math.ceil((n) / (2.0));
               median_pos--;
               let median = X / (n - median_pos);
               return median;
           }
           return 0;
       }
 
       // Driver Code
 
       let n = 1, X = 7;
       document.write(maximizeMedian(n, X));
 
 // This code is contributed by Potta Lokesh
   </script>


 
 

Output

7

 

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

 


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