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Maximize median of Array formed by adding elements of two other Arrays

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  • Last Updated : 02 Sep, 2022
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Given two arrays x[] and y[] of N integers, the task is to form an array z[] of x[i] + y[j], (0 ≤i, j < N) and find the max value of median of z[] formed by optimal rearrangement.

Examples:

Input: N = 5, x[] = {1, 2, 3, 5, 6}, y[] = {4, 7, 0, 9, 8}
Output: 12
Explanation: Let, z[] = { 1+0, 2+4, 5+7, 3+9, 6+8} = {1, 6, 12, 12, 14} 
to get maximum Median i.e., 12.

Input: N = 1, x[] = {11}, y[] = {12}
Output: 23

Approach: The problem can be solved based on the following idea:

The idea is to sort the array, and 

  • For first half (before the median positions) the new array must has the smallest possible elements to just discard this half of elements from both the arrays,  
  • Then from median positions onwards, traverse from right end of 1st array, and from median position of 2nd array, moving in opposite directions,

Find the minimum element formed by sum of these elements, that minimum element will be the median of the new array.

Follow the steps to solve this problem:

  • If N =1, return x[0]+ y[0] as there is only single element,
  • Otherwise, do the following steps
    • First, sort both arrays and then divide the array from min to median-1 and median to max,
    • Then, traverse from median to max and find the sum of opposite elements like, x[Median]+y[n-1], x[median+1]+y[n-2]…. up to Max.
    • Out of these sums calculated above, the minimum one would be the median of the new array.

Below is the implementation of the above approach.

C++




// C++ code to implement the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum median formed
// by addition of two array
double findMedian(int* x, int* y, int n)
{
    // If both arrays contains single element
    if (n == 1) {
        cout << x[0] + y[0] << endl;
    }
    else {
        // Sort both the arrays
        sort(x, x + n);
        sort(y, y + n);
 
        // Initialize counters
        int i = (n - 1) / 2;
        int j = n - 1;
        double Med = INT_MAX;
 
        // Find maximum median
        if (n & 1) {
            while (i < n) {
                Med = min(Med,
                          (double)(x[i++] + y[j--]));
            }
        }
        else {
            vector<int> v;
            while (i < n) {
                v.push_back(x[i++] + y[j--]);
            }
            sort(v.begin(), v.end());
            Med = (double)(v[0] + v[1]) / 2;
        }
        return Med;
    }
}
 
// Driver Code
int main()
{
    int x[] = { 1, 2, 3, 5, 6 };
    int y[] = { 4, 7, 0, 9, 8 };
    int N = 5;
 
    // Function call
    cout << findMedian(x, y, N) << endl;
    return 0;
}


Java




// JAVA program for above approach
import java.util.*;
class GFG
{
 
// Function to find maximum median formed
// by addition of two array
static void findMedian(int[] x, int[] y, int n)
{
    // If both arrays contains single element
    if (n == 1) {
        System.out.print( x[0] + y[0]);
    }
    else {
        // Sort both the arrays
        Arrays.sort(x);
        Arrays.sort(y);
 
        // Initialize counters
        int i = (n - 1) / 2;
        int j = n - 1;
        int Med = Integer.MAX_VALUE;
 
        // Find maximum median
        if ((n & 1) != 0) {
            while (i < n) {
                Med = Math.min(Med,
                          (x[i++] + y[j--]));
            }
        }
        else {
            ArrayList<Integer> v
            = new ArrayList<Integer>(n);
            while (i < n) {
                v.add(x[i++] + y[j--]);
            }
            Collections.sort(v);
            Med = ((v.get(0) + v.get(1)) / 2);
        }
        System.out.print( Med);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int x[] = { 1, 2, 3, 5, 6 };
    int y[] = { 4, 7, 0, 9, 8 };
    int N = 5;
 
    // Function call
    findMedian(x, y, N);
}
}
 
// This code is contributed by code_hunt.


Python3




# Python code to implement the above approach
 
# Function to find maximum median formed
# by addition of two array
def findMedian(x, y, n):
   
    # If both arrays contains single element
    if (n == 1):
        print(x[0] + y[0])
    else:
        # Sort both the arrays
        x.sort()
        y.sort()
         
    # Initialize counters
    i = (n - 1) // 2
    j = n - 1
    Med = 100000000000000000
     
    # Find maximum median
    if (n % 2 == 1):
        while (i < n):
            Med = min(Med, (x[i] + y[j]))
            j -= 1
            i += 1
 
    else:
        v = []
        while (i < n):
            v.append(x[i] + y[j])
            i += 1
            j -= 1
 
        v.sort()
        Med = (v[0] + v[1]) // 2
 
    return Med
 
# Driver Code
if __name__ == "__main__":
    x = [1, 2, 3, 5, 6]
    y = [4, 7, 0, 9, 8]
    N = 5
     
    # Function call
    print(findMedian(x, y, N))
 
# This code is contributed by Rohit Pradhan


C#




// C# program for above approach
 
using System;
using System.Collections;
using System.Collections.Generic;
 
public class GFG {
 
    // Function to find maximum median formed
    // by addition of two array
    static void findMedian(int[] x, int[] y, int n)
    {
        // If both arrays contains single element
        if (n == 1) {
            Console.Write(x[0] + y[0]);
        }
        else {
            // Sort both the arrays
            Array.Sort(x);
            Array.Sort(y);
 
            // Initialize counters
            int i = (n - 1) / 2;
            int j = n - 1;
            int Med = Int32.MaxValue;
 
            // Find maximum median
            if ((n & 1) != 0) {
                while (i < n) {
                    Med = Math.Min(Med, (x[i++] + y[j--]));
                }
            }
            else {
                ArrayList v = new ArrayList(n);
                while (i < n) {
                    v.Add(x[i++] + y[j--]);
                }
                v.Sort();
                Med = ((int)v[0] + (int)v[1]) / 2;
            }
            Console.Write(Med);
        }
    }
 
    static public void Main()
    {
 
        // Code
        int[] x = { 1, 2, 3, 5, 6 };
        int[] y = { 4, 7, 0, 9, 8 };
        int N = 5;
 
        // Function call
        findMedian(x, y, N);
    }
}
 
// This code is contributed by lokeshmvs21.


Javascript




<script>
// JavaScript code for the above approach
 
    // Function to find maximum median formed
    // by addition of two array
    function findMedian(x, y, n)
    {
        // If both arrays contains single element
        if (n == 1) {
            document.write(x[0] + y[0]);
        }
        else {
            // Sort both the arrays
            x.sort();
            y.sort();
 
            // Initialize counters
            let i = (n - 1) / 2;
            let j = n - 1;
            let Med = Number.MAX_VALUE;
 
            // Find maximum median
            if ((n & 1) != 0) {
                while (i < n) {
                    Med = Math.min(Med, (x[i++] + y[j--]));
                }
            }
            else {
                v = [];
                while (i < n) {
                    v.push(x[i++] + y[j--]);
                }
                v.sort();
                Med = (v[0] + v[1]) / 2;
            }
            document.write(Med);
        }
    }
 
        // Driver Code
         
        // Code
        let x = [ 1, 2, 3, 5, 6 ];
        let y = [ 4, 7, 0, 9, 8 ];
        let N = 5;
 
        // Function call
        findMedian(x, y, N);
 
// This code is contributed by sanjoy_62.
</script>


Output

12

Time Complexity: O(N * log N)
Auxiliary Space: O(1)


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