Maximize the sum of X+Y elements by picking X and Y elements from 1st and 2nd array

Namachila November 19, 2025

0 0 9 minutes read

Given two arrays of size N, and two numbers X and Y, the task is to maximize the sum by considering the below points:

Pick x values from the first array and y values from the second array such that the sum of X+Y values is maximum.
It is given that X + Y is equal to N.

Examples:

Input: arr1[] = {1, 4, 1}, arr2[] = {2, 5, 3}, N = 3, X = 2, Y = 1
Output: 8
In order to maximize sum from 2 arrays,
pick 1st and 2nd element from first array and 3rd from second array.

Input: A[] = {1, 4, 1, 2}, B[] = {4, 3, 2, 5}, N = 4, X = 2, Y = 2
Output: 14

Approach: A greedy approach can be used to solve the above problem. Below are the required steps:

Find those elements of arrays first that have maximum value by finding the highest difference between elements of two arrays.
For that, find the absolute difference between the value of the first and second array and then store it in some another array.
Sort this array in decreasing order.
While sorting, track the original positions of elements in the arrays.
Now compare the elements of the two arrays and add the greater value to the maxAmount.
If both have the same value, add an element of the first array if X is not zero else add an element of the second array.
After traversing the arrays completely return the maxAmount calculated.

Below is the implementation of above approach :

C++

// C++ program to print the maximum
// possible sum from two arrays.
#include <bits/stdc++.h>
using namespace std;
 
// class that store values of two arrays
// and also store their absolute difference
class triplet {
public:
    int first;
    int second;
    int diff;
    triplet(int f, int s, int d)
        : first(f), second(s), diff(d)
    {
    }
};
 
// Compare function used to sort array in decreasing order
bool compare(triplet& a, triplet& b)
{
    return a.diff > b.diff; // decreasing order
}
 
/// Function to find the maximum possible
/// sum that can be generated from 2 arrays
int findMaxAmount(int arr1[], int arr2[], int n, int x, int y)
{
    // vector where each index stores 3 things:
    // Value of 1st array
    // Value of 2nd array
    // Their absolute difference
    vector<triplet> v;
 
    for (int i = 0; i < n; i++) {
        triplet t(arr1[i], arr2[i], abs(arr1[i] - arr2[i]));
        v.push_back(t);
    }
 
    // sort according to their absolute difference
    sort(v.begin(), v.end(), compare);
 
    // it will store maximum sum
    int maxAmount = 0;
 
    int i = 0;
 
    // Run loop for N times or
    // value of X or Y becomes zero
    while (i < n && x > 0 && y > 0) {
 
        // if 1st array element has greater
        // value, add it to maxAmount
        if (v[i].first > v[i].second) {
            maxAmount += v[i].first;
            x--;
        }
 
        // if 2nd array element has greater
        // value, add it to maxAmount
        if (v[i].first < v[i].second) {
            maxAmount += v[i].second;
            y--;
        }
 
        // if both have same value, add element
        // of first array if X is not zero
        // else add element of second array
        if (v[i].first == v[i].second) {
            if (x > 0) {
                maxAmount += v[i].first;
                x--;
            }
            else if (y > 0) {
                maxAmount += v[i].second;
                y--;
            }
        }
 
        // increment after picking element
        i++;
    }
 
    // add the remaining values
    // of first array to maxAmount
    while (i < v.size() && x--) {
        maxAmount += v[i++].first;
    }
 
    // add the remaining values of
    // second array to maxAmount
    while (i < v.size() && y--) {
        maxAmount += v[i++].second;
    }
 
    return maxAmount;
}
 
// Driver Code
int main()
{
    int A[] = { 1, 4, 1, 2 };
    int B[] = { 4, 3, 2, 5 };
    int n = sizeof(A) / sizeof(A[0]);
 
    int X = 2, Y = 2;
 
    cout << findMaxAmount(A, B, n, X, Y) << "\n";
}

Java

// Java program to print the maximum 
// possible sum from two arrays. 
import java.util.*;
 
// class that store values of two arrays 
// and also store their absolute difference
class Triplet implements Comparable<Triplet>
{
    int first; 
    int second; 
    int diff; 
 
    Triplet(int f, int s, int d)
    {
        first = f;
        second = s;
        diff = d; 
    }
     
    // CompareTo function used to sort
    // array in decreasing order 
    public int compareTo(Triplet o)
    {
        return o.diff - this.diff;
    }
}
class GFG{
 
// Function to find the maximum possible 
// sum that can be generated from 2 arrays 
public static int findMaxAmount(int arr1[], 
                                int arr2[],
                                int n, int x,
                                int y) 
{ 
     
    // Vector where each index 
    // stores 3 things: 
    // Value of 1st array 
    // Value of 2nd array 
    // Their absolute difference 
    Vector<Triplet> v = new Vector<>(); 
 
    for(int i = 0; i < n; i++) 
    { 
       v.add(new Triplet(arr1[i], arr2[i], 
                         Math.abs(arr1[i] - 
                                  arr2[i]))); 
    } 
 
    // Sort according to their 
    // absolute difference 
    Collections.sort(v); 
 
    // It will store maximum sum 
    int maxAmount = 0; 
 
    int i = 0; 
 
    // Run loop for N times or 
    // value of X or Y becomes zero 
    while (i < n && x > 0 && y > 0)
    { 
         
        // If 1st array element has greater 
        // value, add it to maxAmount 
        if (v.get(i).first > v.get(i).second)
        { 
            maxAmount += v.get(i).first; 
            x--; 
        } 
 
        // If 2nd array element has greater 
        // value, add it to maxAmount 
        if (v.get(i).first < v.get(i).second)
        { 
            maxAmount += v.get(i).second; 
            y--; 
        } 
     
        // If both have same value, add element 
        // of first array if X is not zero 
        // else add element of second array 
        if (v.get(i).first == v.get(i).second)
        { 
            if (x > 0)
            { 
                maxAmount += v.get(i).first; 
                x--; 
            } 
            else if (y > 0)
            { 
                maxAmount += v.get(i).second; 
                y--; 
            } 
        } 
         
        // Increment after picking element 
        i++; 
    } 
 
    // Add the remaining values 
    // of first array to maxAmount 
    while (i < v.size() && x-- > 0)
    { 
        maxAmount += v.get(i++).first; 
    } 
 
    // Add the remaining values of 
    // second array to maxAmount 
    while (i < v.size() && y-- > 0) 
    { 
        maxAmount += v.get(i++).second; 
    } 
     
    return maxAmount; 
} 
 
// Driver Code 
public static void main(String []args) 
{ 
    int A[] = { 1, 4, 1, 2 }; 
    int B[] = { 4, 3, 2, 5 }; 
    int n = A.length; 
 
    int X = 2, Y = 2; 
 
    System.out.println(findMaxAmount(A, B, n, X, Y)); 
}
}
 
// This code is contributed by jrishabh99

Python3

# Python3 program to print the maximum 
# possible sum from two arrays. 
 
# Class that store values of two arrays 
# and also store their absolute difference 
class triplet:
     
    def __init__(self, f, s, d):
        self.first = f
        self.second = s
        self.diff = d
 
# Function to find the maximum possible 
# sum that can be generated from 2 arrays 
def findMaxAmount(arr1, arr2, n, x, y): 
 
    # vector where each index stores 3 things: 
    # Value of 1st array 
    # Value of 2nd array 
    # Their absolute difference 
    v = [] 
 
    for i in range(0, n): 
        t = triplet(arr1[i], arr2[i], 
                abs(arr1[i] - arr2[i])) 
        v.append(t) 
 
    # sort according to their absolute difference 
    v.sort(key = lambda x: x.diff, reverse = True)
 
    # it will store maximum sum 
    maxAmount, i = 0, 0
 
    # Run loop for N times or 
    # value of X or Y becomes zero 
    while i < n and x > 0 and y > 0: 
 
        # if 1st array element has greater 
        # value, add it to maxAmount 
        if v[i].first > v[i].second: 
            maxAmount += v[i].first 
            x -= 1
 
        # if 2nd array element has greater 
        # value, add it to maxAmount 
        if v[i].first < v[i].second: 
            maxAmount += v[i].second 
            y -= 1
 
        # if both have same value, add element 
        # of first array if X is not zero 
        # else add element of second array 
        if v[i].first == v[i].second: 
            if x > 0: 
                maxAmount += v[i].first 
                x -= 1
             
            elif y > 0: 
                maxAmount += v[i].second 
                y -= 1
 
        # increment after picking element 
        i += 1
     
    # add the remaining values 
    # of first array to maxAmount 
    while i < len(v) and x > 0: 
        maxAmount += v[i].first
        i, x = i + 1, x - 1
 
    # add the remaining values of 
    # second array to maxAmount 
    while i < len(v) and y > 0: 
        maxAmount += v[i].second
        i, y = i + 1, y - 1
     
    return maxAmount 
 
# Driver Code 
if __name__ == "__main__": 
 
    A = [1, 4, 1, 2] 
    B = [4, 3, 2, 5] 
    n = len(A) 
 
    X, Y = 2, 2
 
    print(findMaxAmount(A, B, n, X, Y))
 
# This code is contributed by Rituraj Jain 

C#

// C# program to print the maximum
// possible sum from two arrays.
using System;
using System.Collections.Generic;
 
// class that store values of two arrays
// and also store their absolute difference
class Triplet : IComparable<Triplet> {
  public int first;
  public int second;
  public int diff;
 
  public Triplet(int f, int s, int d)
  {
    first = f;
    second = s;
    diff = d;
  }
 
  // CompareTo function used to sort
  // array in decreasing order
  public int CompareTo(Triplet o)
  {
    return o.diff - this.diff;
  }
}
 
class GFG {
  // Function to find the maximum possible
  // sum that can be generated from 2 arrays
  public static int findMaxAmount(int[] arr1, int[] arr2,
                                  int n, int x, int y)
  {
 
    // List where each index
    // stores 3 things:
    // Value of 1st array
    // Value of 2nd array
    // Their absolute difference
    List<Triplet> v = new List<Triplet>();
 
    for (int j = 0; j < n; j++) {
      v.Add(new Triplet(arr1[j], arr2[j],
                        Math.Abs(arr1[j] - arr2[j])));
    }
 
    // Sort according to their
    // absolute difference
    v.Sort();
 
    // It will store maximum sum
    int maxAmount = 0;
 
    int i = 0;
 
    // Run loop for N times or
    // value of X or Y becomes zero
    while (i < n && x > 0 && y > 0) {
 
      // If 1st array element has greater
      // value, add it to maxAmount
      if (v[i].first > v[i].second) {
        maxAmount += v[i].first;
        x--;
      }
 
      // If 2nd array element has greater
      // value, add it to maxAmount
      if (v[i].first < v[i].second) {
        maxAmount += v[i].second;
        y--;
      }
 
      // If both have same value, add element
      // of first array if X is not zero
      // else add element of second array
      if (v[i].first == v[i].second) {
        if (x > 0) {
          maxAmount += v[i].first;
          x--;
        }
        else if (y > 0) {
          maxAmount += v[i].second;
          y--;
        }
      }
 
      // Increment after picking element
      i++;
    }
 
    // Add the remaining values
    // of first array to maxAmount
    while (i < v.Count && x-- > 0) {
      maxAmount += v[i++].first;
    }
 
    // Add the remaining values of
    // second array to maxAmount
    while (i < v.Count && y-- > 0) {
      maxAmount += v[i++].second;
    }
 
    return maxAmount;
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    int[] A = { 1, 4, 1, 2 };
    int[] B = { 4, 3, 2, 5 };
    int n = A.Length;
 
    int X = 2, Y = 2;
 
    Console.WriteLine(findMaxAmount(A, B, n, X, Y));
  }
}

Javascript

<script>
 
// JavaScript program to print the maximum
// possible sum from two arrays.
 
// Class that store values of two arrays
// && also store their absolute difference
class triplet{
     
    constructor(f, s, d){
        this.first = f
        this.second = s
        this.diff = d
    }
}
// Function to find the maximum possible
// sum that can be generated from 2 arrays
function findMaxAmount(arr1, arr2, n, x, y){
 
    // vector where each index stores 3 things:
    // Value of 1st array
    // Value of 2nd array
    // Their absolute difference
    let v = []
 
    for(let i = 0; i < n; i++){
        let t = new triplet(arr1[i], arr2[i],
                Math.abs(arr1[i] - arr2[i]))
        v.push(t)
    }
 
    // sort according to their absolute difference
    v.sort((a,b) => b.diff - a.diff)
 
    // it will store maximum sum
    let maxAmount=0, i = 0
 
    // Run loop for N times or
    // value of X or Y becomes zero
    while(i < n && x > 0 && y > 0){
 
        // if 1st array element has greater
        // value, add it to maxAmount
        if(v[i].first > v[i].second){
            maxAmount += v[i].first
            x -= 1
        }
 
        // if 2nd array element has greater
        // value, add it to maxAmount
        if(v[i].first < v[i].second){
            maxAmount += v[i].second
            y -= 1
        }
        // if both have same value, add element
        // of first array if X is not zero
        // else add element of second array
        if(v[i].first == v[i].second){
            if(x > 0){
                maxAmount += v[i].first
                x--
            }
            else if (y > 0){
                maxAmount += v[i].second
                y--
            }
        }
 
        // increment after picking element
        i++
    }
     
    // add the remaining values
    // of first array to maxAmount
    while(i < v.length && x > 0){
        maxAmount += v[i].first
        i++
        x--
    }
    // add the remaining values of
    // second array to maxAmount
    while(i < v.length && y > 0){
        maxAmount += v[i].second
        i++
        y--
    }
     
    return maxAmount
}
 
// Driver Code
 
let A = [1, 4, 1, 2]
let B = [4, 3, 2, 5]
let n = A.length
 
let X = 2, Y = 2
 
document.write(findMaxAmount(A, B, n, X, Y))
 
// This code is contributed by shinjanpatra
 
</script>

Output

complexity Analysis:

Time complexity: O(N log N)
Auxiliary Space: O(N)

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