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

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 :

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// 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";
}

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Output:

14

Time complexity: O(N log N)



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