# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 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); ` ` `  `    ``int` `X = 2, Y = 2; ` ` `  `    ``cout << findMaxAmount(A, B, n, X, Y) << ``"\n"``; ` `} `

## 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  `

Output:

```14
```

Time complexity: O(N log N)

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