# Maximize the Sum of the given array using given operations

Given two arrays A[] and B[] consisting of N integers and an integer K, the task is to maximize the sum calculated from the array A[] by the following operations:

• For every index in B[] containing 0, the corresponding index in A[] is added to the sum.
• For every index in B[] containing 1, add the value at the corresponding index in A[] to the sum for atmost K such indices. For the remaining indices, subtract from the sum.

Examples:

Input: A[] = {5, 4, 6, 2, 8}, B[] = {1, 0, 1, 1, 0}, K = 2
Output: 21
Explanation:
Add A and A to the sum as B = B = 0
Therefore, sum = 4 + 8 = 12.
Now, add A and A to the sum as K elements can be added.
Finally, subtract 2 from the sum.
Therefore, the maximum possible sum = 12 + 5 + 6 – 2 = 21

Input: A[] = {5, 2, 1, 8, 10, 5}, B[] = {1, 1, 1, 1, 0, 0}, K = 3
Output: 29

Approach:

Follow the steps below to solve the problem:

• Sort the array A[] in decreasing order.
• To maximize the sum, add first K elements from the sorted array corresponding to which the index in B[] contains 1. Subtract the remaining such elements.
• Add to the sum all the values in A[] corresponding to an index in B[] containing 0.

Below is the implementation of the above approach:

## C++

 `// C++ Program to maximise the ` `// sum of the given array ` `#include ` `using` `namespace` `std; ` ` `  `// Comparator to sort the array ` `// in ascending order ` `bool` `compare(pair<``int``, ``int``> p1, ` `             ``pair<``int``, ``int``> p2) ` `{ ` `    ``return` `p1.first > p2.first; ` `} ` ` `  `// Function to maximise the sum of ` `// the given array ` `int` `maximiseScore(``int` `A[], ``int` `B[], ` `                  ``int` `K, ``int` `N) ` `{ ` ` `  `    ``// Stores {A[i], B[i]} pairs ` `    ``vector > pairs(N); ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``pairs[i] = make_pair(A[i], B[i]); ` `    ``} ` ` `  `    ``// Sort in descending order of the ` `    ``// values in the array A[] ` `    ``sort(pairs.begin(), pairs.end(), compare); ` ` `  `    ``// Stores the maximum sum ` `    ``int` `sum = 0; ` `    ``for` `(``int` `i = 0; i < N; i++) { ` ` `  `        ``// If B[i] is equal to 0 ` `        ``if` `(pairs[i].second == 0) { ` ` `  `            ``// Simply add A[i] to the sum ` `            ``sum += pairs[i].first; ` `        ``} ` ` `  `        ``else` `if` `(pairs[i].second == 1) { ` ` `  `            ``// Add the highest K numbers ` `            ``if` `(K > 0) { ` `                ``sum += pairs[i].first; ` `                ``K--; ` `            ``} ` ` `  `            ``// Subtract from the sum ` `            ``else` `{ ` `                ``sum -= pairs[i].first; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Return the sum ` `    ``return` `sum; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` ` `  `    ``int` `A[] = { 5, 4, 6, 2, 8 }; ` `    ``int` `B[] = { 1, 0, 1, 1, 0 }; ` `    ``int` `K = 2; ` `    ``int` `N = ``sizeof``(A) / ``sizeof``(``int``); ` `    ``cout << maximiseScore(A, B, K, N); ` `    ``return` `0; ` `} `

## Java

 `// Java program to maximise the ` `// sum of the given array ` `import` `java.util.*; ` ` `  `class` `Pair ``implements` `Comparable ` `{ ` `    ``int` `first, second; ` `    ``Pair(``int` `x, ``int` `y) ` `    ``{ ` `        ``first = x; ` `        ``second = y; ` `    ``} ` `    ``public` `int` `compareTo(Pair p)  ` `    ``{ ` `        ``return` `p.first - first;  ` `    ``} ` `} ` ` `  `class` `GFG{ ` `     `  `// Function to maximise the sum of ` `// the given array ` `static` `int` `maximiseScore(``int` `A[], ``int` `B[], ` `                         ``int` `K, ``int` `N) ` `{ ` ` `  `    ``// Stores {A[i], B[i]} pairs ` `    ``ArrayList pairs = ``new` `ArrayList<>(); ` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `        ``pairs.add(``new` `Pair(A[i], B[i])); ` `    ``} ` ` `  `    ``// Sort in descending order of the ` `    ``// values in the array A[] ` `    ``Collections.sort(pairs); ` ` `  `    ``// Stores the maximum sum ` `    ``int` `sum = ``0``; ` `    ``for``(``int` `i = ``0``; i < N; i++) ` `    ``{ ` `         `  `        ``// If B[i] is equal to 0 ` `        ``if` `(pairs.get(i).second == ``0``) ` `        ``{ ` `             `  `            ``// Simply add A[i] to the sum ` `            ``sum += pairs.get(i).first; ` `        ``} ` ` `  `        ``else` `if` `(pairs.get(i).second == ``1``)  ` `        ``{ ` `             `  `            ``// Add the highest K numbers ` `            ``if` `(K > ``0``) ` `            ``{ ` `                ``sum += pairs.get(i).first; ` `                ``K--; ` `            ``} ` ` `  `            ``// Subtract from the sum ` `            ``else`  `            ``{ ` `                ``sum -= pairs.get(i).first; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Return the sum ` `    ``return` `sum; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` `  `    ``int` `A[] = { ``5``, ``4``, ``6``, ``2``, ``8` `}; ` `    ``int` `B[] = { ``1``, ``0``, ``1``, ``1``, ``0` `}; ` `    ``int` `K = ``2``; ` `    ``int` `N = A.length; ` `     `  `    ``System.out.print(maximiseScore(A, B, K, N)); ` `} ` `} ` ` `  `// This code is contributed by jrishabh99 `

Output:

```21
```

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

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