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[1] and A[4] to the sum as B[1] = B[4] = 0
Therefore, sum = 4 + 8 = 12.
Now, add A[0] and A[3] 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 maximize the// sum of the given array#include <bits/stdc++.h>using namespace std;// Comparator to sort the array// in ascending orderbool compare(pairs<int, int> p1, pairs<int, int> p2){ return p1.first > p2.first;}// Function to maximize the sum of// the given arrayint maximizeScore(int A[], int B[], int K, int N){ // Stores {A[i], B[i]} pairs vector<pairs<int, int> > pairs(N); for (int i = 0; i < N; i++) { pairs[i] = make_pairs(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 Codeint 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 << maximizeScore(A, B, K, N); return 0;} |
Java
// Java program to maximise the// sum of the given arrayimport java.util.*;class Pairs implements Comparable<Pairs>{ int first, second; Pairs(int x, int y) { first = x; second = y; } public int compareTo(Pairs p) { return p.first - first; }}class GFG{ // Function to maximise the sum of// the given arraystatic int maximiseScore(int A[], int B[], int K, int N){ // Stores {A[i], B[i]} pairs ArrayList<Pairs> pairs = new ArrayList<>(); for(int i = 0; i < N; i++) { pairs.add(new Pairs(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 Codepublic 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 |
Python3
# Python Program to maximise the # sum of the given array # Comparator to sort the array # in ascending order def compare(p1, p2): return p1[0] > p2[0]# Function to maximise the sum of # the given array def maximiseScore(A, B, K, N): # Stores {A[i], B[i]} pairs pairs = [] for i in range(N): pairs.append([A[i], B[i]]) # Sort in descending order of the # values in the array A[] pairs.sort(key = lambda x:x[0], reverse = True) # Stores the maximum sum Sum = 0 for i in range(N): # If B[i] is equal to 0 if(pairs[i][1] == 0): # Simply add A[i] to the sum Sum += pairs[i][0] elif(pairs[i][1] == 1): # Add the highest K numbers if(K > 0): Sum += pairs[i][0] K -= 1 # Subtract from the sum else: Sum -= pairs[i][0] # Return the sum return Sum# Driver Code A = [5, 4, 6, 2, 8]B = [1, 0, 1, 1, 0]K = 2N = len(A)print(maximiseScore(A, B, K, N))# This code is contributed by avanitrachhadiya2155 |
C#
// C# program to maximise the// sum of the given arrayusing System;using System.Collections;using System.Collections.Generic;class Pairs : IComparable{ public int first, second; public Pairs(int x, int y) { first = x; second = y; } public int CompareTo(object obj) { if (obj == null) return 1; Pairs p = obj as Pairs; return p.first - first; }}class GFG{// Function to maximise the sum of// the given arraystatic int maximiseScore(int[] A, int[] B, int K, int N){ // Stores {A[i], B[i]} pairs List<Pairs> pairs = new List<Pairs>(); for(int i = 0; i < N; i++) { pairs.Add(new Pairs(A[i], B[i])); } // Sort in descending order of the // values in the array A[] pairs.Sort(); // 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 Codepublic 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; Console.Write(maximiseScore(A, B, K, N));}}// This code is contributed by phasing17 |
Javascript
<script>// JavaScript Program to maximize the// sum of the given array// Comparator to sort the array// in ascending orderfunction compare(p1,p2){ return p2[0] - p1[0];}// Function to maximize the sum of// the given arrayfunction maximizeScore(A, B, K, N){ // Stores {A[i], B[i]} pairs let pairs = new Array(N); for (let i = 0; i < N; i++) { pairs[i] = [A[i], B[i]]; } // Sort in descending order of the // values in the array A[] pairs.sort(compare); // Stores the maximum sum let sum = 0; for (let i = 0; i < N; i++) { // If B[i] is equal to 0 if (pairs[i][1] == 0) { // Simply add A[i] to the sum sum += pairs[i][0]; } else if (pairs[i][1] == 1) { // Add the highest K numbers if (K > 0) { sum += pairs[i][0]; K--; } // Subtract from the sum else { sum -= pairs[i][0]; } } } // Return the sum return sum;}// Driver Codelet A = [ 5, 4, 6, 2, 8 ];let B = [ 1, 0, 1, 1, 0 ];let K = 2;let N = A.length;document.write(maximizeScore(A, B, K, N),"</br>");// This code is contributed by shinjanpatra.</script> |
Output:
21
Time complexity: O(N*log(N))
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
