Maximize array sum by replacing at most L elements to R for Q queries

Last Updated : 15 Feb, 2023

Given an array arr[] consisting of N integers and an array Query[][] consisting of M pairs of the type {L, R}, the task is to find the maximum sum of the array by performing the queries Query[][] such that for each query {L, R} replace at most L array elements to the value R.

Examples:

Input: arr[]= {5, 1, 4}, Query[][] = {{2, 3}, {1, 5}}
Output: 14
Explanation:
Following are the operations performed:
Query 1: For the Query {2, 3}, do nothing.
Query 2: For the Query {1, 5}, replace at most L(= 1) array element with value R(= 5), replace arr[1] with value 5.
After the above steps, array modifies to {5, 5, 4}. The sum of array element is 14, which is maximum.

Input: arr[] = {1, 2, 3, 4}, Query[][] = {{3, 1}, {2, 5}}
Output: 17

Approach: The given problem can be solved with the help of the Greedy Approach. The main idea to maximize the array sum is to perform the query to increase the minimum number to a maximum value as the order of the operations does not matter as they are independent of each other. Follow the steps below to solve the given problem:

Maintain a min-heap priority queue and store all the array elements.
Traverse the given array of queries Q[][] and for each query {L, R} perform the following steps:
- Change the value of at most L elements smaller than R to the value R, starting from the smallest.
- Perform the above operation, pop the elements smaller than R and push R at their places in the priority queue.
After completing the above steps, print the sum of values stored in the priority queue as the maximum sum.

Below is the implementation of the above approach:

C++

// C++ program for the above approach  #include <bits/stdc++.h> using namespace std;  // Function to find the maximum array // sum after performing M queries void maximumArraySumWithMQuery(     int arr[], vector<vector<int> >& Q,     int N, int M) {      // Maintain a min-heap Priority-Queue     priority_queue<int, vector<int>,                    greater<int> >         pq;      // Push all the elements in the     // priority queue     for (int i = 0; i < N; i++) {         pq.push(arr[i]);     }      // Iterate through M Operations     for (int i = 0; i < M; i++) {          // Iterate through the total         // possible changes allowed         // and maximize the array sum         int l = Q[i][0];         int r = Q[i][1];          for (int j = 0; j < l; j++) {              // Change the value of elements             // less than r to r, starting             // from the smallest             if (pq.top() < r) {                 pq.pop();                 pq.push(r);             }              // Break if current element >= R             else {                 break;             }         }     }      // Find the resultant maximum sum     int ans = 0;      while (!pq.empty()) {         ans += pq.top();         pq.pop();     }      // Print the sum     cout << ans; }  // Driver Code int main() {     int N = 3, M = 2;     int arr[] = { 5, 1, 4 };     vector<vector<int> > Query         = { { 2, 3 }, { 1, 5 } };      maximumArraySumWithMQuery(         arr, Query, N, M);      return 0; }

Java

// Java program for the above approach  import java.util.PriorityQueue;  class GFG {      // Function to find the maximum array     // sum after performing M queries     public static void maximumArraySumWithMQuery(int arr[], int[][] Q, int N, int M) {          // Maintain a min-heap Priority-Queue         PriorityQueue<Integer> pq = new PriorityQueue<Integer>();          // Push all the elements in the         // priority queue         for (int i = 0; i < N; i++) {             pq.add(arr[i]);         }          // Iterate through M Operations         for (int i = 0; i < M; i++) {              // Iterate through the total             // possible changes allowed             // and maximize the array sum             int l = Q[i][0];             int r = Q[i][1];              for (int j = 0; j < l; j++) {                  // Change the value of elements                 // less than r to r, starting                 // from the smallest                 if (pq.peek() < r) {                     pq.remove();                     pq.add(r);                 }                  // Break if current element >= R                 else {                     break;                 }             }         }          // Find the resultant maximum sum         int ans = 0;          while (!pq.isEmpty()) {             ans += pq.peek();             pq.remove();         }          // Print the sum         System.out.println(ans);     }      // Driver Code     public static void main(String args[]) {         int N = 3, M = 2;         int arr[] = { 5, 1, 4 };         int[][] Query = { { 2, 3 }, { 1, 5 } };          maximumArraySumWithMQuery(arr, Query, N, M);      } }  // This code is contributed by saurabh_jaiswal.

Python3

# Python program for the above approach from queue import PriorityQueue  # Function to find the maximum array # sum after performing M queries def maximumArraySumWithMQuery(arr, Q, N, M):      # Maintain a min-heap Priority-Queue     pq = PriorityQueue()      # Push all the elements in the     # priority queue     for i in range(N):         pq.put(arr[i])      # Iterate through M Operations     for i in range(M):          # Iterate through the total         # possible changes allowed         # and maximize the array sum         l = Q[i][0];         r = Q[i][1];          for j in range(l):              # Change the value of elements             # less than r to r, starting             # from the smallest             if (pq.queue[0] < r):                 pq.get();                 pq.put(r);               # Break if current element >= R             else:                 break              # Find the resultant maximum sum     ans = 0;          while ( not pq.empty() ):         ans += pq.queue[0];         pq.get();       # Print the sum     print(ans)   # Driver Code N = 3 M = 2 arr = [5, 1, 4] Query = [[2, 3], [1, 5]]  maximumArraySumWithMQuery(arr, Query, N, M)  # This code is contributed by gfgking.

// C# Program for the above approach using System; using System.Collections.Generic;  public class GFG {    // Function to find the maximum array   // sum after performing M queries   public static void maximumArraySumWithMQuery(int[] arr,                                                int[, ] Q,                                                int N,                                                int M)   {      // Maintain a min-heap Priority-Queue     List<int> pq = new List<int>();      // Push all the elements in the     // priority queue     for (int i = 0; i < N; i++) {       pq.Add(arr[i]);     }     pq.Sort();      // Iterate through M Operations     for (int i = 0; i < M; i++) {        // Iterate through the total       // possible changes allowed       // and maximize the array sum       int l = Q[i, 0];       int r = Q[i, 1];        for (int j = 0; j < l; j++) {          // Change the value of elements         // less than r to r, starting         // from the smallest         if (pq[0] < r) {           pq.Remove(pq[0]);           pq.Add(r);           pq.Sort();         }          // Break if current element >= R         else {           break;         }       }     }      // Find the resultant maximum sum     int ans = 0;      while (pq.Count > 0) {       ans += pq[0];       pq.Remove(pq[0]);     }      // Print the sum     Console.WriteLine(ans);   }    // Driver Code   public static void Main(string[] args)   {     int N = 3, M = 2;     int[] arr = { 5, 1, 4 };     int[, ] Query = { { 2, 3 }, { 1, 5 } };      maximumArraySumWithMQuery(arr, Query, N, M);   } }  // This code is contributed by akashish__

JavaScript

function maximumArraySumWithMQuery(arr, Q, N, M) {     let pq = [];      // Push all the elements in the     // priority queue     for (let i = 0; i < N; i++) {         pq.push(arr[i]);     }      // Iterate through M Operations     for (let i = 0; i < M; i++) {          // Iterate through the total         // possible changes allowed         // and maximize the array sum         let l = Q[i][0];         let r = Q[i][1];          for (let j = 0; j < l; j++) {              // Change the value of elements             // less than r to r, starting             // from the smallest             if (pq[0] < r) {                 pq.shift();                 pq.push(r);                 pq.sort((a, b) => a - b);             }              // Break if current element >= R             else {                 break;             }         }     }      // Find the resultant maximum sum     let ans = 1;      while (pq.length > 0) {         ans += pq.shift() + 1;     }      // Print the sum     console.log(ans); }  // Driver Code let N = 3, M = 2; let arr = [5, 1, 4]; let Query = [[2, 3], [1, 5]];  maximumArraySumWithMQuery(arr, Query, N, M);

Output:

Time Complexity: O(M*N*log N)
Auxiliary Space: O(N)

Minimize operations to reduce Array sum by half by reducing any elements by half

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Maximize array sum by replacing at most L elements to R for Q queries

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