# Maximum possible Array sum after performing given operations

• Difficulty Level : Easy
• Last Updated : 17 Feb, 2023

Given array arr[] of positive integers, an integer Q, and arrays X[] and Y[] of size Q. For each element in arrays X[] and Y[], we can perform the below operations:

• For each query from array X[] and Y[], select at most X[i] elements from array arr[] and replace all the selected elements with integer Y[i].
• After performing Q operations, the task is to obtain maximum sum from the array arr[].

Examples:

Input: arr[] = {5, 2, 6, 3, 8, 5, 4, 7, 9, 10}, Q = 3, X[] = {2, 4, 1}, Y[] = {4, 3, 10}
Output: 68
Explanation:
For i = 1,
We can replace atmost 2 elements from array arr[] with integer 4. Here 2 element of array arr[] are smaller than 4 so we will replace elements 2 and 3 from array arr[] with 4 and arr[] becomes {5, 4, 6, 4, 8, 5, 4, 7, 9, 10}.
For i = 2,
We can replace at most 4 elements from array ar[] with integer 3, but no element of array arr[] is smaller than 3. So we will not replace anything.
For i = 3,
We can replace at most 1 element from array arr[] with integer 10, 9 elements of array arr[] are smaller than 10. To get the maximum sum, we will replace the smallest element from array arr[] with 10. Array arr[] after 3rd operation = {5, 10, 6, 4, 8, 5, 10, 7, 9, 10 }. The maximum possible sum is 68.
Input: ar[] = {200, 100, 200, 300}, Q = 2, X[] = {2, 3}, Y[] = {100, 90}
Output: 800
Explanation:
For i = 1,
We can replace atmost 2 elements from array arr[] with integer 100, no element of array arr[] is smaller than 100. So we will replace 0 elements.
For i = 2,
We can replace at most 3 elements from array arr[] with integer 90, no element of array arr[] is smaller than 90. So we will replace 0 elements. So the maximum sum we can obtain after q operation is 800.

Naive Approach: The naive idea is to pick X[i] number elements from the array arr[]. If the elements in the array are less than Y[i] then update X[i] of such elements.
Time Complexity: (N2), as we will be using nested loops for traversing N*N times. Where N is the number of elements in the array.
Auxiliary Space: O(1), as we will not be using any extra space.

Efficient Approach: The idea is to use a priority queue to get the element with higher value before the element with lower value, precisely priority queue of pairs to store value with its frequency. Below are the steps:

• Insert each element of the array arr[] with their occurrence in the priority queue.
• For each element(say X[i]) in the array X[] do the following:
1. Choose at most X[i] number of minimum element from the priority queue.
2. Replace it with Y[i] if choose element is less than Y[i].
3. Insert back the replaced element into the priority queue with their corresponding frequency.
• After the above operations the array arr[] will have elements such that sum of all element is maximum. Print the sum.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the``// maximum possible sum of array``// after performing given operations``#include ``using` `namespace` `std;` `// Function to get maximum``// sum after q operations``void` `max_sum(``int` `ar[], ``int` `n,``             ``int` `q, ``int` `x[], ``int` `y[])``{``    ``int` `ans = 0, i;` `    ``// priority queue to``    ``// get maximum sum``    ``priority_queue > pq;` `    ``// Push pair, value and 1``    ``// in the priority queue``    ``for` `(i = 0; i < n; i++)``        ``pq.push({ ar[i], 1 });` `    ``// Push pair, value (to be replaced)``    ``// and number of elements (to be replaced)``    ``for` `(i = 0; i < q; i++)``        ``pq.push({ y[i], x[i] });` `    ``// Add top n elements from``    ``// the priority queue``    ``// to get max sum``    ``while` `(n > 0) {` `        ``// pr is the pair``        ``// pr.first is the value and``        ``// pr.second is the occurrence``        ``auto` `pr = pq.top();` `        ``// pop from the priority queue``        ``pq.pop();` `        ``// Add value to answer``        ``ans += pr.first * min(n, pr.second);` `        ``// Update n``        ``n -= pr.second;``    ``}` `    ``cout << ans << ``"\n"``;``}` `// Driver code``int` `main()``{``    ``int` `ar[] = { 200, 100, 200, 300 };``    ``int` `n = (``sizeof` `ar) / (``sizeof` `ar[0]);``    ``int` `q = 2;``    ``int` `x[] = { 2, 3 };``    ``int` `y[] = { 100, 90 };``    ``max_sum(ar, n, q, x, y);` `    ``return` `0;``}`

## Java

 `// Java implementation to find the``// maximum possible sum of array``// after performing given operations``import` `java.util.*;``import` `java.lang.*;` `class` `GFG{` `static` `class` `pair``{``    ``int` `first, second;``    ``pair(``int` `first, ``int` `second)``    ``{``        ``this``.first = first;``        ``this``.second = second;``    ``}``}` `// Function to get maximum``// sum after q operations``static` `void` `max_sum(``int` `ar[], ``int` `n, ``int` `q,``                    ``int` `x[], ``int` `y[])``{``    ``int` `ans = ``0``, i;` `    ``// priority queue to``    ``// get maximum sum``    ``PriorityQueue pq = ``new` `PriorityQueue<>(``        ``(a, b) -> Integer.compare(a.second, b.second));` `    ``// Push pair, value and 1``    ``// in the priority queue``    ``for``(i = ``0``; i < n; i++)``        ``pq.add(``new` `pair(ar[i], ``1` `));` `    ``// Push pair, value (to be replaced)``    ``// and number of elements (to be replaced)``    ``for``(i = ``0``; i < q; i++)``        ``pq.add(``new` `pair(y[i], x[i]));` `    ``// Add top n elements from``    ``// the priority queue``    ``// to get max sum``    ``while` `(n > ``0``)``    ``{``        ` `        ``// pr is the pair``        ``// pr.first is the value and``        ``// pr.second is the occurrence``        ``pair pr = pq.peek();` `        ``// pop from the priority queue``        ``pq.poll();` `        ``// Add value to answer``        ``ans += pr.first * Math.min(n, pr.second);` `        ``// Update n``        ``n -= pr.second;``    ``}``    ``System.out.println(ans);``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `ar[] = { ``200``, ``100``, ``200``, ``300` `};``    ``int` `n = ar.length;``    ``int` `q = ``2``;``    ``int` `x[] = { ``2``, ``3` `};``    ``int` `y[] = { ``100``, ``90` `};``    ` `    ``max_sum(ar, n, q, x, y);``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python implementation to find the``# maximum possible sum of array``# after performing given operations` `from` `queue ``import` `PriorityQueue`  `def` `max_sum(arr, n, q, x, y):``    ``ans ``=` `0``    ``i ``=` `0``    ``#  priority queue to``    ``#  get maximum sum``    ``pq ``=` `PriorityQueue()``    ``# Push pair, value and 1``    ``# in the priority queue``    ``for` `i ``in` `range``(n):``        ``pq.put((``-``arr[i], ``1``))` `    ``# Push pair, value (to be replaced)``    ``# and number of elements (to be replaced)``    ``for` `i ``in` `range``(q):``        ``pq.put((``-``y[i], x[i]))` `    ``# Add top n elements from``    ``# the priority queue``    ``# to get max sum` `    ``while` `n > ``0``:``        ``# pr is the pair``        ``# pr.first is the value and``        ``# pr.second is the occurrence``        ``pr ``=` `pq.get()``        ``# Add value to answer``        ``ans ``+``=` `abs``(pr[``0``]) ``*` `min``(n, pr[``1``])``        ``# Update n``        ``n ``-``=` `pr[``1``]``    ``print``(ans)`  `ar ``=` `[``200``, ``100``, ``200``, ``300``]``n ``=` `len``(ar)``q ``=` `2``x ``=` `[``2``, ``3``]``y ``=` `[``100``, ``90``]``max_sum(ar, n, q, x, y)` `# This code is provided by sdeadityasharma`

## C#

 `// C# implementation to find the``// maximum possible sum of array``// after performing given operations``using` `System;``using` `System.Linq;``using` `System.Collections.Generic;` `public` `class` `Pair {``  ``public` `int` `first;``  ``public` `int` `second;``  ``public` `Pair(``int` `first, ``int` `second)``  ``{``    ``this``.first = first;``    ``this``.second = second;``  ``}``}` `public` `class` `GFG {` `  ``// Function to get maximum``  ``// sum after q operations``  ``static` `void` `MaxSum(``int``[] ar, ``int` `n, ``int` `q, ``int``[] x,``                     ``int``[] y)``  ``{``    ``int` `ans = 0;``    ``int` `i, j;` `    ``// Push pair, value and 1``    ``// in the array``    ``List pq = ``new` `List();``    ``for` `(i = 0; i < n; i++)``      ``pq.Add(``new` `Pair(ar[i], 1));` `    ``// Push pair, value (to be replaced)``    ``// and number of elements (to be replaced)``    ``for` `(i = 0; i < q; i++)``      ``pq.Add(``new` `Pair(y[i], x[i]));` `    ``pq = pq.OrderBy(p => p.second).ToList();` `    ``// Add top n elements from``    ``// the priority queue``    ``// to get max sum``    ``while` `(n > 0) {` `      ``// pr is the pair``      ``// pr.first is the value and``      ``// pr.second is the occurrence``      ``var` `pr = pq[0];` `      ``// pop from the priority queue``      ``pq.RemoveAt(0);` `      ``// Add value to answer``      ``ans += pr.first * Math.Min(n, pr.second);` `      ``// Update n``      ``n -= pr.second;``    ``}``    ``Console.WriteLine(ans);``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(``string``[] args)``  ``{``    ``int``[] ar = { 200, 100, 200, 300 };``    ``int` `n = ar.Length;``    ``int` `q = 2;``    ``int``[] x = { 2, 3 };``    ``int``[] y = { 100, 90 };` `    ``MaxSum(ar, n, q, x, y);``  ``}``}` `// This code is contributed by phasing17`

## Javascript

 `// Javascript implementation to find the``// maximum possible sum of array``// after performing given operations` `// Function to get maximum``// sum after q operations``    ``function` `max_sum(arr, n, q, x, y) {``    ``let ans = 0;``    ` `     ``// priority queue to``    ``// get maximum sum``    ``let pq = [];``      ` `     ``// Push pair, value and 1``    ``// in the priority queue``    ``for` `(let i = 0; i < n; i++) {``        ``pq.push([-arr[i], 1]);``    ``}``    ` `    ``// Push pair, value (to be replaced)``    ``// and number of elements (to be replaced)``    ``for` `(let i = 0; i < q; i++) {``        ``pq.push([-y[i], x[i]]);``    ``}``    ``pq.sort((a, b) => a[0] - b[0]);``    ` `    ``// Add top n elements from``    ``// the priority queue``    ``// to get max sum``    ``while` `(n > 0)``    ``{``    ` `        ``// pr is the pair``        ``// pr.first is the value and``        ``// pr.second is the occurrence``        ``let pr = pq.shift();``        ` `        ``// Add value to answer``        ``ans += Math.abs(pr[0]) * Math.min(n, pr[1]);``        ` `        ``// Update n``        ``n -= pr[1];``    ``}``      ` `    ``console.log(ans);``    ``}``    ` `    ``// Driver code``    ``let ar = [200, 100, 200, 300];``    ``let n = ar.length;``    ``let q = 2;``    ``let x = [2, 3];``    ``let y = [100, 90];``    ` `    ``max_sum(ar, n, q, x, y);``    ` `    ``// This code is contributed by Aman Kumar.`

Output:

`800`

Time Complexity: O(N*log2N), as we are using a loop to traverse N times and priority queue operation will take log2N times. Where N is the number of elements in the array.
Auxiliary Space: O(N), as we are using extra space for the priority queue. Where N is the number of elements in the array.

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