Maximum possible Array sum after performing given operations
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:
- Choose at most X[i] number of minimum element from the priority queue.
- Replace it with Y[i] if choose element is less than Y[i].
- 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 <bits/stdc++.h>
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<pair<int, int> > 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 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<pair> 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
# 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# 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<Pair> pq = new List<Pair>();
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 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.
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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