给定正整数数组arr [] ，整数Q以及大小Q的数组X []和Y [] 。对于数组X []和Y []中的每个元素，我们可以执行以下操作：
- 对于数组X []和Y []的每个查询，请从数组arr []中最多选择X [i]个元素，并将所有选定元素替换为整数Y [i] 。
- 执行Q操作后，任务是从数组arr []获得最大和。
例子：
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.
天真的方法：天真的想法是从数组arr []中选择X [i]个数字元素。如果数组中的元素小于Y [i]，则更新此类元素的X [i] 。
时间复杂度： (N 2 )
辅助空间： O(1)
高效的方法：这个想法是使用优先级队列来获得具有较高价值的元素，然后再使用具有较低价值的元素，而正是通过成对的优先级队列来存储具有其频率的价值。步骤如下：
- 将数组arr []的每个元素及其出现的位置插入优先级队列。
- 对于数组X []中的每个元素(例如X [i] )，执行以下操作：
- 从优先级队列中选择最多X [i]个最小元素。
- 如果选择元素小于Y [i]，则将其替换为Y [i]。
- 将替换的元素及其对应的频率插入到优先级队列中。
- 完成上述操作后，数组arr []将具有所有元素的总和最大的元素。打印总和。
下面是上述方法的实现：
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 800
时间复杂度： O(N * log 2 N)
辅助空间： O(N)