通过选择 K 个数组元素然后将它们递减 1 来最大化和

给定一个由N 个正整数和一个整数K组成的数组arr[] 。在一个操作中，选择一个数组元素，将其添加到总和中，然后将其减1 。任务是打印执行K次操作可以获得的最大和。

例子：

Input: arr[] = {2, 5}, K = 4
Output: 14
Explanation:
Perform the following operations to maximize the sum:
Operation 1: Select 5, then reduce it so new array become {2, 4}.
Operation 2: Select 4, then reduce it so new array become {2, 3}.
Operation 3: Select 3, then reduce it so new array become {2, 2}.
Operation 4: Select 2, then reduce it so new array become {2, 1}.
Therefore, the maximum sum is 5 + 4 + 3 + 2 = 14.    

Input: arr[] = {2, 8, 4, 10, 6}, K = 2
Output: 19
Explanation:
Perform the following operations to maximize the sum:
Operation 1: Select 10, then reduce it so new array become {2, 8, 4, 9, 6}.
Operation 2: Select 9, then reduce it so new array become {2, 8, 4, 8, 6}.
Therefore, the maximum sum is 10 + 9 = 19.

天真的方法：最简单的方法是每次使用Max Heap 来选择最大元素。将最大元素添加到总和并将其从最大堆中删除并添加(maximum element – 1) 。执行此操作K并打印总和。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find maximum possible
// after adding elements K times and
// decrementing each added value by 1
long maxSum(vector arr, int k)
{
     
    // Stores the maximum sum
    long max_sum = 0;
 
    // Create max_heap to get
    // the maximum element
    priority_queue max_heap;
 
    // Update the max_heap
    for(int t : arr)
        max_heap.push(t);
 
    // Calculate the max_sum
    while (k-- > 0)
    {
        int tmp = max_heap.top();
        max_heap.pop();
        max_sum += tmp;
        max_heap.push(tmp - 1);
    }
 
    // Return the maximum sum
    return max_sum;
}
 
// Driver code
int main()
{
     
    // Given an array arr[]
    vector arr = { 2, 5 };
     
    // Given K
    int K = 4;
     
    // Function Call
    cout << maxSum(arr, K);
}
 
// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find maximum possible
    // after adding elements K times and
    // decrementing each added value by 1
    public static long maxSum(int[] arr,
                              int k)
    {
        // Stores the maximum sum
        long max_sum = 0;
 
        // Create max_heap to get
        // the maximum element
        PriorityQueue max_heap
            = new PriorityQueue<>(
                Collections.reverseOrder());
 
        // Update the max_heap
        for (int t : arr)
            max_heap.add(t);
 
        // Calculate the max_sum
        while (k-- > 0) {
            int tmp = max_heap.poll();
            max_sum += tmp;
            max_heap.add(tmp - 1);
        }
 
        // Return the maximum sum
        return max_sum;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given an array arr[]
        int[] arr = { 2, 5 };
 
        // Given K
        int K = 4;
 
        // Function Call
        System.out.println(maxSum(arr, K));
    }
}

Python3

# Python3 program for the above approach
 
# Function to find maximum possible
# after adding elements K times and
# decrementing each added value by 1
from heapq import heappop, heappush, heapify
 
def maxSum(arr, k):
     
    # Stores the maximum sum
    max_sum = 0
 
    # Create max_heap to get
    # the maximum element
    max_heap = []
    heapify(max_heap)
 
    # Update the max_heap
    for i in range(len(arr) - 1, 0, -1):
        heappush(max_heap, arr[i])
         
    # Calculate the max_sum
    while (k > 0):
        tmp = heappop(max_heap)
        max_sum += tmp
        #max_heap.put(tmp - 1);
        heappush(max_heap, tmp - 1)
        k -= 1
 
    # Return the maximum sum
    return max_sum
 
# Driver Code
if __name__ == '__main__':
     
    # Given an array arr
    arr = [ 2, 5 ]
 
    # Given K
    K = 4
 
    # Function Call
    print(maxSum(arr, K))
 
# This code is contributed by gauravrajput1

C#

// C# program for the
// above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find maximum possible
// after adding elements K times and
// decrementing each added value by 1
public static long maxSum(int[] arr, int k)
{
     
    // Stores the maximum sum
    long max_sum = 0;
 
    // Create max_heap to get
    // the maximum element
    List max_heap = new List();
 
    // Update the max_heap
    foreach(int t in arr) max_heap.Add(t);
    max_heap.Sort();
    max_heap.Reverse();
     
    // Calculate the max_sum
    while (k-- > 1)
    {
        int tmp = max_heap[0];
        max_heap.Remove(0);
        max_sum += tmp;
        max_heap.Add(tmp - 1);
        max_heap.Sort();
        max_heap.Reverse();
    }
     
    // Return the maximum sum
    return max_sum - 1;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given an array []arr
    int[] arr = { 2, 5 };
     
    // Given K
    int K = 4;
 
    // Function Call
    Console.WriteLine(maxSum(arr, K));
}
}
 
// This code is contributed by gauravrajput1

Javascript

 
// Javascript program for the above approach
 
// Function to find maximum possible
// after adding elements K times and
// decrementing each added value by 1
function maxSum(arr, k)
{
     
    // Stores the maximum sum
    let max_sum = 0;
 
    // Create max_heap to get
    // the maximum element
    let max_heap = [];
 
    // Update the max_heap
    for(let t = 0; t < arr.length; t++)
        max_heap.push(arr[t]);
      
    max_heap.sort(function(a, b){return b - a;});
     
    // Calculate the max_sum
    while (k-- > 0)
    {
        let tmp = max_heap.shift();
        max_sum += tmp;
        max_heap.push(tmp - 1);
        max_heap.sort(function(a, b){return b - a;});
    }
 
    // Return the maximum sum
    return max_sum;
}
 
// Driver Code
 
// Given an array arr[]
let arr = [ 2, 5 ];
 
// Given K
let K = 4;
 
// Function Call
document.write(maxSum(arr, K));
 
// This code is contributed by avanitrachhadiya2155
 

输出：

14

时间复杂度： O(K*log(N))，其中 N 是给定数组的长度，K 是给定的操作次数。
辅助空间： O(N)

高效的方法：这个想法是通过存储给定数组的每个元素的频率来使用哈希概念。请按照以下步骤解决问题：

• 创建大小为M + 1 的freq[] ，其中M是给定数组中存在的最大元素和一个变量max_sum，分别用于存储arr[]和最大可能总和的每个元素的频率。
• 从右到左遍历数组freq[] ，即从i = M到0 。
• 通过频率增量max_sum [I] * 1，会降低K由频率[i]和增量频率[I – 1]由频率[I]，如果K≥频率[i]中。
• 否则将max_sum增加i*K并中断循环，因为K变为0 。
• 返回max_sum作为最大可能总和。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find maximum possible
// after adding elements K times and
// decrementing each added value by 1
long maxSum(int arr[], int k, int n)
{
     
    // Stores the maximum sum
    long max_sum = 0;
 
    // Stores freqency of element
    int freq[100000 + 1] = {0};
 
    // Update freqency of array element
    for(int i = 0; i < n; i++)
        freq[arr[i]]++;
 
    // Traverse from right to left in
    // freq[] to find maximum sum
    for(int i = 100000; i > 0; i--)
    {
        if (k >= freq[i])
        {
             
            // Update max_sum
            max_sum += i * freq[i];
 
            // Decrement k
            k -= freq[i];
            freq[i - 1] += freq[i];
        }
 
        // Update max_sum and break
        else
        {
            max_sum += i * k;
            break;
        }
    }
 
    // Return the maximum sum
    return max_sum;
}
 
// Driver Code
int main()
{
     
    // Given array arr[]
    int arr[] = { 2, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Given K
    int K = 4;
 
    // Function Call
    cout << (maxSum(arr, K, n));
    return 0;
}
 
// This code is contributed by chitranayal

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find maximum possible
    // after adding elements K times and
    // decrementing each added value by 1
    public static long maxSum(int[] arr,
                              int k)
    {
        // Stores the maximum sum
        long max_sum = 0;
 
        // Stores freqency of element
        int[] freq = new int[100000 + 1];
 
        // Update freqency of array element
        for (int t : arr)
            freq[t]++;
 
        // Traverse from right to left in
        // freq[] to find maximum sum
        for (int i = 100000; i > 0; i--) {
 
            if (k >= freq[i]) {
 
                // Update max_sum
                max_sum += i * freq[i];
 
                // Decrement k
                k -= freq[i];
                freq[i - 1] += freq[i];
            }
 
            // Update max_sum and break
            else {
                max_sum += i * k;
                break;
            }
        }
 
        // Return the maximum sum
        return max_sum;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array arr[]
        int[] arr = { 2, 5 };
 
        // Given K
        int K = 4;
 
        // Function Call
        System.out.println(maxSum(arr, K));
    }
}

蟒蛇3

# Python program for the above approach
 
# Function to find maximum possible
# after adding elements K times and
# decrementing each added value by 1
def maxSum(arr, k):
   
    # Stores the maximum sum
    max_sum = 0;
 
    # Stores freqency of element
    freq = [0]*(100000 + 1);
 
    # Update freqency of array element
    for i in range(len(arr)):
        freq[arr[i]] += 1;
 
    # Traverse from right to left in
    # freq to find maximum sum
    for i in range(100000, 1, -1):
 
        if (k >= freq[i]):
 
            # Update max_sum
            max_sum += i * freq[i];
 
            # Decrement k
            k -= freq[i];
            freq[i - 1] += freq[i];
 
        # Update max_sum and break
        else:
            max_sum += i * k;
            break;
 
    # Return the maximum sum
    return max_sum;
 
# Driver Code
if __name__ == '__main__':
  
    # Given array arr
    arr = [2, 5];
 
    # Given K
    K = 4;
 
    # Function Call
    print(maxSum(arr, K));
 
    # This code contributed by gauravrajput1

C＃

// C# program for the
// above approach
using System;
class GFG{
 
// Function to find maximum possible
// after adding elements K times and
// decrementing each added value by 1
public static long maxSum(int[] arr,
                          int k)
{
  // Stores the maximum
  // sum
  long max_sum = 0;
 
  // Stores freqency of
  // element
  int[] freq =
        new int[100000 + 1];
 
  // Update freqency of
  // array element
  foreach (int t in arr)
  {
    freq[t]++;
  }
 
  // Traverse from right to
  // left in []freq to find
  // maximum sum
  for (int i = 100000; i > 0; i--)
  {
    if (k >= freq[i])
    {
      // Update max_sum
      max_sum += i * freq[i];
 
      // Decrement k
      k -= freq[i];
      freq[i - 1] += freq[i];
    }
 
    // Update max_sum and
    // break
    else
    {
      max_sum += i * k;
      break;
    }
  }
 
  // Return the maximum
  // sum
  return max_sum;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given array []arr
  int[] arr = {2, 5};
 
  // Given K
  int K = 4;
 
  // Function Call
  Console.WriteLine(
  maxSum(arr, K));
}
}
 
// This code is contributed by gauravrajput1

Javascript

//Javascript program for the above approach
 
 
// Function to find maximum possible
// after adding elements K times and
// decrementing each added value by 1
function maxSum(arr, k, n)
{
     
    // Stores the maximum sum
    var max_sum = 0;
 
    // Stores freqency of element
    var freq = new Array(100000 + 1);
    freq.fill(0);
 
    // Update freqency of array element
    for(var i = 0; i < n; i++)
        freq[arr[i]]++;
 
    // Traverse from right to left in
    // freq[] to find maximum sum
    for(var i = 100000; i > 0; i--)
    {
        if (k >= freq[i])
        {
             
            // Update max_sum
            max_sum += i * freq[i];
 
            // Decrement k
            k -= freq[i];
            freq[i - 1] += freq[i];
        }
 
        // Update max_sum and break
        else
        {
            max_sum += i * k;
            break;
        }
    }
 
    // Return the maximum sum
    return max_sum;
}
 
 
var arr = [ 2, 5 ];
var n = arr.length;
// Given K
var K = 4;
// Function Call
document.write (maxSum(arr, K, n));
 
 
// This code is contributed by SoumikMondal

输出：

14

时间复杂度： O(N)，其中 N 是给定数组的长度。
辅助空间： O(N)

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