通过用任何正整数取模替换任何 K 个元素来最大化数组总和

给定一个正整数数组arr[]和一个数字K 。任务是通过用任何小于arr[i]的正整数取模来替换数组的任何K个元素来最大化数组的总和，即(arr[i] = arr[i]%X 其中 X ≤ arr[i])。

例子：

Input: arr[] = {5, 7, 18, 12, 11, 3}, K = 4
Output: 41
Explanation: The replacement should be {5%3, 7%4, 18, 12, 11%6, 3%2}

Input: arr[] = {8, 2, 28, 12, 7, 9}, K = 4
Output: 55
Explanation: The replacement should be {8%5, 2%2, 28, 12, 7%4, 9}

方法：对于每个元素arr[i] 在数组arr[]中，使用(arr[i]/2 +1)对其进行模块化，这将在操作后给出arr[i]的最高可能值。以下是解决问题的步骤

• 对数组arr[]进行排序。
• 使用变量i遍历范围[0, K)并执行以下任务：
  • 对于每个元素arr[i]，用(arr[i]/2 +1)对其进行模块化并更新结果。
• 求更新数组的总和并输出。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the maximum possible sum
int find(int arr[], int K, int N)
{
    // Sorting the array
    sort(arr, arr + N);
    int sum = 0;
 
    // Loop to take update K
    for (int i = 0; i < K; i++) {
 
        // Smallest number in array
        arr[i] %= (arr[i] / 2) + 1;
    }
 
    // Loop to find sum
    for (int i = 0; i < N; i++) {
        sum += arr[i];
    }
    return sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 7, 18, 12, 11, 3 };
    int K = 4;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << find(arr, K, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.Arrays;
class GFG {
 
  // Function to find the maximum possible sum
  static int find(int arr[], int K, int N) {
    // Sorting the array
    Arrays.sort(arr);
    int sum = 0;
 
    // Loop to take update K
    for (int i = 0; i < K; i++) {
 
      // Smallest number in array
      arr[i] %= (arr[i] / 2) + 1;
    }
 
    // Loop to find sum
    for (int i = 0; i < N; i++) {
      sum += arr[i];
    }
    return sum;
  }
 
  // Driver Code
  public static void main(String args[]) {
    int arr[] = { 5, 7, 18, 12, 11, 3 };
    int K = 4;
    int N = arr.length;
 
    System.out.println(find(arr, K, N));
  }
}
 
// This code is contributed by saurabh_jaiswal.

Python3

# python3 program for the above approach
 
# Function to find the maximum possible sum
def find(arr, K, N):
 
    # Sorting the array
    arr.sort()
    sum = 0
 
    # Loop to take update K
    for i in range(0, K):
 
        # Smallest number in array
        arr[i] %= (arr[i] // 2) + 1
 
    # Loop to find sum
    for i in range(0, N):
        sum += arr[i]
 
    return sum
 
# Driver Code
if __name__ == "__main__":
 
    arr = [5, 7, 18, 12, 11, 3]
    K = 4
    N = len(arr)
 
    print(find(arr, K, N))
 
# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to find the maximum possible sum
  static int find(int[] arr, int K, int N)
  {
    // Sorting the array
    Array.Sort(arr);
    int sum = 0;
 
    // Loop to take update K
    for (int i = 0; i < K; i++) {
 
      // Smallest number in array
      arr[i] %= (arr[i] / 2) + 1;
    }
 
    // Loop to find sum
    for (int i = 0; i < N; i++) {
      sum += arr[i];
    }
    return sum;
  }
 
  // Driver Code
  public static void Main()
  {
    int[] arr = { 5, 7, 18, 12, 11, 3 };
    int K = 4;
    int N = arr.Length;
 
    Console.Write(find(arr, K, N));
  }
}
 
// This code is contributed by ukasp.

Javascript

      // JavaScript code for the above approach
 
 
      // Function to find the maximum possible sum
      function find(arr, K, N) {
          // Sorting the array
          arr.sort(function (a, b) { return a - b })
          let sum = 0;
 
          // Loop to take update K
          for (let i = 0; i < K; i++) {
 
              // Smallest number in array
              arr[i] %= Math.floor(arr[i] / 2) + 1;
          }
 
          // Loop to find sum
          for (let i = 0; i < N; i++) {
              sum += arr[i];
          }
          return sum;
      }
 
      // Driver Code
 
      let arr = [5, 7, 18, 12, 11, 3];
      let K = 4;
      let N = arr.length;
 
      document.write(find(arr, K, N));
 
// This code is contributed by Potta Lokesh
  

输出

41

时间复杂度： O(N * logN) 其中 N 是数组的大小
辅助空间： O(1)