至少有k个距离元素的最大和子序列

给定一个数组和一个数字k，找到一个子序列，使得

1. 子序列中的元素总数最大
2. 子序列元素的索引至少相差k

   例子

   Input : arr[] = {4, 5, 8, 7, 5, 4, 3, 4, 6, 5}
              k = 2
   Output: 19
   Explanation: The highest value is obtained 
   if you pick indices 1, 4, 7, 10 giving 
   4 + 7 + 3 + 5 = 19
   
   Input: arr[] = {50, 70, 40, 50, 90, 70, 60, 
                                 40, 70, 50}
              k = 2
   Output: 230
   Explanation: There are 10 elements and k = 2. 
   If you select 2, 5, and 9 you get a total 
   value of 230, which is the maximum possible.
   

   一个简单的解决方案是一一考虑所有子序列。在每个子序列中，检查距离条件并返回最大总和子序列。

   一个有效的解决方案是使用动态编程。

   有两种情况：

   1. 如果我们选择索引i处的元素，使得i + k + 1> = N，那么我们就不能选择任何其他元素作为子序列的一部分。因此，我们需要决定是选择该元素还是选择其后的元素之一。
   2. 如果我们在索引i处选择元素，使i + k + 1

   这两种情况可以写成：

   Let MS[i] denotes the maximum sum of subsequence 
   from i = N-2 to 0. 
   
   Base Case: 
      MS[N-1] = arr[N-1]
   
   If  i + 1 + k >= N
      MS[i] = max(arr[i], MS[i+1]),  
   Else
      MS[i] = max(arr[i] + MS[i+k+1], MS[i+1])
   
   Evidently, the solution to the problem
   is to find MS[0].

   下面是实现：

   C++

   // CPP program to find maximum sum subsequence
   // such that elements are at least k distance
   // away.
   #include 
   using namespace std;
     
   int maxSum(int arr[], int N, int k)
   {
       // MS[i] is going to store maximum sum
       // subsequence in subarray from arr[i]
       // to arr[n-1]
       int MS[N];
     
       // We fill MS from right to left.
       MS[N - 1] = arr[N - 1];
       for (int i = N - 2; i >= 0; i--) {
           if (i + k + 1 >= N)
               MS[i] = max(arr[i], MS[i + 1]);
           else
               MS[i] = max(arr[i] + MS[i + k + 1], MS[i + 1]);
       }
     
       return MS[0];
   }
     
   // Driver code
   int main()
   {
       int N = 10, k = 2;
       int arr[] = { 50, 70, 40, 50, 90, 70, 60, 40, 70, 50 };
       cout << maxSum(arr, N, k);
       return 0;
   }

   Java

   // Java program to find maximum sum subsequence
   // such that elements are at least k distance
   // away.
   import java.io.*;
     
   class GFG {
     
       static int maxSum(int arr[], int N, int k)
       {
           // MS[i] is going to store maximum sum
           // subsequence in subarray from arr[i]
           // to arr[n-1]
           int MS[] = new int[N];
     
           // We fill MS from right to left.
           MS[N - 1] = arr[N - 1];
           for (int i = N - 2; i >= 0; i--) {
               if (i + k + 1 >= N)
                   MS[i] = Math.max(arr[i], MS[i + 1]);
               else
                   MS[i] = Math.max(arr[i] + MS[i + k + 1],
                                   MS[i + 1]);
           }
     
           return MS[0];
       }
     
       // Driver code
       public static void main(String[] args)
       {
           int N = 10, k = 2;
           int arr[] = { 50, 70, 40, 50, 90, 70, 60,
                         40, 70, 50 };
           System.out.println(maxSum(arr, N, k));
       }
   }
   // This code is contributed by Prerna Saini

   Python3

   # Python3 program to find maximum
   # sum subsequence such that elements
   # are at least k distance away.
     
   def maxSum(arr, N, k):
     
       # MS[i] is going to store maximum sum
       # subsequence in subarray from arr[i]
       # to arr[n-1]
       MS = [0 for i in range(N)]
     
       # We fill MS from right to left.
       MS[N - 1] = arr[N - 1]
       for i in range(N - 2, -1, -1):
           if (i + k + 1 >= N):
               MS[i] = max(arr[i], MS[i + 1])
           else:
               MS[i] = max(arr[i] + MS[i + k + 1],
                                    MS[i + 1])
         
       return MS[0]
     
   # Driver code
   N = 10; k = 2
   arr = [ 50, 70, 40, 50, 90, 70, 60, 40, 70, 50 ]
   print(maxSum(arr, N, k))
         
   # This code is contributed by Anant Agarwal.

   C#

   // C# program to find maximum sum
   // subsequence such that elements
   // are at least k distance away.
   using System;
      
   class GFG {
      
       static int maxSum(int []arr, int N, int k)
       {
           // MS[i] is going to store maximum sum
           // subsequence in subarray from arr[i]
           // to arr[n-1]
           int []MS = new int[N];
      
           // We fill MS from right to left.
           MS[N - 1] = arr[N - 1];
           for (int i = N - 2; i >= 0; i--) {
                 
               if (i + k + 1 >= N)
                   MS[i] = Math.Max(arr[i], MS[i + 1]);
               else
                   MS[i] = Math.Max(arr[i] + MS[i + k + 1],
                                   MS[i + 1]);
           }
      
           return MS[0];
       }
      
       // Driver code
       public static void Main()
       {
           int N = 10, k = 2;
           int []arr = { 50, 70, 40, 50, 90, 70, 60,
                                       40, 70, 50 };
           Console.WriteLine(maxSum(arr, N, k));
       }
   }
     
   // This code is contributed by Anant Agarwal.

   PHP

   = 0; $i--) 
       {
           if ($i + $k + 1 >= $N)
               $MS[$i] = max($arr[$i], 
                         $MS[$i + 1]);
           else
               $MS[$i] = max($arr[$i] + 
                         $MS[$i + $k + 1], 
                         $MS[$i + 1]);
       }
     
       return $MS[0];
   }
     
   // Driver code
   $N = 10; $k = 2;
   $arr = array(50, 70, 40, 50, 90, 
                70, 60, 40, 70, 50);
   echo(maxSum($arr, $N, $k));
     
   // This code is contributed by Ajit.
   ?>

   输出：

   230
   

   时间复杂度：O(n)
   辅助空间：O(n)