相邻元素的最大子序列总和在索引上至少有 K 个差异

给定一个由长度为N的整数和一个整数K (1 ≤ k ≤ N) 组成的数组arr[] ，任务是找到数组中的最大子序列和，使得该子序列中的相邻元素至少相差K在原始数组中的索引中。
例子：

Input: arr[] = {1, 2, -2, 4, 3, 1}, K = 4 
Output: 4 
Explanation: 
Such Subsequences that can be selected: {1, 3}, {1, 1}, {2, 1} 
Subsequence with maximum sum = (1 + 3) = 4 
Selected elements are a[0] and a[4] (difference between indices = 4)
Input: arr[] = {1, 2, 72, 4, 3}, K = 2 
Output: 76 
Explanation: 
Such Subsequences that can be selected – {{1, 72, 3}, {2, 4}, {2, 3}, {1, 4}, {1, 3}} 
Subsequence with maximum sum = (1 + 72 + 3) = 76 
Selected elements are a[0], a[2] and a[4] (difference between each indices = 2) 
 

朴素的方法：生成数组的所有可能子集，并检查每个子集是否满足条件，使得两个相邻元素的索引至少相差 K。如果是，则将其总和与迄今为止获得的最大总和进行比较，如果它大于迄今为止获得的总和，则更新总和。
高效的方法：这个问题可以使用动态规划来解决。其中对于数组中的每个元素，考虑这样如果我们采用这个元素，那么它是否有助于获得的最终总和 如果是，则将其添加到该元素的 DP 数组中。
让我们决定“dp”的状态。令 dp[i] 是从索引 ‘0’ 开始到索引 ‘i’ 结束的子数组的最大可能总和。现在，我们必须找到这个状态和低阶状态之间的递推关系。
现在，对于每个索引 i，数组的循环关系将有两个选择。

• 选择当前索引：
  在这种情况下，可以选择的元素位于索引 ik。
  所以，

dp[i] = arr[i] + dp[i - k]

• 跳过当前索引。
  在这种情况下，可以选择的元素位于索引 i-1。
  所以，

dp[i] = dp[i - 1]

对于每个索引，选择在该索引处给出最大总和的条件，所以最终的递推关系将是：

dp[i] = max(dp[i – 1], arr[i] + dp[i – k]) 
 

下面是上述方法的实现。

C++

// C++ implementation to find the
// maximum sum subsequence such that
// two adjacent element have atleast
// difference of K in their indices
 
#include 
using namespace std;
 
// Function to find the maximum
// between two elements a and b
int maxi(int a, int b)
{
    if (a > b) {
        return a;
    }
    else {
        return b;
    }
}
 
// Function to find the maximum sum
// subsequence such that two adjacent
// element have atleast difference
// of K in their indices
int max_sum(int arr[], int n, int k)
{
    // DP Array to store the maximum
    // sum obtained till now
    int dp[n];
 
    // Either select the first element
    // or Nothing
    dp[0] = maxi(0, arr[0]);
    int i = 1;
 
    // Either Select the (i - 1) element
    // or let the previous best answer be
    // the current best answer
    while (i < k) {
        dp[i] = maxi(dp[i - 1], arr[i]);
        i++;
    }
    i = k;
 
    // Either select the best sum
    // till previous_index or select the
    // current element + best_sum till index-k
    while (i < n) {
        dp[i] = maxi(dp[i - 1], arr[i] + dp[i - k]);
        i++;
    }
    return dp[n - 1];
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, -2, 4, 3, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;
    cout << max_sum(arr, n, k);
    return 0;
}

Java

// Java implementation to find the
// maximum sum subsequence such that
// two adjacent element have atleast
// difference of K in their indices
class GFG
{
 
    // Function to find the maximum sum
    // subsequence such that two adjacent
    // element have atleast difference
    // of K in their indices
    static int max_sum(int arr[], int n, int k)
    {
        // DP Array to store the maximum
        // sum obtained till now
        int dp[] = new int[n];
     
        // Either select the first element
        // or Nothing
        dp[0] = Math.max(0, arr[0]);
        int i = 1;
     
        // Either Select the (i - 1) element
        // or let the previous best answer be
        // the current best answer
        while (i < k)
        {
            dp[i] = Math.max(dp[i - 1], arr[i]);
            i++;
        }
        i = k;
     
        // Either select the best sum
        // till previous_index or select the
        // current element + best_sum till index-k
        while (i < n)
        {
            dp[i] = Math.max(dp[i - 1],
                    arr[i] + dp[i - k]);
            i++;
        }
        return dp[n - 1];
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int arr[] = { 1, 2, -2, 4, 3, 1 };
        int n = arr.length;
        int k = 4;
        System.out.println(max_sum(arr, n, k));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 implementation to find the
# maximum sum subsequence such that
# two adjacent element have atleast
# difference of K in their indices
 
# Function to find the maximum sum
# subsequence such that two adjacent
# element have atleast difference
# of K in their indices
def max_sum(arr, n, k) :
 
    # DP Array to store the maximum
    # sum obtained till now
    dp = [0] * n;
 
    # Either select the first element
    # or Nothing
    dp[0] = max(0, arr[0]);
    i = 1;
 
    # Either Select the (i - 1) element
    # or let the previous best answer be
    # the current best answer
    while (i < k) :
        dp[i] = max(dp[i - 1], arr[i]);
        i += 1;
 
    i = k;
 
    # Either select the best sum
    # till previous_index or select the
    # current element + best_sum till index-k
    while (i < n) :
        dp[i] = max(dp[i - 1], arr[i] + dp[i - k]);
        i += 1;
     
    return dp[n - 1];
 
# Driver Code
if __name__ == "__main__" :
 
    arr = [ 1, 2, -2, 4, 3, 1 ];
    n = len(arr)
    k = 4;
    print(max_sum(arr, n, k));
 
# This code is contributed by AnkitRai01

C#

// C# implementation to find the
// maximum sum subsequence such that
// two adjacent element have atleast
// difference of K in their indices
using System;
 
class GFG
{
 
    // Function to find the maximum sum
    // subsequence such that two adjacent
    // element have atleast difference
    // of K in their indices
    static int max_sum(int []arr, int n, int k)
    {
        // DP Array to store the maximum
        // sum obtained till now
        int []dp = new int[n];
     
        // Either select the first element
        // or Nothing
        dp[0] = Math.Max(0, arr[0]);
        int i = 1;
     
        // Either Select the (i - 1) element
        // or let the previous best answer be
        // the current best answer
        while (i < k)
        {
            dp[i] = Math.Max(dp[i - 1], arr[i]);
            i++;
        }
        i = k;
     
        // Either select the best sum
        // till previous_index or select the
        // current element + best_sum till index-k
        while (i < n)
        {
            dp[i] = Math.Max(dp[i - 1],
                    arr[i] + dp[i - k]);
            i++;
        }
        return dp[n - 1];
    }
     
    // Driver Code
    public static void Main()
    {
        int []arr = { 1, 2, -2, 4, 3, 1 };
        int n = arr.Length;
        int k = 4;
        Console.WriteLine(max_sum(arr, n, k));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

 
// javascript implementation to find the
// maximum sum subsequence such that
// two adjacent element have atleast
// difference of K in their indices
 
// Function to find the maximum sum
// subsequence such that two adjacent
// element have atleast difference
// of K in their indices
function max_sum(arr , n , k)
{
    // DP Array to store the maximum
    // sum obtained till now
    var dp = Array.from({length: n}, (_, i) => 0);
 
    // Either select the first element
    // or Nothing
    dp[0] = Math.max(0, arr[0]);
    var i = 1;
 
    // Either Select the (i - 1) element
    // or let the previous best answer be
    // the current best answer
    while (i < k)
    {
        dp[i] = Math.max(dp[i - 1], arr[i]);
        i++;
    }
    i = k;
 
    // Either select the best sum
    // till previous_index or select the
    // current element + best_sum till index-k
    while (i < n)
    {
        dp[i] = Math.max(dp[i - 1],
                arr[i] + dp[i - k]);
        i++;
    }
    return dp[n - 1];
}
 
// Driver Code
var arr = [ 1, 2, -2, 4, 3, 1 ];
var n = arr.length;
var k = 4;
document.write(max_sum(arr, n, k));
 
// This code is contributed by 29AjayKumar

输出：

4

时间复杂度： O(N)
辅助空间： O(N)

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