# Maximum sum possible for a sub-sequence such that no two elements appear at a distance < K in the array

Given an array arr[] of n integers and an integer k, the task is to find the maximum sum possible for a sub-sequence such that no two elements of the sub-sequence appear at a distance ≤ k in the original array.

Examples:

Input: arr[] = {5, 3, 4, 11, 2}, k=1
Output: 16
All possible sub-sequences are {5, 4, 2}, {5, 11}, {5, 2}, {3, 11}, {3, 2}, {4, 2} and {11}
Out of which 5 + 11 = 16 gives the maximum sum.

Input: arr[] = {6, 7, 1, 3, 8, 2, 4}, k = 2
Output: 15

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: While chosing an element at index i, we have two options, either we include the current element in the sub-sequence or we don’t. Let dp[i] represents the maximum sum so far on reaching element at index i. We can calculate the value of dp[i] as follows:

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

dp[i – (k + 1)] + arr[i] is the case when element at index i is included. In that situation, maximum value will be arr[i] + maximum value till the last included element from the array.

dp[i – 1] is the case when current element is not included and the maximum value till now will be the maximum value till the previous element.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum sum possible ` `int` `maxSum(``int``* arr, ``int` `k, ``int` `n) ` `{ ` `    ``if` `(n == 0) ` `        ``return` `0; ` `    ``if` `(n == 1) ` `        ``return` `arr; ` `    ``if` `(n == 2) ` `        ``return` `max(arr, arr); ` ` `  `    ``// dp[i] represent the maximum sum so far ` `    ``// after reaching current position i ` `    ``int` `dp[n]; ` ` `  `    ``// Initialize dp ` `    ``dp = arr; ` ` `  `    ``// Initialize the dp values till k since any ` `    ``// two elements included in the sub-sequence ` `    ``// must be atleast k indices apart, and thus ` `    ``// first element and second element ` `    ``// will be k indices apart ` `    ``for` `(``int` `i = 1; i <= k; i++) ` `        ``dp[i] = max(arr[i], dp[i - 1]); ` ` `  `    ``// Fill remaining positions ` `    ``for` `(``int` `i = k + 1; i < n; i++) ` `        ``dp[i] = max(arr[i], dp[i - (k + 1)] + arr[i]); ` ` `  `    ``// Return the maximum sum ` `    ``int` `max = *(std::max_element(dp, dp + n)); ` `    ``return` `max; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 7, 1, 3, 8, 2, 4 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 2; ` `    ``cout << maxSum(arr, k, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `     `  `// Function to return the maximum sum possible ` `static` `int` `maxSum(``int` `[]arr, ``int` `k, ``int` `n) ` `{ ` `    ``if` `(n == ``0``) ` `        ``return` `0``; ` `    ``if` `(n == ``1``) ` `        ``return` `arr[``0``]; ` `    ``if` `(n == ``2``) ` `        ``return` `Math.max(arr[``0``], arr[``1``]); ` ` `  `    ``// dp[i] represent the maximum sum so far ` `    ``// after reaching current position i ` `    ``int``[] dp = ``new` `int``[n]; ` ` `  `    ``// Initialize dp ` `    ``dp[``0``] = arr[``0``]; ` ` `  `    ``// Initialize the dp values till k since any ` `    ``// two elements included in the sub-sequence ` `    ``// must be atleast k indices apart, and thus ` `    ``// first element and second element ` `    ``// will be k indices apart ` `    ``for` `(``int` `i = ``1``; i <= k; i++) ` `        ``dp[i] = Math.max(arr[i], dp[i - ``1``]); ` ` `  `    ``// Fill remaining positions ` `    ``for` `(``int` `i = k + ``1``; i < n; i++) ` `        ``dp[i] = Math.max(arr[i], dp[i - (k + ``1``)] + arr[i]); ` ` `  `    ``// Return the maximum sum ` `    ``return` `maximum(dp); ` `} ` ` `  `static` `int` `maximum(``int``[] arr) ` `{ ` `    ``int` `max = Integer.MIN_VALUE; ` `    ``for``(``int` `i = ``0``; i < arr.length; i++)  ` `    ``{ ` `        ``if``(arr[i] > max)  ` `        ``{ ` `            ``max = arr[i]; ` `        ``} ` `    ``} ` `    ``return` `max; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `[]arr = { ``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``2``; ` `    ``System.out.println(maxSum(arr, k, n)); ` `} ` `} ` ` `  `// This code is contributed by mits `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the  ` `# maximum sum possible  ` `def` `maxSum(arr, k, n) :  ` `     `  `    ``if` `(n ``=``=` `0``) : ` `        ``return` `0``;  ` `    ``if` `(n ``=``=` `1``) : ` `        ``return` `arr[``0``];  ` `    ``if` `(n ``=``=` `2``) : ` `        ``return` `max``(arr[``0``], arr[``1``]);  ` ` `  `    ``# dp[i] represent the maximum sum so far  ` `    ``# after reaching current position i  ` `    ``dp ``=` `[``0``] ``*` `n ;  ` ` `  `    ``# Initialize dp  ` `    ``dp[``0``] ``=` `arr[``0``];  ` ` `  `    ``# Initialize the dp values till k since any  ` `    ``# two elements included in the sub-sequence  ` `    ``# must be atleast k indices apart, and thus  ` `    ``# first element and second element  ` `    ``# will be k indices apart  ` `    ``for` `i ``in` `range``(``1``, k ``+` `1``) :  ` `        ``dp[i] ``=` `max``(arr[i], dp[i ``-` `1``]);  ` ` `  `    ``# Fill remaining positions  ` `    ``for` `i ``in` `range``(k ``+` `1``, n) :  ` `        ``dp[i] ``=` `max``(arr[i],  ` `                    ``dp[i ``-` `(k ``+` `1``)] ``+` `arr[i]);  ` ` `  `    ``# Return the maximum sum  ` `    ``max_element ``=` `max``(dp);  ` `    ``return` `max_element;  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` `    ``arr ``=` `[ ``6``, ``7``, ``1``, ``3``, ``8``, ``2``, ``4` `];  ` `    ``n ``=` `len``(arr);  ` `    ``k ``=` `2``;  ` `     `  `    ``print``(maxSum(arr, k, n));  ` `     `  `# This code is contributed by Ryuga `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Linq; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the maximum sum possible ` `static` `int` `maxSum(``int` `[]arr, ``int` `k, ``int` `n) ` `{ ` `    ``if` `(n == 0) ` `        ``return` `0; ` `    ``if` `(n == 1) ` `        ``return` `arr; ` `    ``if` `(n == 2) ` `        ``return` `Math.Max(arr, arr); ` ` `  `    ``// dp[i] represent the maximum sum so far ` `    ``// after reaching current position i ` `    ``int``[] dp = ``new` `int``[n]; ` ` `  `    ``// Initialize dp ` `    ``dp = arr; ` ` `  `    ``// Initialize the dp values till k since any ` `    ``// two elements included in the sub-sequence ` `    ``// must be atleast k indices apart, and thus ` `    ``// first element and second element ` `    ``// will be k indices apart ` `    ``for` `(``int` `i = 1; i <= k; i++) ` `        ``dp[i] = Math.Max(arr[i], dp[i - 1]); ` ` `  `    ``// Fill remaining positions ` `    ``for` `(``int` `i = k + 1; i < n; i++) ` `        ``dp[i] = Math.Max(arr[i], dp[i - (k + 1)] + arr[i]); ` ` `  `    ``// Return the maximum sum ` `    ``int` `max = dp.Max(); ` `    ``return` `max; ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 6, 7, 1, 3, 8, 2, 4 }; ` `    ``int` `n = arr.Length; ` `    ``int` `k = 2; ` `    ``Console.WriteLine(maxSum(arr, k, n)); ` `} ` `} ` ` `  `// This code is contributed by mits `

Output:

```15
```

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