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 

Approach: While choosing 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 <bits/stdc++.h>
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[0];
    if (n == 2)
        return max(arr[0], arr[1]);

    // dp[i] represent the maximum sum so far
    // after reaching current position i
    int dp[n];

    // Initialize dp[0]
    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] = 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[0]);
    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[0]
    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[0]
    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[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[0]
    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
    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

Javascript

<script>

    // JavaScript implementation of the approach

    // Function to return the maximum sum possible
    function maxSum(arr, k, 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
        let dp = new Array(n);

        // Initialize dp[0]
        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 (let i = 1; i <= k; i++)
            dp[i] = Math.max(arr[i], dp[i - 1]);

        // Fill remaining positions
        for (let i = k + 1; i < n; i++)
            dp[i] = Math.max(arr[i], dp[i - (k + 1)] + arr[i]);

        // Return the maximum sum
        let max = Number.MIN_VALUE;
        for(let i = 0; i < dp.length; i++)
        {
            max = Math.max(max, dp[i]);
        }
        return max;
    }

    let arr = [ 6, 7, 1, 3, 8, 2, 4 ];
    let n = arr.length;
    let k = 2;
    document.write(maxSum(arr, k, n));

</script>

Output:

15

Time Complexity: O(N) 
Auxiliary Space: O(N)