Maximum sub-sequence sum such that indices of any two adjacent elements differs at least by 3

Given an array arr[] of integers, the task is to find the maximum sum of any sub-sequence in the array such that any two adjacent elements in the selected sequence have at least a difference of 3 in their indices in the given array. 
In other words, if you select arr[i] then the next element you can select is arr[i + 3], arr[i + 4], and so on… but you cannot select arr[i + 1] and arr[i + 2].
Examples:

Input: arr[] = {1, 2, -2, 4, 3} 
Output: 5 
{1, 4} and {2, 3} are the only sub-sequences 
with maximum sum.
Input: arr[] = {1, 2, 72, 4, 3, 9} 
Output: 81 

Naive approach: We generate all possible subsets of the array and check if the current subset satisfies the condition.If yes, then we compare its sum with the largest sum we have obtained till now.It is not an efficient approach as it takes exponential time .
Efficient approach: This problem can be solved using dynamic programming. Let’s decide the states of the dp. Let dp[i] be the largest possible sum for the sub-sequence starting from index 0 and ending at index i. Now, we have to find a recurrence relation between this state and a lower-order state. 
In this case for an index i, we will have two choices: 

1. Choose the current index: In this case, the relation will be dp[i] = arr[i] + dp[i – 3].
2. Skip the current index: Relation will be dp[i] = dp[i – 1].

We will choose the path that maximizes our result. Thus the final relation will be: 
dp[i] = max(dp[i – 3] + arr[i], dp[i – 1])
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
// of the sub-sequence such that two
// consecutive elements have a difference of
// at least 3 in their indices
// in the given array
int max_sum(int a[], int n)
{

    int dp[n];

    // If there is a single element in the array
    if (n == 1) {

        // Either select it or don't
        dp[0] = max(0, a[0]);
    }

    // If there are two elements
    else if (n == 2) {

        // Either select the first
        // element or don't
        dp[0] = max(0, a[0]);

        // Either select the first or the second element
        // or don't select any element
        dp[1] = max(a[1], dp[0]);
    }
    else if (n >= 3) {

        // Either select the first
        // element or don't
        dp[0] = max(0, a[0]);

        // Either select the first or the second element
        // or don't select any element
        dp[1] = max(a[1], max(0, a[0]));

        // Either select first, second, third or nothing
        dp[2] = max(a[2], max(a[1], max(0, a[0])));

        int i = 3;

        // For the rest of the elements
        while (i < n) {

            // Either select the best sum till
            // previous_index or select the current
            // element + best_sum till index-3
            dp[i] = max(dp[i - 1], a[i] + dp[i - 3]);
            i++;
        }
    }

    return dp[n - 1];
}

// Driver code
int main()
{
    int arr[] = { 1, 2, -2, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << max_sum(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the maximum sum
// of the sub-sequence such that two
// consecutive elements have a difference of
// at least 3 in their indices
// in the given array
static int max_sum(int a[], int n)
{

    int []dp = new int[n];

    // If there is a single element in the array
    if (n == 1)
    {

        // Either select it or don't
        dp[0] = Math.max(0, a[0]);
    }

    // If there are two elements
    else if (n == 2)
    {

        // Either select the first
        // element or don't
        dp[0] = Math.max(0, a[0]);

        // Either select the first or the second element
        // or don't select any element
        dp[1] = Math.max(a[1], dp[0]);
    }
    else if (n >= 3)
    {

        // Either select the first
        // element or don't
        dp[0] = Math.max(0, a[0]);

        // Either select the first or the second element
        // or don't select any element
        dp[1] = Math.max(a[1], Math.max(0, a[0]));

        // Either select first, second, third or nothing
        dp[2] = Math.max(a[2], Math.max(a[1], Math.max(0, a[0])));

        int i = 3;

        // For the rest of the elements
        while (i < n)
        {

            // Either select the best sum till
            // previous_index or select the current
            // element + best_sum till index-3
            dp[i] = Math.max(dp[i - 1], a[i] + dp[i - 3]);
            i++;
        }
    }

    return dp[n - 1];
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, -2, 4, 3 };
    int n = arr.length;

    System.out.println(max_sum(arr, n));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach

# Function to return the maximum sum
# of the sub-sequence such that two
# consecutive elements have a difference of
# at least 3 in their indices
# in the given array
def max_sum(a, n) :

    dp = [0]*n;

    # If there is a single element in the array
    if (n == 1) :

        # Either select it or don't
        dp[0] = max(0, a[0]);

    # If there are two elements
    elif (n == 2) :

        # Either select the first
        # element or don't
        dp[0] = max(0, a[0]);

        # Either select the first or the second element
        # or don't select any element
        dp[1] = max(a[1], dp[0]);

    elif (n >= 3) :

        # Either select the first
        # element or don't
        dp[0] = max(0, a[0]);

        # Either select the first or the second element
        # or don't select any element
        dp[1] = max(a[1], max(0, a[0]));

        # Either select first, second, third or nothing
        dp[2] = max(a[2], max(a[1], max(0, a[0])));

        i = 3;

        # For the rest of the elements
        while (i < n) :

            # Either select the best sum till
            # previous_index or select the current
            # element + best_sum till index-3
            dp[i] = max(dp[i - 1], a[i] + dp[i - 3]);
            i += 1;

    return dp[n - 1];


# Driver code
if __name__ == "__main__" :

    arr = [ 1, 2, -2, 4, 3 ];
    n = len(arr);

    print(max_sum(arr, n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the maximum sum
// of the sub-sequence such that two
// consecutive elements have a difference of
// at least 3 in their indices
// in the given array
static int max_sum(int []a, int n)
{

    int []dp = new int[n];

    // If there is a single element in the array
    if (n == 1)
    {

        // Either select it or don't
        dp[0] = Math.Max(0, a[0]);
    }

    // If there are two elements
    else if (n == 2)
    {

        // Either select the first
        // element or don't
        dp[0] = Math.Max(0, a[0]);

        // Either select the first or the second element
        // or don't select any element
        dp[1] = Math.Max(a[1], dp[0]);
    }
    else if (n >= 3)
    {

        // Either select the first
        // element or don't
        dp[0] = Math.Max(0, a[0]);

        // Either select the first or the second element
        // or don't select any element
        dp[1] = Math.Max(a[1], Math.Max(0, a[0]));

        // Either select first, second, third or nothing
        dp[2] = Math.Max(a[2], Math.Max(a[1], Math.Max(0, a[0])));

        int i = 3;

        // For the rest of the elements
        while (i < n)
        {

            // Either select the best sum till
            // previous_index or select the current
            // element + best_sum till index-3
            dp[i] = Math.Max(dp[i - 1], a[i] + dp[i - 3]);
            i++;
        }
    }

    return dp[n - 1];
}

// Driver code
static public void Main ()
{

    int []arr = { 1, 2, -2, 4, 3 };
    int n = arr.Length;

    Console.Write(max_sum(arr, n));
}
}

// This code is contributed by ajit..

Javascript

<script>
    // Javascript implementation of the approach

    // Function to return the maximum sum
    // of the sub-sequence such that two
    // consecutive elements have a difference of
    // at least 3 in their indices
    // in the given array
    function max_sum(a, n)
    {

        let dp = new Array(n);

        // If there is a single element in the array
        if (n == 1)
        {

            // Either select it or don't
            dp[0] = Math.max(0, a[0]);
        }

        // If there are two elements
        else if (n == 2)
        {

            // Either select the first
            // element or don't
            dp[0] = Math.max(0, a[0]);

            // Either select the first or the second element
            // or don't select any element
            dp[1] = Math.max(a[1], dp[0]);
        }
        else if (n >= 3)
        {

            // Either select the first
            // element or don't
            dp[0] = Math.max(0, a[0]);

            // Either select the first or the second element
            // or don't select any element
            dp[1] = Math.max(a[1], Math.max(0, a[0]));

            // Either select first, second, third or nothing
            dp[2] = Math.max(a[2], Math.max(a[1], Math.max(0, a[0])));

            let i = 3;

            // For the rest of the elements
            while (i < n)
            {

                // Either select the best sum till
                // previous_index or select the current
                // element + best_sum till index-3
                dp[i] = Math.max(dp[i - 1], a[i] + dp[i - 3]);
                i++;
            }
        }

        return dp[n - 1];
    }

    let arr = [ 1, 2, -2, 4, 3 ];
    let n = arr.length;

    document.write(max_sum(arr, n));

</script>

Output:

5

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