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 staring 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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..

chevron_right


Output:

5

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



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Rajput-Ji, jit_t, AnkitRai01