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 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++

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 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;
}

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 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

15


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 : AnkitRai01, Mithun Kumar