Maximise array sum after taking non-overlapping sub-arrays of length K

Given an integer array arr[] of length N and an integer K, the task is to select some non-overlapping sub-arrays such that each sub-array is exactly of length K, no two sub-arrays are adjacent and sum of all the elements of the selected sub-arrays is maximum.
Examples:

Input: arr[] = {1, 2, 3, 4, 5}, K = 2 
Output: 12 
Sub-arrays that maximizes sum will be {{1, 2}, {4, 5}}. 
Thus, the answer will be 12.
Input: arr[] = {1, 1, 1, 1, 1}, K = 1 
Output: 3 

Approach: This problem can be solved using dynamic programming. Let’s suppose we are at an index i. Let dp[i] be defined as the maximum sum of elements of all possible subsets of sub-array arr[i…n-1] satisfying above conditions. 
We will have two possible choices i.e. select the sub-array arr[i…i+k-1] and solve for dp[i + k + 1] or reject it and solve for dp[i + 1].
Thus, recurrence relation will be 

dp[i] = max(dp[i + 1], arr[i] + arr[i + 1] + arr[i + 2] + … + arr[i + k – 1] + dp[i + k + 1]) 

Since, the values of K can be large, we will use prefix-sum array to find the sum of all the elements of the sub-array arr[i…i + k – 1] in O(1). 
Overall, time complexity of the algorithm will be O(N).
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
#define maxLen 10
using namespace std;

// To store the states of dp
int dp[maxLen];

// To check if a given state
// has been solved
bool v[maxLen];

// To store the prefix-sum
int prefix_sum[maxLen];

// Function to fill the prefix_sum[] with
// the prefix sum of the given array
void findPrefixSum(int arr[], int n)
{
    prefix_sum[0] = arr[0];
    for (int i = 1; i < n; i++)
        prefix_sum[i] = arr[i] + prefix_sum[i - 1];
}

// Function to find the maximum sum subsequence
// such that no two elements are adjacent
int maxSum(int arr[], int i, int n, int k)
{
    // Base case
    if (i + k > n)
        return 0;

    // To check if a state has
    // been solved
    if (v[i])
        return dp[i];
    v[i] = 1;

    int x;

    if (i == 0)
        x = prefix_sum[k - 1];
    else
        x = prefix_sum[i + k - 1] - prefix_sum[i - 1];

    // Required recurrence relation
    dp[i] = max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));

    // Returning the value
    return dp[i];
}

// Driver code
int main()
{
    int arr[] = { 1, 3, 7, 6 };
    int n = sizeof(arr) / sizeof(int);
    int k = 1;

    // Finding prefix-sum
    findPrefixSum(arr, n);

    // Finding the maximum possible sum
    cout << maxSum(arr, 0, n, k);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

    static int maxLen = 10;

    // To store the states of dp
    static int[] dp = new int[maxLen];

    // To check if a given state
    // has been solved
    static boolean[] v = new boolean[maxLen];

    // To store the prefix-sum
    static int[] prefix_sum = new int[maxLen];

    // Function to fill the prefix_sum[] with
    // the prefix sum of the given array
    static void findPrefixSum(int arr[], int n)
    {
        prefix_sum[0] = arr[0];
        for (int i = 1; i < n; i++)
        {
            prefix_sum[i] = arr[i] + prefix_sum[i - 1];
        }
    }

    // Function to find the maximum sum subsequence
    // such that no two elements are adjacent
    static int maxSum(int arr[], int i, int n, int k)
    {
        // Base case
        if (i + k > n)
        {
            return 0;
        }

        // To check if a state has
        // been solved
        if (v[i])
        {
            return dp[i];
        }
        v[i] = true;

        int x;

        if (i == 0)
        {
            x = prefix_sum[k - 1];
        }
        else
        {
            x = prefix_sum[i + k - 1] - prefix_sum[i - 1];
        }

        // Required recurrence relation
        dp[i] = Math.max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));

        // Returning the value
        return dp[i];
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 3, 7, 6};
        int n = arr.length;
        int k = 1;

        // Finding prefix-sum
        findPrefixSum(arr, n);

        // Finding the maximum possible sum
        System.out.println(maxSum(arr, 0, n, k));
    }
}

// This code contributed by Rajput-Ji

Python3

# Python3 implementation of the approach

maxLen = 10

# To store the states of dp
dp = [0]*maxLen;

# To check if a given state
# has been solved
v = [0]*maxLen;

# To store the prefix-sum
prefix_sum = [0]*maxLen;

# Function to fill the prefix_sum[] with
# the prefix sum of the given array
def findPrefixSum(arr, n) :

    prefix_sum[0] = arr[0];
    for i in range(n) :
        prefix_sum[i] = arr[i] + prefix_sum[i - 1];


# Function to find the maximum sum subsequence
# such that no two elements are adjacent
def maxSum(arr, i, n, k) :

    # Base case
    if (i + k > n) :
        return 0;

    # To check if a state has
    # been solved
    if (v[i]) :
        return dp[i];

    v[i] = 1;

    if (i == 0) :
        x = prefix_sum[k - 1];
    else :
        x = prefix_sum[i + k - 1] - prefix_sum[i - 1];

    # Required recurrence relation
    dp[i] = max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));

    # Returning the value
    return dp[i];


# Driver code
if __name__ == "__main__" :

    arr = [ 1, 3, 7, 6 ];

    n = len(arr);
    k = 1;

    # Finding prefix-sum
    findPrefixSum(arr, n);

    # Finding the maximum possible sum
    print(maxSum(arr, 0, n, k));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

    static int maxLen = 10;

    // To store the states of dp
    static int[] dp = new int[maxLen];

    // To check if a given state
    // has been solved
    static bool[] v = new bool[maxLen];

    // To store the prefix-sum
    static int[] prefix_sum = new int[maxLen];

    // Function to fill the prefix_sum[] with
    // the prefix sum of the given array
    static void findPrefixSum(int []arr, int n)
    {
        prefix_sum[0] = arr[0];
        for (int i = 1; i < n; i++)
        {
            prefix_sum[i] = arr[i] + prefix_sum[i - 1];
        }
    }

    // Function to find the maximum sum subsequence
    // such that no two elements are adjacent
    static int maxSum(int []arr, int i, int n, int k)
    {
        // Base case
        if (i + k > n)
        {
            return 0;
        }

        // To check if a state has
        // been solved
        if (v[i])
        {
            return dp[i];
        }
        v[i] = true;

        int x;

        if (i == 0)
        {
            x = prefix_sum[k - 1];
        }
        else
        {
            x = prefix_sum[i + k - 1] - prefix_sum[i - 1];
        }

        // Required recurrence relation
        dp[i] = Math.Max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));

        // Returning the value
        return dp[i];
    }

    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 3, 7, 6};
        int n = arr.Length;
        int k = 1;

        // Finding prefix-sum
        findPrefixSum(arr, n);

        // Finding the maximum possible sum
        Console.Write(maxSum(arr, 0, n, k));
    }
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript implementation of the approach
var maxLen = 10

// To store the states of dp
var dp = Array(maxLen);

// To check if a given state
// has been solved
var v = Array(maxLen);

// To store the prefix-sum
var prefix_sum = Array(maxLen);;

// Function to fill the prefix_sum[] with
// the prefix sum of the given array
function findPrefixSum(arr, n)
{
    prefix_sum[0] = arr[0];
    for (var i = 1; i < n; i++)
        prefix_sum[i] = arr[i] + prefix_sum[i - 1];
}

// Function to find the maximum sum subsequence
// such that no two elements are adjacent
function maxSum(arr, i, n, k)
{
    // Base case
    if (i + k > n)
        return 0;

    // To check if a state has
    // been solved
    if (v[i])
        return dp[i];
    v[i] = 1;

    var x;

    if (i == 0)
        x = prefix_sum[k - 1];
    else
        x = prefix_sum[i + k - 1] - prefix_sum[i - 1];

    // Required recurrence relation
    dp[i] = Math.max(maxSum(arr, i + 1, n, k),
                x + maxSum(arr, i + k + 1, n, k));

    // Returning the value
    return dp[i];
}

// Driver code
var arr = [1, 3, 7, 6];
var n = arr.length;
var k = 1;
// Finding prefix-sum
findPrefixSum(arr, n);
// Finding the maximum possible sum
document.write( maxSum(arr, 0, n, k));

</script>

Output:

9