# Maximum sum of non-overlapping subarrays of length atmost K

Given an integer array ‘arr’ of length N and an integer ‘k’, select some non-overlapping subarrays such that each sub-array if of length at most ‘k’, no two sub-arrays are adjacent and sum of all the elements of the selected sub-arrays are maximum.

Examples:

```Input : arr[] = {-1, 2, -3, 4, 5}, k = 2
Output : 11
Sub-arrays that maximizes sum will be {{2}, {4, 5}}.
Thus, the answer will be 2+4+5 = 11.

Input :arr[] = {1, 1, 1, 1, 1}, k = 1
Output : 3
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach : A simple way is to generate all possible subsets of elements satisfying above conditions recursively and find the subset with maximum sum.
Time Complexity : O(2N)

Efficient Approach: A better approach is to use 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 {i, n-1} satisfying above conditions.

We will have ‘K+1’ possible choices i.e.

1. Reject ‘i’ and solve for ‘i+1’.
2. Select sub-array {i} and solve for ‘i+2’
3. Select sub-array {i, i+1} and solve for ‘i+3’
4. .
.
k+1) Select sub-array {i, i+1, i+2, .., i+k-1} and solve for ‘i+k+1’.

Thus, recurrence relation will be:

```dp[i] = max(dp[i+1], arr[i]+dp[i+2], arr[i]+arr[i+1]+dp[i+3],
...arr[i]+arr[i+1]+arr[i+2]...+arr[i+k-1] + dp[i+k+1])
```

Below is the implementation of the above approach:

## C++

 `// C++ program to implement above approach ` `#include ` `#define maxLen 10 ` `using` `namespace` `std; ` ` `  `// Variable to store states of dp ` `int` `dp[maxLen]; ` ` `  `// Variable to check if a given state has been solved ` `bool` `visit[maxLen]; ` ` `  `// 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 >= n) ` `        ``return` `0; ` ` `  `    ``// To check if a state has been solved ` `    ``if` `(visit[i]) ` `        ``return` `dp[i]; ` `    ``visit[i] = 1; ` ` `  `    ``// Variable to store ` `    ``// prefix sum for sub-array ` `    ``// {i, j} ` `    ``int` `tot = 0; ` `    ``dp[i] = maxSum(arr, i + 1, n, k); ` ` `  `    ``// Required recurrence relation ` `    ``for` `(``int` `j = i; j < i + k and j < n; j++) { ` `        ``tot += arr[j]; ` `        ``dp[i] = max(dp[i], tot + ` `                     ``maxSum(arr, j + 2, n, k)); ` `    ``} ` ` `  `    ``// Returning the value ` `    ``return` `dp[i]; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// Input array ` `    ``int` `arr[] = { -1, 2, -3, 4, 5 }; ` ` `  `    ``int` `k = 2; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``); ` ` `  `    ``cout << maxSum(arr, 0, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to implement above approach ` `import` `java.io.*; ` ` `  `class` `GFG ` `{ ` `     `  `    ``static` `int` `maxLen = ``10``; ` `     `  `    ``// Variable to store states of dp ` `    ``static` `int` `dp[] = ``new` `int``[maxLen]; ` `     `  `    ``// Variable to check  ` `    ``// if a given state has been solved ` `    ``static` `boolean` `[]visit = ``new` `boolean``[maxLen]; ` `     `  `    ``// 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 >= n) ` `            ``return` `0``; ` `     `  `        ``// To check if a state has been solved ` `        ``if` `(visit[i]) ` `            ``return` `dp[i]; ` `        ``visit[i] = ``true``; ` `     `  `        ``// Variable to store ` `        ``// prefix sum for sub-array ` `        ``// {i, j} ` `        ``int` `tot = ``0``; ` `        ``dp[i] = maxSum(arr, i + ``1``, n, k); ` `     `  `        ``// Required recurrence relation ` `        ``for` `(``int` `j = i; j < (i + k) && ` `                            ``(j < n); j++) ` `        ``{ ` `            ``tot += arr[j]; ` `            ``dp[i] = Math.max(dp[i], tot + ` `                    ``maxSum(arr, j + ``2``, n, k)); ` `        ``} ` `     `  `        ``// Returning the value ` `        ``return` `dp[i]; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` ` `  `        ``// Input array ` `        ``int` `arr[] = { -``1``, ``2``, -``3``, ``4``, ``5` `}; ` `         `  `        ``int` `k = ``2``; ` `         `  `        ``int` `n = arr.length; ` `         `  `        ``System.out.println(maxSum(arr, ``0``, n, k)); ` `    ``} ` `} ` ` `  `// This code is contributed by ajit. `

## Python3

 `# Python3 program to implement above approach  ` `maxLen ``=` `10` ` `  `# Variable to store states of dp  ` `dp ``=` `[``0``]``*``maxLen;  ` ` `  `# Variable to check if a given state has been solved  ` `visit ``=` `[``0``]``*``maxLen;  ` ` `  `# Function to find the maximum sum subsequence  ` `# such that no two elements are adjacent  ` `def` `maxSum(arr, i, n, k) :  ` ` `  `    ``# Base case  ` `    ``if` `(i >``=` `n) : ` `        ``return` `0``;  ` ` `  `    ``# To check if a state has been solved  ` `    ``if` `(visit[i]) :  ` `        ``return` `dp[i];  ` `         `  `    ``visit[i] ``=` `1``;  ` ` `  `    ``# Variable to store  ` `    ``# prefix sum for sub-array  ` `    ``# {i, j}  ` `    ``tot ``=` `0``;  ` `    ``dp[i] ``=` `maxSum(arr, i ``+` `1``, n, k);  ` ` `  `    ``# Required recurrence relation  ` `    ``j ``=` `i ` `    ``while` `(j < i ``+` `k ``and` `j < n) : ` `        ``tot ``+``=` `arr[j];  ` `        ``dp[i] ``=` `max``(dp[i], tot ``+` `                    ``maxSum(arr, j ``+` `2``, n, k));  ` `                     `  `        ``j ``+``=` `1` `     `  `    ``# Returning the value  ` `    ``return` `dp[i];  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``# Input array  ` `    ``arr ``=` `[ ``-``1``, ``2``, ``-``3``, ``4``, ``5` `];  ` ` `  `    ``k ``=` `2``;  ` ` `  `    ``n ``=` `len``(arr);  ` ` `  `    ``print``(maxSum(arr, ``0``, n, k));  ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# program to implement above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `static` `int` `maxLen = 10; ` ` `  `// Variable to store states of dp ` `static` `int` `[]dp = ``new` `int``[maxLen]; ` ` `  `// Variable to check  ` `// if a given state has been solved ` `static` `bool` `[]visit = ``new` `bool``[maxLen]; ` ` `  `// 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 >= n) ` `        ``return` `0; ` ` `  `    ``// To check if a state has been solved ` `    ``if` `(visit[i]) ` `        ``return` `dp[i]; ` `    ``visit[i] = ``true``; ` ` `  `    ``// Variable to store ` `    ``// prefix sum for sub-array ` `    ``// {i, j} ` `    ``int` `tot = 0; ` `    ``dp[i] = maxSum(arr, i + 1, n, k); ` ` `  `    ``// Required recurrence relation ` `    ``for` `(``int` `j = i; j < (i + k) && ` `                        ``(j < n); j++) ` `    ``{ ` `        ``tot += arr[j]; ` `        ``dp[i] = Math.Max(dp[i], tot + ` `                  ``maxSum(arr, j + 2, n, k)); ` `    ``} ` ` `  `    ``// Returning the value ` `    ``return` `dp[i]; ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` ` `  `    ``// Input array ` `    ``int` `[]arr = { -1, 2, -3, 4, 5 }; ` `     `  `    ``int` `k = 2; ` `     `  `    ``int` `n = arr.Length; ` `     `  `    ``Console.WriteLine (maxSum(arr, 0, n, k)); ` `} ` `} ` ` `  `// This code is contributed by ajit. `

Output:

```11
```

Time Complexity: O(n*k)

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