# Maximum subsequence sum such that all elements are K distance apart

Given an array arr[] of N integers and another integer K. The task is to find the maximum sum of a subsequence such that the difference of the indices of all consecutive elements in the subsequence in the original array is exactly K. For example, if arr[i] is the first element of the subsequence then the next element has to be arr[i + k] then arr[i + 2k] and so on.

Examples:

Input: arr[] = {2, -3, -1, -1, 2}, K = 2
Output: 3

Input: arr[] = {2, 3, -1, -1, 2}, K = 3
Output: 5

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

Approach: There are K sequences possible where the elememts are K distance apart and these can be of the forms:

0, 0 + k, 0 + 2k, …, 0 + n*k
1, 1 + k, 1 + 2k, …, 1 + n*k
2, 2 + k, 2 + 2k, …, 2 + n*k
k-1, k-1 + k, k-1 + 2k, …, k-1 + n*k

Now, any subarray of the sequences is a subsequence of the original array where elements are K distance apart from each other.

So, the task now reduces to find the maximum subarray sum of these sequences which can be found by Kadane’s algorithm.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum subarray sum ` `// for the array {a[i], a[i + k], a[i + 2k], ...} ` `int` `maxSubArraySum(``int` `a[], ``int` `n, ``int` `k, ``int` `i) ` `{ ` `    ``int` `max_so_far = INT_MIN, max_ending_here = 0; ` ` `  `    ``while` `(i < n) { ` `        ``max_ending_here = max_ending_here + a[i]; ` `        ``if` `(max_so_far < max_ending_here) ` `            ``max_so_far = max_ending_here; ` ` `  `        ``if` `(max_ending_here < 0) ` `            ``max_ending_here = 0; ` ` `  `        ``i += k; ` `    ``} ` `    ``return` `max_so_far; ` `} ` ` `  `// Function to return the sum of ` `// the maximum required subsequence ` `int` `find(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``// To store the result ` `    ``int` `maxSum = 0; ` ` `  `    ``// Run a loop from 0 to k ` `    ``for` `(``int` `i = 0; i <= min(n, k); i++) { ` `        ``int` `sum = 0; ` ` `  `        ``// Find the maximum subarray sum for the ` `        ``// array {a[i], a[i + k], a[i + 2k], ...} ` `        ``maxSum = max(maxSum, ` `                     ``maxSubArraySum(arr, n, k, i)); ` `    ``} ` ` `  `    ``// Return the maximum value ` `    ``return` `maxSum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, -3, -1, -1, 2 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 2; ` ` `  `    ``cout << find(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the maximum subarray sum  ` `// for the array {a[i], a[i + k], a[i + 2k], ...}  ` `static` `int` `maxSubArraySum(``int` `a[], ``int` `n,  ` `                          ``int` `k, ``int` `i)  ` `{  ` `    ``int` `max_so_far = Integer.MIN_VALUE, ` `        ``max_ending_here = ``0``;  ` ` `  `    ``while` `(i < n)  ` `    ``{  ` `        ``max_ending_here = max_ending_here + a[i];  ` `        ``if` `(max_so_far < max_ending_here)  ` `            ``max_so_far = max_ending_here;  ` ` `  `        ``if` `(max_ending_here < ``0``)  ` `            ``max_ending_here = ``0``;  ` ` `  `        ``i += k;  ` `    ``}  ` `    ``return` `max_so_far;  ` `}  ` ` `  `// Function to return the sum of  ` `// the maximum required subsequence  ` `static` `int` `find(``int` `arr[], ``int` `n, ``int` `k)  ` `{  ` `    ``// To store the result  ` `    ``int` `maxSum = ``0``;  ` ` `  `    ``// Run a loop from 0 to k  ` `    ``for` `(``int` `i = ``0``; i <= Math.min(n, k); i++) ` `    ``{  ` `        ``int` `sum = ``0``;  ` ` `  `        ``// Find the maximum subarray sum for the  ` `        ``// array {a[i], a[i + k], a[i + 2k], ...}  ` `        ``maxSum = Math.max(maxSum,  ` `                      ``maxSubArraySum(arr, n, k, i));  ` `    ``}  ` ` `  `    ``// Return the maximum value  ` `    ``return` `maxSum;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `arr[] = {``2``, -``3``, -``1``, -``1``, ``2``}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``2``; ` ` `  `    ``System.out.println(find(arr, n, k)); ` `} ` `}  ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` `# Function to return the maximum subarray sum ` `# for the array {a[i], a[i + k], a[i + 2k], ...} ` `import` `sys ` `def` `maxSubArraySum(a, n, k, i): ` ` `  `    ``max_so_far ``=` `-``sys.maxsize; ` `    ``max_ending_here ``=` `0``; ` ` `  `    ``while` `(i < n): ` `        ``max_ending_here ``=` `max_ending_here ``+` `a[i]; ` `        ``if` `(max_so_far < max_ending_here): ` `            ``max_so_far ``=` `max_ending_here; ` ` `  `        ``if` `(max_ending_here < ``0``): ` `            ``max_ending_here ``=` `0``; ` ` `  `        ``i ``+``=` `k; ` `     `  `    ``return` `max_so_far; ` ` `  `# Function to return the sum of ` `# the maximum required subsequence ` `def` `find(arr, n, k): ` ` `  `    ``# To store the result ` `    ``maxSum ``=` `0``; ` ` `  `    ``# Run a loop from 0 to k ` `    ``for` `i ``in` `range``(``0``, ``min``(n, k) ``+` `1``): ` `        ``sum` `=` `0``; ` ` `  `        ``# Find the maximum subarray sum for the ` `        ``# array {a[i], a[i + k], a[i + 2k], ...} ` `        ``maxSum ``=` `max``(maxSum, ` `                     ``maxSubArraySum(arr, n, k, i)); ` `     `  `    ``# Return the maximum value ` `    ``return` `maxSum; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr ``=` `[ ``2``, ``-``3``, ``-``1``, ``-``1``, ``2` `]; ` `    ``n ``=` `len``(arr); ` `    ``k ``=` `2``; ` ` `  `    ``print``(find(arr, n, k)); ` ` `  `# This code is contributed by Princi Singh `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to return the maximum subarray sum  ` `    ``// for the array {a[i], a[i + k], a[i + 2k], ...}  ` `    ``static` `int` `maxSubArraySum(``int` `[]a, ``int` `n,  ` `                              ``int` `k, ``int` `i)  ` `    ``{  ` `        ``int` `max_so_far = ``int``.MinValue, ` `            ``max_ending_here = 0;  ` `     `  `        ``while` `(i < n)  ` `        ``{  ` `            ``max_ending_here = max_ending_here + a[i];  ` `            ``if` `(max_so_far < max_ending_here)  ` `                ``max_so_far = max_ending_here;  ` `     `  `            ``if` `(max_ending_here < 0)  ` `                ``max_ending_here = 0;  ` `     `  `            ``i += k;  ` `        ``}  ` `        ``return` `max_so_far;  ` `    ``}  ` `     `  `    ``// Function to return the sum of  ` `    ``// the maximum required subsequence  ` `    ``static` `int` `find(``int` `[]arr, ``int` `n, ``int` `k)  ` `    ``{  ` `        ``// To store the result  ` `        ``int` `maxSum = 0;  ` `     `  `        ``// Run a loop from 0 to k  ` `        ``for` `(``int` `i = 0; i <= Math.Min(n, k); i++) ` `        ``{  ` ` `  `            ``// Find the maximum subarray sum for the  ` `            ``// array {a[i], a[i + k], a[i + 2k], ...}  ` `            ``maxSum = Math.Max(maxSum,  ` `                              ``maxSubArraySum(arr, n, k, i));  ` `        ``}  ` `     `  `        ``// Return the maximum value  ` `        ``return` `maxSum;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]arr = {2, -3, -1, -1, 2}; ` `        ``int` `n = arr.Length; ` `        ``int` `k = 2; ` `     `  `        ``Console.WriteLine(find(arr, n, k)); ` `    ``}     ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

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