# Pick maximum sum M elements such that contiguous repetitions do not exceed K

Given an array arr[] of distinct elements and two integers M and K, the task is to generate an array from the given array elements (elements can repeat in the generated array) such that the size of the generated array is M and the length of any sub-array with all same elements must not exceed K. Print the maximum sum of the elements among all the possible arrays that can be generated.

Examples:

Input: arr[] = {1, 3, 6, 7, 4, 5}, M = 9, K = 2
Output: 60
The maxim sum arrangement is 7 7 6 7 7 6 7 7 6. Note that there is no subarray of size more than 2 with all same elements.

Input: arr[] = {8, 13, 9, 17, 4, 12}, M = 5, K = 1
Output: 77
The maxim sum arrangement is 17, 13, 17, 13, 17

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

Approach: If we want the maximum sum we have to take maximum value from the array but we can repeat this maximum value at most K times so we have to separate it by second maximum value only once and after that we again take first maximum value up to K times and this cycle goes on until we take total M values.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum required sum ` `long` `int` `maxSum(``int` `arr[], ``int` `n, ``int` `m, ``int` `k) ` `{ ` ` `  `    ``int` `max1 = -1, max2 = -1; ` ` `  `    ``// All the elements in the array are distinct ` `    ``// Finding the maximum and the second maximum ` `    ``// element from the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(arr[i] > max1) { ` `            ``max2 = max1; ` `            ``max1 = arr[i]; ` `        ``} ` `        ``else` `if` `(arr[i] > max2) ` `            ``max2 = arr[i]; ` `    ``} ` ` `  `    ``// Total times the second maximum element ` `    ``// will appear in the generated array ` `    ``int` `counter = m / (k + 1); ` ` `  `    ``long` `int` `sum = counter * max2 + (m - counter) * max1; ` ` `  `    ``// Return the required sum ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 1, 3, 6, 7, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `m = 9, k = 2; ` `    ``cout << maxSum(arr, n, m, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the maximum required sum ` `static` `int` `maxSum(``int` `arr[], ``int` `n, ``int` `m, ``int` `k) ` `{ ` ` `  `    ``int` `max1 = -``1``, max2 = -``1``; ` ` `  `    ``// All the elements in the array are distinct ` `    ``// Finding the maximum and the second maximum ` `    ``// element from the array ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``if` `(arr[i] > max1)  ` `        ``{ ` `            ``max2 = max1; ` `            ``max1 = arr[i]; ` `        ``} ` `        ``else` `if` `(arr[i] > max2) ` `            ``max2 = arr[i]; ` `    ``} ` ` `  `    ``// Total times the second maximum element ` `    ``// will appear in the generated array ` `    ``int` `counter = m / (k + ``1``); ` ` `  `    ``int` `sum = counter * max2 + (m - counter) * max1; ` ` `  `    ``// Return the required sum ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `arr[] = { ``1``, ``3``, ``6``, ``7``, ``4``, ``5` `}; ` `    ``int` `n = arr.length; ` `    ``int` `m = ``9``, k = ``2``; ` `    ``System.out.println(maxSum(arr, n, m, k)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Surendra Gangwar `

## Python3

 `# Python3 implementation of the approach ` `def` `maxSum(arr, n, m, k): ` ` `  `    ``max1 ``=` `-``1` `    ``max2 ``=` `-``1` `     `  `    ``# All the elements in the array are distinct ` `    ``# Finding the maximum and the second maximum  ` `    ``# element from the array ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``if``(arr[i] > max1): ` `            ``max2 ``=` `max1 ` `            ``max1 ``=` `arr[i] ` `        ``elif``(arr[i] > max2): ` `            ``max2 ``=` `arr[i] ` `     `  `    ``# Total times the second maximum element  ` `    ``# will appear in the generated array ` `    ``counter ``=` `int``(m ``/` `(k ``+` `1``)) ` `     `  `    ``sum` `=` `counter ``*` `max2 ``+` `(m ``-` `counter) ``*` `max1 ` `     `  `    ``# Return the required sum ` `    ``return` `int``(``sum``) ` ` `  `# Driver code ` `arr ``=` `[``1``, ``3``, ``6``, ``7``, ``4``, ``5``]  ` `n ``=` `len``(arr) ` `m ``=` `9` `k ``=` `2` ` `  `print``(maxSum(arr, n, m, k)) `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to return the maximum required sum ` `static` `int` `maxSum(``int` `[]arr, ``int` `n, ``int` `m, ``int` `k) ` `{ ` ` `  `    ``int` `max1 = -1, max2 = -1; ` ` `  `    ``// All the elements in the array are distinct ` `    ``// Finding the maximum and the second maximum ` `    ``// element from the array ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``if` `(arr[i] > max1)  ` `        ``{ ` `            ``max2 = max1; ` `            ``max1 = arr[i]; ` `        ``} ` `        ``else` `if` `(arr[i] > max2) ` `            ``max2 = arr[i]; ` `    ``} ` ` `  `    ``// Total times the second maximum element ` `    ``// will appear in the generated array ` `    ``int` `counter = m / (k + 1); ` ` `  `    ``int` `sum = counter * max2 + (m - counter) * max1; ` ` `  `    ``// Return the required sum ` `    ``return` `sum; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``int` `[]arr = { 1, 3, 6, 7, 4, 5 }; ` `    ``int` `n = arr.Length; ` `    ``int` `m = 9, k = 2; ` `    ``Console.WriteLine(maxSum(arr, n, m, k)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

Output:

```60
```

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