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# Maximum sum of elements divisible by K from the given array

• Difficulty Level : Hard
• Last Updated : 06 Jul, 2021

Given an array of integers and a number K. The task is to find the maximum sum which is divisible by K from the given array.
Examples:

Input: arr[] = {3, 6, 5, 1, 8}, k = 3
Output: 18
Explanation: 18 is formed by the elements 3, 6, 1, 8.
Input: arr = { 43, 1, 17, 26, 15 } , k = 16
Output: 32
Explanation: 32 is formed by the elements 17, 15.

Naive Approach: Recursively check all the possible combinations to find the solution. The solution is of exponential time complexity and thus inefficient.
Efficient Approach: A dynamic programming approach by maintaining a 2-D array dp which stores the state of variable sum and i (where sum is the current sum and i is the ith index of integer array). By recurring over all elements, calculate the sum including the element at index i as well as excluding it and check if divisible by k. If so, store the maximum of them in dp[i][sum] and return.
Below code is the implementation of the above approach:

## CPP

 `#include ``using` `namespace` `std;` `int` `dp[1001][1001];` `// Function to return the maximum sum``// divisible by k from elements of v``int` `find_max(``int` `i, ``int` `sum, vector<``int``>& v,``int` `k)``{` `    ``if` `(i == v.size())``        ``return` `0;` `    ``if` `(dp[i][sum] != -1)``        ``return` `dp[i][sum];` `    ``int` `ans = 0;``    ``// check if sum of elements excluding the``    ``// current one is divisible by k``    ``if` `((sum + find_max(i + 1, sum, v, k)) % k == 0)``        ``ans = find_max(i + 1, sum, v, k);``    ` `    ``// check if sum of elements including the``    ``// current one is divisible by k``    ``if``((sum + v[i] + find_max(i + 1,(sum + v[i]) % k,``                                   ``v, k)) % k == 0)``        ``// Store the maximum``        ``ans = max(ans, v[i] + find_max(i + 1,``                            ``(sum + v[i]) % k,v, k));``    `  `    ``return` `dp[i][sum] = ans;``}` `// Driver code``int` `main()``{``    ``vector<``int``> arr = { 43, 1, 17, 26, 15 };``    ``int` `k = 16;``    ``memset``(dp, -1, ``sizeof``(dp));``    ``cout << find_max(0, 0, arr, k);``}`

## Java

 `class` `GFG{`` ` `static` `int` `[][]dp = ``new` `int``[``1001``][``1001``];`` ` `// Function to return the maximum sum``// divisible by k from elements of v``static` `int` `find_max(``int` `i, ``int` `sum, ``int` `[]v, ``int` `k)``{`` ` `    ``if` `(i == v.length)``        ``return` `0``;`` ` `    ``if` `(dp[i][sum] != -``1``)``        ``return` `dp[i][sum];`` ` `    ``int` `ans = ``0``;` `    ``// check if sum of elements excluding the``    ``// current one is divisible by k``    ``if` `((sum + find_max(i + ``1``, sum, v, k)) % k == ``0``)``        ``ans = find_max(i + ``1``, sum, v, k);``     ` `    ``// check if sum of elements including the``    ``// current one is divisible by k``    ``if``((sum + v[i] + find_max(i + ``1``,(sum + v[i]) % k,``                                   ``v, k)) % k == ``0``)``        ``// Store the maximum``        ``ans = Math.max(ans, v[i] + find_max(i + ``1``,``                            ``(sum + v[i]) % k, v, k));``     ` `    ``return` `dp[i][sum] = ans;``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `[]arr = { ``43``, ``1``, ``17``, ``26``, ``15` `};``    ``int` `k = ``16``;``    ``for` `(``int` `i = ``0``; i < ``1001``; i++)``        ``for` `(``int` `j = ``0``; j < ``1001``; j++)``            ``dp[i][j] = -``1``;``    ``System.out.print(find_max(``0``, ``0``, arr, k));``}``}` `// This code is contributed by 29AjayKumar`

## Python 3

 `# Python3 implementation``dp ``=` `[[``-``1` `for` `i ``in` `range``(``1001``)] ``for` `j ``in` `range``(``1001``)]` `# Function to return the maximum sum``# divisible by k from elements of v``def` `find_max(i, ``sum``, v, k):``    ``if` `(i ``=``=` `len``(v)):``        ``return` `0` `    ``if` `(dp[i][``sum``] !``=` `-``1``):``        ``return` `dp[i][``sum``]` `    ``ans ``=` `0``    ` `    ``# check if sum of elements excluding the``    ``# current one is divisible by k``    ``if` `((``sum` `+` `find_max(i ``+` `1``, ``sum``, v, k)) ``%` `k ``=``=` `0``):``        ``ans ``=` `find_max(i ``+` `1``, ``sum``, v, k)``    ` `    ``# check if sum of elements including the``    ``# current one is divisible by k``    ``if``((``sum` `+` `v[i] ``+` `find_max(i ``+` `1``,(``sum` `+` `v[i]) ``%` `k, v, k)) ``%` `k ``=``=` `0``):``        ` `        ``# Store the maximum``        ``ans ``=` `max``(ans, v[i] ``+` `find_max(i ``+` `1``,(``sum` `+` `v[i]) ``%` `k, v, k))``    ` `    ``dp[i][``sum``] ``=` `ans` `    ``return` `dp[i][``sum``]` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``43``, ``1``, ``17``, ``26``, ``15``]``    ``k ``=` `16``    ``print``(find_max(``0``, ``0``, arr, k))` `# This code is contributed by Surendra_Gangwar`

## C#

 `using` `System;` `class` `GFG{``  ` `static` `int` `[,]dp = ``new` `int``[1001,1001];``  ` `// Function to return the maximum sum``// divisible by k from elements of v``static` `int` `find_max(``int` `i, ``int` `sum, ``int` `[]v, ``int` `k)``{``  ` `    ``if` `(i == v.Length)``        ``return` `0;``  ` `    ``if` `(dp[i,sum] != -1)``        ``return` `dp[i,sum];``  ` `    ``int` `ans = 0;`` ` `    ``// check if sum of elements excluding the``    ``// current one is divisible by k``    ``if` `((sum + find_max(i + 1, sum, v, k)) % k == 0)``        ``ans = find_max(i + 1, sum, v, k);``      ` `    ``// check if sum of elements including the``    ``// current one is divisible by k``    ``if``((sum + v[i] + find_max(i + 1,(sum + v[i]) % k,``                                   ``v, k)) % k == 0)``        ``// Store the maximum``        ``ans = Math.Max(ans, v[i] + find_max(i + 1,``                            ``(sum + v[i]) % k, v, k));``      ` `    ``return` `dp[i, sum] = ans;``}``  ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 43, 1, 17, 26, 15 };``    ``int` `k = 16;``    ``for` `(``int` `i = 0; i < 1001; i++)``        ``for` `(``int` `j = 0; j < 1001; j++)``            ``dp[i,j] = -1;``    ``Console.Write(find_max(0, 0, arr, k));``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``
Output

`32`

Iterative implementation using top down dp:

We will be using the index and the modulus value of the sum as our our states of dp. dp[i][j] would store the maximum sum of the array till ith index whose modulus is j.

## C++14

 `#include ``using` `namespace` `std;``int` `main()``{``    ``int` `k=16;``    ``vector<``int``>arr={ 43, 1, 17, 26, 15 } ;``    ``int` `n=arr.size();``    ``vector> dp(n+2, vector<``int``>(k, 0));``    ``for` `(``int` `i = 1; i <= n; i++) {``        ` `        ``for` `(``int` `j = 0; j < k ; j++) {``            ``dp[i][j] = dp[i - 1][j];``        ``}``        ` `        ``dp[i][arr[i - 1] % k] = max(dp[i][arr[i - 1] % k], arr[i - 1]);``      ` `        ``for` `(``int` `j = 0; j < k; j++) {``            ``int` `m = (j + arr[i - 1]) % k;``            ``if` `(dp[i - 1][j] != 0)``                ``dp[i][m] = max(dp[i][m],arr[i - 1] + dp[i - 1][j]);``        ``}``      ` `    ``}``    ``cout <
Output
`32`

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