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Maximum sum of elements divisible by K from the given array

  • Difficulty Level : Hard
  • Last Updated : 06 Jul, 2021
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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 <bits/stdc++.h>
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




<script>
 
// JavaScript program to implement
// the above approach
let dp = new Array(1000 + 1);
 
// Function to return the maximum sum
// divisible by k from elements of v
function find_max(i, sum, v, k)
{
     
    if (i == v.length)
        return 0;
 
    if (dp[i][sum] != -1)
        return dp[i][sum];
 
    let 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
    let arr = [ 43, 1, 17, 26, 15 ];
    let k = 16;
 
    // Loop to create 2D array using 1D array
    for (var i = 0; i < dp.length; i++) {
        dp[i] = new Array(2);
    }
 
    for (var i = 0; i < dp.length; i++) {
        for (var j = 0; j < dp.length; j++) {
 
            dp[i][j] = -1;
        }
    }
 
    document.write(find_max(0, 0, arr, k));
 
    // This code is contributed by Dharanendra L V.
</script>
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 <bits/stdc++.h>
using namespace std;
int main()
{
    int k=16;
    vector<int>arr={ 43, 1, 17, 26, 15 } ;
    int n=arr.size();
    vector<vector<int>> 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 <<dp[n][0];
    return 0;
}
Output
32

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