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Find maximum subset-sum divisible by D by taking at most K elements from given array

  • Last Updated : 05 Aug, 2021
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Given an array A[] of size N, and two numbers K and D, the task is to calculate the maximum subset-sum divisible by D possible by taking at most K elements from A.

Examples:

Input: A={11, 5, 5, 1, 18}, N=5, K=3, D=7
Output:
28
Explanation:
The subset {5, 5, 18} gives the maximum sum=(5+5+18)=28 that is divisible by 7 and also has contains atmost 3 elements

Input: A={7, 7, 7, 7, 7}, N=5, K=2, D=7
Output:
14

Naive Approach: The Naive approach would be to generate all subsets of A(using bit masking), and for each subset, calculate the sum, and check whether the length of the subset is not greater than K, and the sum is divisible by D, and calculating the maximum among them.



Below is the implementation of the above approach:

C++




// C++ program for tha above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate maximum sum possible by taking at
// most K elements that is divisibly by D
int maximumSum(vector<int> A, int N, int K, int D)
{
    // variable to store final answer
    int ans = 0;
    // Traverse all subsets
    for (int i = 0; i < (1 << N); i++) {
        int sum = 0;
        int c = 0;
        for (int j = 0; j < N; j++) {
            if (i >> j & 1) {
                sum += A[j];
                c++;
            }
        }
        // Update ans if necessary
        // conditions are satisfied
        if (sum % D == 0 && c <= K)
            ans = max(ans, sum);
    }
    return ans;
}
// Driver code
int main()
{
    // Input
    int N = 5, K = 3, D = 7;
    vector<int> A = { 1, 11, 5, 5, 18 };
 
    // Function call
    cout << maximumSum(A, N, K, D) << endl;
 
    return 0;
}

Java




// Java program for tha above approach
import java.util.*;
 
class GFG{
 
// Function to calculate maximum sum
// possible by taking at most K
// elements that is divisibly by D
static int maximumSum(int[] A, int N,
                      int K, int D)
{
     
    // Variable to store final answer
    int ans = 0;
     
    // Traverse all subsets
    for(int i = 0; i < (1 << N); i++)
    {
        int sum = 0;
        int c = 0;
        for(int j = 0; j < N; j++)
        {
            if ((i >> j & 1) != 0)
            {
                sum += A[j];
                c++;
            }
        }
         
        // Update ans if necessary
        // conditions are satisfied
        if (sum % D == 0 && c <= K)
            ans = Math.max(ans, sum);
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Input
    int N = 5, K = 3, D = 7;
    int[] A = { 1, 11, 5, 5, 18 };
 
    // Function call
    System.out.print(maximumSum(A, N, K, D));
}   
}
 
// This code is contributed by susmitakundugoaldanga

Python3




# Python3 program for tha above approach
 
# Function to calculate maximum sum
# possible by taking at most K elements
# that is divisibly by D
def maximumSum(A, N, K, D):
     
    # Variable to store final answer
    ans = 0
     
    # Traverse all subsets
    for i in range((1 << N)):
        sum = 0
        c = 0
         
        for j in range(N):
            if (i >> j & 1):
                sum += A[j]
                c += 1
                 
        # Update ans if necessary
        # conditions are satisfied
        if (sum % D == 0 and c <= K):
            ans = max(ans, sum)
             
    return ans
 
# Driver code
if __name__ == '__main__':
     
    # Input
    N = 5
    K = 3
    D = 7
    A = [ 1, 11, 5, 5, 18 ]
 
    # Function call
    print(maximumSum(A, N, K, D))
 
# This code is contributed by mohit kumar 29

C#




// C# program for tha above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to calculate maximum sum possible
// by taking at most K elements that is divisibly by D
static int maximumSum(List<int> A, int N,
                           int K, int D)
{
     
    // Variable to store final answer
    int ans = 0;
     
    // Traverse all subsets
    for(int i = 0; i < (1 << N); i++)
    {
        int sum = 0;
        int c = 0;
        for(int j = 0; j < N; j++)
        {
            if ((i >> j & 1) != 0)
            {
                sum += A[j];
                c++;
            }
        }
         
        // Update ans if necessary
        // conditions are satisfied
        if (sum % D == 0 && c <= K)
            ans = Math.Max(ans, sum);
    }
    return ans;
}
 
// Driver code
public static void Main()
{
     
    // Input
    int N = 5, K = 3, D = 7;
    List<int> A = new List<int>(){ 1, 11, 5, 5, 18 };
 
    // Function call
    Console.Write(maximumSum(A, N, K, D));
}
}
 
// This code is contributed by SURENDRA_GANGWAR

Javascript




<script>
 
// JavaScript program for tha above approach
 
 
// Function to calculate maximum sum possible by taking at
// most K elements that is divisibly by D
function maximumSum(A, N, K, D) {
    // variable to store final answer
    let ans = 0;
    // Traverse all subsets
    for (let i = 0; i < (1 << N); i++) {
        let sum = 0;
        let c = 0;
        for (let j = 0; j < N; j++) {
            if (i >> j & 1) {
                sum += A[j];
                c++;
            }
        }
        // Update ans if necessary
        // conditions are satisfied
        if (sum % D == 0 && c <= K)
            ans = Math.max(ans, sum);
    }
    return ans;
}
// Driver code
 
// Input
let N = 5, K = 3, D = 7;
let A = [1, 11, 5, 5, 18];
 
// Function call
document.write(maximumSum(A, N, K, D) + "<br>");
 
</script>
Output: 
28

 

Time Complexity: O(N.2N)
Auxiliary Space: O(1)

Efficient Approach: This problem can be solved with the help of dynamic programming, with a 3D dp array, where dp[i][j][p] stores the maximum sum possible if j elements are taken till the ith index and its modulo D is p. Follow the steps to solve the problem: 

  1. Create a 3D array dp[][][] of size (N+1)x(K+1)x(D), and initialize it with -1.
  2. Iterate from 1 to N, and for each current index i, do the following:
    1. Initialize two variables element and mod to A[i-1] and A[i-1]%D.
    2. Copy dp[i-1] to dp[i].
    3. Iterate from 1 to K, and for each current index j, do the following:
      1. Update dp[i][j][mod] as the maximum of dp[i][j][mod] and element.
      2. Iterate from 0 to D-1, and for each current index p, do the following:
        1. If dp[i-1][j-1][p] is not equal to -1, Update dp[i][j][(p+mod)%D] as the maximum of dp[i][j][(p+mod)%D] and dp[i-1][j-1][p]+element.
  3. If dp[N][K][0] is -1, the answer is 0.
  4. Otherwise, the answer is dp[N][K][0].

Below is the implementation of the above approach: 

C++




#include <bits/stdc++.h>
using namespace std;
int maximumSum(vector<int> A, int N, int K, int D)
{
    // Dp vector
    vector<vector<vector<int> > > dp(
        N + 1, vector<vector<int> >(
                   K + 1, vector<int>(D + 1, -1)));
    for (int i = 1; i <= N; i++) {
        // current element
        int element = A[i - 1];
        // current element modulo D
        int mod = A[i - 1] % D;
        // copy previous state
        dp[i] = dp[i - 1];
        for (int j = 1; j <= K; j++) {
            // Transitions
            dp[i][j][mod] = max(dp[i][j][mod], element);
            for (int p = 0; p < D; p++) {
                if (dp[i - 1][j - 1][p] != -1) {
                    dp[i][j][(p + mod) % D] = max(
                        dp[i][j][(p + mod) % D],
                        dp[i - 1][j - 1][p] + element);
                }
            }
        }
    }
    // return answer
    if (dp[N][K][0] == -1)
        return 0;
    return dp[N][K][0];
}
// Driver code
int main()
{
    // Input
    int N = 5, K = 3, D = 7;
    vector<int> A = { 1, 11, 5, 5, 18 };
 
    // Function call
    cout << maximumSum(A, N, K, D) << endl;
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
class GFG {
 
static int maximumSum(int[] A, int N, int K, int D)
{
   
    // Dp vector
    int[][][] dp = new int[N+1][K+1][D+1];
     
    for(int i = 0; i < N + 1; i++)
    {
        for(int j = 0; j < K + 1; j++)
        {
            for(int k = 0; k < D + 1; k++)
            {
                dp[i][j][k] = -1;
            }
        }
    }
 
    for (int i = 1; i <= N; i++)
    {
       
        // current element
        int element = A[i - 1];
       
        // current element modulo D
        int mod = A[i - 1] % D;
       
        // copy previous state
        dp[i] = dp[i - 1];
        for (int j = 1; j <= K; j++)
        {
           
            // Transitions
            dp[i][j][mod] = Math.max(dp[i][j][mod], element);
            for (int p = 0; p < D; p++) {
                if (dp[i - 1][j - 1][p] != -1) {
                    dp[i][j][(p + mod) % D] = Math.max(
                        dp[i][j][(p + mod) % D],
                        dp[i - 1][j - 1][p] + element);
                }
            }
        }
    }
   
    // return answer
    if (dp[N][K][0] == -1)
        return 0;
    return dp[N][K][0];
}
 
// Driver Code
public static void main(String[] args)
{
    // Input
    int N = 5, K = 3, D = 7;
    int[] A = { 1, 11, 5, 5, 18 };
 
    // Function call
    System.out.print(maximumSum(A, N, K, D));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.
Output: 
28

 

Time Complexity: O(NKD)
Auxiliary Space: O(NKD)

 

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