Find maximum subset-sum divisible by D by taking at most K elements from given array
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 elementsInput: 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 the 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 the 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 the 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 the 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 the 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> |
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:
- Create a 3D array dp[][][] of size (N+1)x(K+1)x(D), and initialize it with -1.
- Iterate from 1 to N, and for each current index i, do the following:
- Initialize two variables element and mod to A[i-1] and A[i-1]%D.
- Copy dp[i-1] to dp[i].
- Iterate from 1 to K, and for each current index j, do the following:
- Update dp[i][j][mod] as the maximum of dp[i][j][mod] and element.
- Iterate from 0 to D-1, and for each current index p, do the following:
- 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.
- If dp[N][K][0] is -1, the answer is 0.
- 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. |
Python3
# Python3 code to implement the approach def maximumSum(A, N, K, D): # Dp vector dp = list () for i in range (N + 1 ): dp.append([ list () for _ in range (K + 1 )]) for j in range (K + 1 ): dp[i][j] = ([ - 1 for _ in range (D + 1 )]) for i in range ( 1 , N + 1 ): # current element element = A[i - 1 ]; # current element modulo D mod = A[i - 1 ] % D; # copy previous state dp[i] = dp[i - 1 ]; for j in range ( 1 , K + 1 ): # Transitions dp[i][j][mod] = max (dp[i][j][mod], element); for p in range (D): 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 # Input N, K, D = 5 , 3 , 7 A = [ 1 , 11 , 5 , 5 , 18 ]; # Function call print (maximumSum(A, N, K, D)); # This code is contributed by phasing17 |
C#
// C# program to implement above approach using System; using System.Collections.Generic; class GFG { static int maximumSum( int [] A, int N, int K, int D) { // Dp vector int [][][] dp = new int [N+1][][]; for ( int i = 0 ; i < N + 1 ; i++) { dp[i] = new int [K+1][]; for ( int j = 0 ; j < K + 1 ; j++) { dp[i][j] = new int [D+1]; 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 = new int []{ 1, 11, 5, 5, 18 }; // Function call Console.Write(maximumSum(A, N, K, D)); } } // This code is contributed by subhamgoyal2014. |
Javascript
// JavaScript code to implement the approach function maximumSum(A, N, K, D) { // Dp vector let dp = []; for ( var i = 0; i < N; i++) { dp.push([]); for ( var j = 0; j < K + 1; j++) dp[i].push( new Array(D + 1).fill(-1)); } for ( var i = 1; i <= N; i++) { // current element var element = A[i - 1]; // current element modulo D var mod = A[i - 1] % D; // copy previous state dp[i] = dp[i - 1]; for ( var j = 1; j <= K; j++) { // Transitions dp[i][j][mod] = Math.max(dp[i][j][mod], element); for ( var 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 // Input let N = 5, K = 3, D = 7; let A = [ 1, 11, 5, 5, 18 ]; // Function call console.log(maximumSum(A, N, K, D)); // This code is contributed by phasing17 |
28
Time Complexity: O(NKD)
Auxiliary Space: O(NKD)