Maximum subset sum such that no two elements in set have same digit in them

Last Updated : 18 Apr, 2023

Given an array of N elements. Find the subset of elements which has maximum sum such that no two elements in the subset has common digit present in them.
Examples:

Input : array[] = {22, 132, 4, 45, 12, 223}
Output : 268
Maximum Sum Subset will be = {45, 223} .
All possible digits are present except 1.
But to include 1 either 2 or both 2 and 3 have
to be removed which result in smaller sum value.
Input : array[] = {1, 21, 32, 4, 5 }
Output : 42

Here we can use Dynamic Programming and Bit Masking to solve this question.
Consider a 10-bit representation of every number where each bit is 1 if digit corresponding to that bit is present in that number.
Now maintain a dp[i], which stores the maximum possible sum which can be achieved with all those digits present in the set, corresponding to the bit positions which are 1 in Binary Representation of i.
Recurrence Relation will be of the form dp[i] = max(dp[i], dp[i^mask] + a[j]) , for all those j from 1 to n such that mask (10-bit Representation of a[j]) satisfy i || mask = i. (Since then only we can assure that all digit available in i are satisfied).

Below is the implementation of the above approach:

C++

   // C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
int dp[1024];
  
// Function to create mask for every number
int get_binary(int u)
{
    int ans = 0;
    while (u) {
        int rem = u % 10;
        ans |= (1 << rem);
        u /= 10;
    }
  
    return ans;
}
  
// Recursion for Filling DP array
int recur(int u, int array[], int n)
{
    // Base Condition
    if (u == 0)
        return 0;
    if (dp[u] != -1)
        return dp[u];
  
    int temp = 0;
    for (int i = 0; i < n; i++) {
        int mask = get_binary(array[i]);
  
        // Recurrence Relation
        if ((mask | u) == u) {
            dp[u] = max(max(0,
                    dp[u ^ mask]) + array[i], dp[u]);
        }
    }
  
    return dp[u];
}
  
// Function to find Maximum Subset Sum
int solve(int array[], int n)
{
    // Initialize DP array
    for (int i = 0; i < (1 << 10); i++) {
        dp[i] = -1;
    }
  
    int ans = 0;
  
    // Iterate over all possible masks of 10 bit number
    for (int i = 0; i < (1 << 10); i++) {
        ans = max(ans, recur(i, array, n));
    }
  
    return ans;
}
  
// Driver Code
int main()
{
    int array[] = { 22, 132, 4, 45, 12, 223 };
    int n = sizeof(array) / sizeof(array[0]);
     
    cout << solve(array, n);
}
 
  

Java

   // Java implementation of above approach
import java.io.*;
 
class GFG
{
     
static int []dp = new int [1024];
 
// Function to create mask for every number
static int get_binary(int u)
{
    int ans = 0;
    while (u > 0) 
     
    {
        int rem = u % 10;
        ans |= (1 << rem);
        u /= 10;
    }
 
    return ans;
}
 
// Recursion for Filling DP array
static int recur(int u, int []array, int n)
{
    // Base Condition
    if (u == 0)
        return 0;
    if (dp[u] != -1)
        return dp[u];
 
    for (int i = 0; i < n; i++) 
    {
        int mask = get_binary(array[i]);
 
        // Recurrence Relation
        if ((mask | u) == u)
        {
            dp[u] = Math.max(Math.max(0,
                    dp[u ^ mask]) + array[i], dp[u]);
        }
    }
 
    return dp[u];
}
 
// Function to find Maximum Subset Sum
static int solve(int []array, int n)
{
    // Initialize DP array
    for (int i = 0; i < (1 << 10); i++) 
    {
        dp[i] = -1;
    }
 
    int ans = 0;
 
    // Iterate over all possible masks of 10 bit number
    for (int i = 0; i < (1 << 10); i++) 
    {
        ans = Math.max(ans, recur(i, array, n));
    }
 
    return ans;
}
 
// Driver Code
static public void main (String[] args)
{
    int []array = { 22, 132, 4, 45, 12, 223 };
    int n = array.length;
     
    System.out.println(solve(array, n));
}
}
 
// This code is contributed by anuj_67..
 
  

Python3

   # Python3 implementation of above approach 
 
dp = [0]*1024; 
 
# Function to create mask for every number 
def get_binary(u) : 
 
    ans = 0; 
    while (u) :
        rem = u % 10; 
        ans |= (1 << rem); 
        u //= 10; 
    return ans; 
 
 
# Recursion for Filling DP array 
def recur(u, array, n) : 
 
    # Base Condition 
    if (u == 0) :
        return 0; 
         
    if (dp[u] != -1) : 
        return dp[u]; 
 
    temp = 0; 
    for i in range(n) : 
        mask = get_binary(array[i]); 
 
        # Recurrence Relation 
        if ((mask | u) == u) :
            dp[u] = max(max(0, dp[u ^ mask]) + array[i], dp[u]); 
 
    return dp[u]; 
 
 
# Function to find Maximum Subset Sum 
def solve(array, n)  :
    i = 0
     
    # Initialize DP array 
    while(i < (1 << 10)) :
        dp[i] = -1;
        i += 1
     
    ans = 0; 
 
    i = 0
    # Iterate over all possible masks of 10 bit number
    while(i < (1 << 10)) :
        ans = max(ans, recur(i, array, n));
         
        i += 1
     
    return ans; 
 
# Driver Code 
if __name__ ==  "__main__" : 
 
    array = [ 22, 132, 4, 45, 12, 223 ] ; 
    n = len(array); 
     
    print(solve(array, n)); 
     
    # This code is contributed by AnkitRai01
 
  

Javascript

   <script>
    // Javascript implementation of above approach
     
    let dp = new Array(1024);
    dp.fill(-1);
   
    // Function to create mask for every number
    function get_binary(u)
    {
        let ans = 0;
        while (u > 0) 
 
        {
            let rem = u % 10;
            ans |= (1 << rem);
            u = parseInt(u / 10, 10);
        }
 
        return ans;
    }
 
    // Recursion for Filling DP array
    function recur(u, array, n)
    {
        // Base Condition
        if (u == 0)
            return 0;
        if (dp[u] != -1)
            return dp[u];
 
        for (let i = 0; i < n; i++) 
        {
            let mask = get_binary(array[i]);
 
            // Recurrence Relation
            if ((mask | u) == u)
            {
                dp[u] = Math.max(Math.max(0,
                        dp[u ^ mask]) + array[i], dp[u]);
            }
        }
 
        return dp[u];
    }
 
    // Function to find Maximum Subset Sum
    function solve(array, n)
    {
        // Initialize DP array
        for (let i = 0; i < (1 << 10); i++) 
        {
            dp[i] = -1;
        }
 
        let ans = 0;
 
        // Iterate over all possible masks of 10 bit number
        for (let i = 0; i < (1 << 10); i++) 
        {
            ans = Math.max(ans, recur(i, array, n));
        }
 
        return ans;
    }
     
    let array = [ 22, 132, 4, 45, 12, 223 ];
    let n = array.length;
       
    document.write(solve(array, n));
 
</script>
 
  

C#

   // C# implementation of above approach
using System;
 
class GFG
{
     
static int []dp = new int [1024];
 
// Function to create mask for every number
static int get_binary(int u)
{
    int ans = 0;
    while (u > 0) 
     
    {
        int rem = u % 10;
        ans |= (1 << rem);
        u /= 10;
    }
 
    return ans;
}
 
// Recursion for Filling DP array
static int recur(int u, int []array, int n)
{
    // Base Condition
    if (u == 0)
        return 0;
    if (dp[u] != -1)
        return dp[u];
 
    for (int i = 0; i < n; i++) 
    {
        int mask = get_binary(array[i]);
 
        // Recurrence Relation
        if ((mask | u) == u)
        {
            dp[u] = Math.Max(Math.Max(0,
                    dp[u ^ mask]) + array[i], dp[u]);
        }
    }
 
    return dp[u];
}
 
// Function to find Maximum Subset Sum
static int solve(int []array, int n)
{
    // Initialize DP array
    for (int i = 0; i < (1 << 10); i++) 
    {
        dp[i] = -1;
    }
 
    int ans = 0;
 
    // Iterate over all possible masks of 10 bit number
    for (int i = 0; i < (1 << 10); i++) 
    {
        ans = Math.Max(ans, recur(i, array, n));
    }
 
    return ans;
}
 
// Driver Code
static public void Main ()
{
    int []array = { 22, 132, 4, 45, 12, 223 };
    int n = array.Length;
     
    Console.WriteLine (solve(array, n));
}
}
 
// This code is contributed by ajit.
 
  

Output:

Time Complexity : O(N*(2^10))

Auxiliary Space: O(1024)

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

Create a DP to store the solution of the subproblems.
Initialize the DP with base cases.
Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
Create a variable ans to store the final result.
Iterate over Dp and update ans.
At last return and print ans.

Implementation :

C++

   #include <bits/stdc++.h>
using namespace std;
 
int dp[1024];
 
// Function to create mask for every number
int get_binary(int u)
{
    int ans = 0;
    while (u) {
        int rem = u % 10;
        ans |= (1 << rem);
        u /= 10;
    }
    return ans;
}
 
// Function to find Maximum Subset Sum
int solve(int array[], int n)
{
    // Initialize DP array
    for (int i = 0; i < (1 << 10); i++) {
        dp[i] = 0;
    }
 
    // Fill DP table using bottom-up approach
    for (int i = 0; i < n; i++) {
        int mask = get_binary(array[i]);
        for (int j = (1 << 10) - 1; j >= 0; j--) {
            if ((mask | j) == j) {
                dp[j] = max(dp[j], dp[j ^ mask] + array[i]);
            }
        }
    }
 
    // Find maximum sum from DP array
    int ans = 0;
    for (int i = 0; i < (1 << 10); i++) {
        ans = max(ans, dp[i]);
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int array[] = { 22, 132, 4, 45, 12, 223 };
    int n = sizeof(array) / sizeof(array[0]);
 
    cout << solve(array, n);
}
 
  

Java

   import java.util.Arrays;
 
class GFG {
  static int dp[] = new int[1024];
 
  // Function to create mask for every number
  static int get_binary(int u)
  {
    int ans = 0;
    while (u != 0) {
      int rem = u % 10;
      ans |= (1 << rem);
      u /= 10;
    }
    return ans;
  }
 
  // Function to find Maximum Subset Sum
  static int solve(int array[], int n)
  {
    // Initialize DP array
    Arrays.fill(dp, 0);
 
    // Fill DP table using bottom-up approach
    for (int i = 0; i < n; i++) {
      int mask = get_binary(array[i]);
      for (int j = (1 << 10) - 1; j >= 0; j--) {
        if ((mask | j) == j) {
          dp[j] = Math.max(dp[j], dp[j ^ mask]
                           + array[i]);
        }
      }
    }
 
    // Find maximum sum from DP array
    int ans = 0;
    for (int i = 0; i < (1 << 10); i++) {
      ans = Math.max(ans, dp[i]);
    }
 
    return ans;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int array[] = { 22, 132, 4, 45, 12, 223 };
    int n = array.length;
 
    System.out.println(solve(array, n));
  }
}
 
  

Python3

   # Function to create mask for every number
def get_binary(u):
    ans = 0
    while u:
        rem = u % 10
        ans |= (1 << rem)
        u //= 10
    return ans
 
 # Function to find Maximum Subset Sum
 
 
def solve(array, n):
    # Initialize DP array
    dp = [0] * (1 << 10)
 
    # Fill DP table using bottom-up approach
    for i in range(n):
        mask = get_binary(array[i])
        for j in range((1 << 10) - 1, -1, -1):
            if (mask | j) == j:
                dp[j] = max(dp[j], dp[j ^ mask] + array[i])
 
    # Find maximum sum from DP array
    ans = 0
    for i in range(1 << 10):
        ans = max(ans, dp[i])
 
    return ans
 
 
# Driver Code
array = [22, 132, 4, 45, 12, 223]
n = len(array)
print(solve(array, n))
 
  

C#

   using System;
 
class Program
{
    static int[] dp = new int[1024];
 
    // Function to create mask for every number
    static int GetBinary(int u)
    {
        int ans = 0;
        while (u != 0)
        {
            int rem = u % 10;
            ans |= (1 << rem);
            u /= 10;
        }
        return ans;
    }
 
    // Function to find Maximum Subset Sum
    static int Solve(int[] array, int n)
    {
        // Initialize DP array
        for (int i = 0; i < (1 << 10); i++)
        {
            dp[i] = 0;
        }
 
        // Fill DP table using bottom-up approach
        for (int i = 0; i < n; i++)
        {
            int mask = GetBinary(array[i]);
            for (int j = (1 << 10) - 1; j >= 0; j--)
            {
                if ((mask | j) == j)
                {
                    dp[j] = Math.Max(dp[j], dp[j ^ mask] + array[i]);
                }
            }
        }
 
        // Find maximum sum from DP array
        int ans = 0;
        for (int i = 0; i < (1 << 10); i++)
        {
            ans = Math.Max(ans, dp[i]);
        }
 
        return ans;
    }
 
    // Driver Code
    static void Main()
    {
        int[] array = { 22, 132, 4, 45, 12, 223 };
        int n = array.Length;
 
        Console.WriteLine(Solve(array, n));
    }
}
 
  

Javascript

   // Function to create mask for every number
function getBinary(u) {
  let ans = 0;
  while (u) {
    let rem = u % 10;
    ans |= (1 << rem);
    u = Math.floor(u / 10);
  }
  return ans;
}
 
// Function to find Maximum Subset Sum
function solve(array, n) {
  // Initialize DP array
  let dp = new Array(1 << 10).fill(0);
 
  // Fill DP table using bottom-up approach
  for (let i = 0; i < n; i++) {
    let mask = getBinary(array[i]);
    for (let j = (1 << 10) - 1; j >= 0; j--) {
      if ((mask | j) == j) {
        dp[j] = Math.max(dp[j], dp[j ^ mask] + array[i]);
      }
    }
  }
 
  // Find maximum sum from DP array
  let ans = 0;
  for (let i = 0; i < 1 << 10; i++) {
    ans = Math.max(ans, dp[i]);
  }
 
  return ans;
}
 
// Driver Code
let array = [22, 132, 4, 45, 12, 223];
let n = array.length;
console.log(solve(array, n));
 
  

Output

Time Complexity : O(N*(2^10))

Auxiliary Space: O(2^10)

Find all distinct subset (or subsequence) sums of an array

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Maximum subset sum such that no two elements in set have same digit in them

C++

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Python3

Javascript

C#

C++

Java

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C#

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