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

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 ` `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 `

## 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:

```268
```

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

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Improved By : jit_t, vt_m, AnkitRai01