Count ways to assign unique cap to every person

https://www.geeksforgeeks.org/bitmasking-and-dynamic-programming-set-1-count-ways-to-assign-unique-cap-to-every-person/

There are 100 different types of caps each having a unique id from 1 to 100. Also, there are ‘n’ persons each having a collection of a variable number of caps. One day all of these persons decide to go in a party wearing a cap but to look unique they decided that none of them will wear the same type of cap. So, count the total number of arrangements or ways such that none of them is wearing the same type of cap.

Constraints: 1 <= n <= 10 Example:

The first line contains the value of n, next n lines contain collections 
of all the n persons.
Input: 
3
5 100 1     // Collection of the first person.
2           // Collection of the second person.
5 100       // Collection of the third person.

Output:
4
Explanation: All valid possible ways are (5, 2, 100),  (100, 2, 5),
            (1, 2, 5) and  (1, 2, 100)

A Simple Solution is to try all possible combinations. Start by picking the first element from the first set, marking it as visited and recur for remaining sets. It is basically a Backtracking based solution.

A better solution is to use Bitmasking and DP. Let us first introduce Bitmasking.

What is Bitmasking?
Suppose we have a collection of elements which are numbered from 1 to N. If we want to represent a subset of this set then it can be encoded by a sequence of N bits (we usually call this sequence a “mask”). In our chosen subset the i-th element belongs to it if and only if the i-th bit of the mask is set i.e., it equals to 1. For example, the mask 10000101 means that the subset of the set [1… 8] consists of elements 1, 3 and 8. We know that for a set of N elements there are total 2N subsets thus 2N masks are possible, one representing each subset. Each mask is, in fact, an integer number written in binary notation.

Our main methodology is to assign a value to each mask (and, therefore, to each subset) and thus calculate the values for new masks using values of the already computed masks. Usually our main target is to calculate value/solution for the complete set i.e., for mask 11111111. Normally, to find the value for a subset X we remove an element in every possible way and use values for obtained subsets X’1, X’2… ,X’k to compute the value/solution for X. This means that the values for X’i must have been computed already, so we need to establish an ordering in which masks will be considered. It’s easy to see that the natural ordering will do: go over masks in increasing order of corresponding numbers. Also, We sometimes, start with the empty subset X and we add elements in every possible way and use the values of obtained subsets X’1, X’2… ,X’k to compute the value/solution for X.

We mostly use the following notations/operations on masks:
bit(i, mask) – the i-th bit of mask
count(mask) – the number of non-zero bits in the mask
first(mask) – the number of the lowest non-zero bit in the mask
set(i, mask) – set the ith bit in the mask
check(i, mask) – check the ith bit in the mask

   // capList[i]'th vector contains the list of persons having a cap with id i

    // id is between 1 to 100 so we declared an array of 101 vectors as indexing

    // starts from 0.

    static Vector<Integer> capList[] = new Vector[101];

      

       

    // dp[2^10][101] .. in dp[i][j], i denotes the mask i.e., it tells that

    // how many and which persons are wearing cap. j denotes the first j caps

    // used. So, dp[i][j] tells the number ways we assign j caps to mask i

    // such that none of them wears the same cap

    static int dp[][] = new int[1025][101];

       

    // This is used for base case, it has all the N bits set

    // so, it tells whether all N persons are wearing a cap.

    static int allmask;

       

    // Mask is the set of persons, i is cap-id (OR the 

    // number of caps processed starting from first cap).

    static long countWaysUtil(int mask, int i)

    {

        // If all persons are wearing a cap so we

        // are done and this is one way so return 1

        if (mask == allmask) return 1;

       

        // If not everyone is wearing a cap and also there are no more

        // caps left to process, so there is no way, thus return 0;

        if (i > 100) return 0;

       

        // If we already have solved this subproblem, return the answer.

        if (dp[mask][i] != -1) return dp[mask][i];

       

        // Ways, when we don't include this cap in our arrangement

        // or solution set.

        long ways = countWaysUtil(mask, i+1);

       

        // size is the total number of persons having cap with id i.

        int size = capList[i].size();

       

        // So, assign one by one ith cap to all the possible persons

        // and recur for remaining caps.

        for (int j = 0; j < size; j++)

        {

            // if person capList[i][j] is already wearing a cap so continue as

            // we cannot assign him this cap

            if ((mask & (1 << capList[i].get(j))) != 0) continue;

       

            // Else assign him this cap and recur for remaining caps with

            // new updated mask vector

            else ways += countWaysUtil(mask | (1 << capList[i].get(j)), i+1);

            ways %= MOD;

        }

       

        // Save the result and return it.

        return dp[mask][i] = (int) ways;

    }

       

    // Reads n lines from standard input for current test case

    static void countWays(int n) throws Exception

    {

        //----------- READ INPUT --------------------------

        String str;

        String split[];

        int x;

                

        for (int i=0; i<n; i++)

        {

            str = br.readLine();

            split = str.split(" ");

            

            // while there are words in the split[]

            for (int j = 0; j < split.length; j++) {

                 // add the ith person in the list of cap if with id x

                x = Integer.parseInt(split[j]);

                capList[x].add(i);

            }

            

        }

        //----------------------------------------------------

       

        // All mask is used to check of all persons

        // are included or not, set all n bits as 1

        allmask = (1 << n) - 1;

       

        // Initialize all entries in dp as -1

        for (int[] is : dp) {

            for (int i = 0; i < is.length; i++) {

                is[i] = -1;

            }

        }

       

        // Call recursive function count ways

        System.out.println(countWaysUtil(0, 1));

    }