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# Bitmasking and Dynamic Programming | Set 1 (Count ways to assign unique cap to every person)

Consider the below problems statement. 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)

Since, number of ways could be large, so output modulo 1000000007

We strongly recommend you to minimize your browser and try this yourself first.

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

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.

How is this problem solved using Bitmasking + DP? The idea is to use the fact that there are upto 10 persons. So we can use an integer variable as a bitmask to store which person is wearing a cap and which is not.

Let i be the current cap number (caps from 1 to i-1 are already
processed). Let integer variable mask indicates that the persons w
earing and not wearing caps.  If i'th bit is set in mask, then
i'th person is wearing a cap, else not.

// consider the case when ith cap is not included
// in the arrangement

// when ith cap is included in the arrangement
// so, assign this cap to all possible persons
// one by one and recur for remaining persons.
∑ countWays(mask | (1 << j), i+1)
for every person j that can wear cap i

Note that the expression "mask | (1 << j)" sets j'th bit in mask.
And a person can wear cap i if it is there in the person's cap list
provided as input.

If we draw the complete recursion tree, we can observe that many subproblems are solved again and again. So we use Dynamic Programming. A table dp[][] is used such that in every entry dp[i][j], i is mask and j is cap number. Since we want to access all persons that can wear a given cap, we use an array of vectors, capList[101]. A value capList[i] indicates the list of persons that can wear cap i.

Below is the implementation of above idea.

## Python

 #Python program to find number of ways to wear hatsfrom collections import defaultdict class AssignCap:     # Initialize variables    def __init__(self):             self.allmask = 0             self.total_caps = 100             self.caps = defaultdict(list)      #  Mask is the set of persons, i is the current cap number.    def countWaysUtil(self,dp, mask, cap_no):                 # If all persons are wearing a cap so we        # are done and this is one way so return 1        if mask == self.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 cap_no > self.total_caps:            return 0         # If we have already solved this subproblem, return the answer.        if dp[mask][cap_no]!= -1 :            return dp[mask][cap_no]         # Ways, when we don't include this cap in our arrangement        # or solution set        ways = self.countWaysUtil(dp, mask, cap_no + 1)                 # assign ith cap one by one  to all the possible persons        # and recur for remaining caps.        if cap_no in self.caps:             for ppl in self.caps[cap_no]:                                 # if person 'ppl' is already wearing a cap then continue                if mask & (1 << ppl) : continue                                 # Else assign him this cap and recur for remaining caps with                # new updated mask vector                ways += self.countWaysUtil(dp, mask | (1 << ppl), cap_no + 1)                 ways = ways % (10**9 + 7)         # Save the result and return it        dp[mask][cap_no] = ways         return dp[mask][cap_no]       def countWays(self,N):         # Reads n lines from standard input for current test case        # create dictionary for cap. cap[i] = list of person having        # cap no i        for ppl in range(N):             cap_possessed_by_person = map(int, raw_input().strip().split())             for i in cap_possessed_by_person:                 self.caps[i].append(ppl)         # allmask is used to check if all persons        # are included or not, set all n bits as 1        self.allmask = (1 << N) -1         # Initialize all entries in dp as -1        dp = [[-1 for j in range(self.total_caps + 1)] for i in range(2 ** N)]         # Call recursive function countWaysUtil        # result will be in dp[0][1]        print self.countWaysUtil(dp, 0, 1,) #Driver Programdef main():    No_of_people = input() # number of persons in every test case     AssignCap().countWays(No_of_people)  if __name__ == '__main__':    main() # This code is contributed by Neelam Yadav

## C#

Input:

3
5 100 1
2
5 100

Output:

4

Time Complexity: O(n*2^n), where n is the number of persons. This is because the recursive function is called at most 2^n times and the for loop is executed at most n times.
Auxiliary Space: O(n*2^n), as we are using a dp 2D array of size n*2^n.