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Recurrence relation for bitmasking

kavinderrana121 asked:

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Basically I was trying to calculate the time complexity of below solution ,I took this problem from [Bit masking cap problem][1] and understand it very well But unable to calculate the time complexity as recursion is going inside loop
Even my professor didn't get this how to do it?
How can i write recurrence relation for this?


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:

What I have tried:

// C++ program to find number of ways to wear hats
#define MOD 1000000007
using namespace std;

// 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.
vector<int> capList[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
int dp[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.
int allmask;

// Mask is the set of persons, i is cap-id (OR the
// number of caps processed starting from first cap).
long long int 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 long int 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][j])) continue;

        // Else assign him this cap and recur for remaining caps with
        // new updated mask vector
        else ways += countWaysUtil(mask | (1 << capList[i][j]), i+1);
        ways %= MOD;

    // Save the result and return it.
    return dp[mask][i] = ways;

// Reads n lines from standard input for current test case
void countWays(int n)
    //----------- READ INPUT --------------------------
    string temp, str;
    int x;
    getline(cin, str); // to get rid of newline character
    for (int i=0; i<n; i++)
        getline(cin, str);
        stringstream ss(str);

        // while there are words in the streamobject ss
        while (ss >> temp)
            stringstream s;
            s << temp;
            s >> x;

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

    // All mask is used to check whether all persons
    // are included or not, set all n bits as 1
    allmask = (1 << n) - 1;

    // Initialize all entries in dp as -1
    memset(dp, -1, sizeof dp);

    // Call recursive function count ways
    cout << countWaysUtil(0, 1) << endl;

// Driver Program
int main()
    int n; // number of persons in every test case
    cin >> n;
    return 0;
Tags: C++14, Algorithms, analysis


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