Microsoft - MaxEmployeeAttendence (R repititions - DP solution bitmask), C++ (gcc)

Microsoft - MaxEmployeeAttendence (R repititions - DP solution bitmask)

//g++  7.4.0
// https://leetcode.com/discuss/interview-experience/1930164/Microsoft-or-SDE2-or-Hyderabad-or-April-2022-Reject

#include <bits/stdc++.h>

using namespace std;

const int MAX_DAYS = 10;
const int MAX_REPITITIONS = 10;
const int MAX_EMPLOYEES = 10;

int dp[MAX_DAYS][MAX_REPITITIONS][1 << MAX_EMPLOYEES];

vector<vector<int>> possible_attendants_at_day;

int CountSetBits (int n) {
    int ans = 0;
    for (int j = 0; j < 30; j ++) {
        if (n&(1 << j)) ans ++;
    }
    return ans;
}

int MaxEmployeeAttendence (int current_day, int repititions_left, int employees_mask) {
    if (repititions_left == 0) return CountSetBits (employees_mask);
    if (current_day < 0) return 0;           // There are no more days & we haven't organised the course for required `repititions`.
    
    int &result = dp[current_day][repititions_left][employees_mask];
    if (result != -1) return result;
    
    result = 0;
    
    // take the current day
    int new_mask = employees_mask;
    for (int employee : possible_attendants_at_day[current_day]) {
        new_mask |= (1 << employee);
    }
    result = max (result, MaxEmployeeAttendence(current_day-1, repititions_left-1, new_mask));
    
    // don't take the current day
    result = max (result, MaxEmployeeAttendence(current_day-1, repititions_left, employees_mask));
    
    return result;
}

/*
 * availability[i] - list of days on which ith employee will be available to take the training. [0 <= availability[i][j] < D]
 *
 * total_days (D)  - total number of days in which the training can be conducted.
 *
 * repititions (R) - number of times the training will be repeated in 'total_days' duration.
 *
**/
int MaxEmployeeAttendence (vector<vector<int>>& availability, int total_days, int repititions) {
    memset(dp, -1, sizeof(dp));
    possible_attendants_at_day.resize(total_days, {});
    
    for (int employee = 0; employee < availability.size(); employee ++) {
        for (int day : availability[employee]) {
            possible_attendants_at_day [day].push_back(employee);
        }
    }
    
    return MaxEmployeeAttendence (total_days-1, repititions, 0);
}

int main() {
    vector<vector<int>> availability = {
        {0, 3, 9},
        {4},
        {1, 4},
        {3, 2},
        {3, 4},
        {7}
    };
    
    cout << MaxEmployeeAttendence (availability, 10, 3) << endl;
    return 0;
}


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