Number of Ways to Wear Different Hats to Each Other

class Solution {
    public int findWays(List<List<Integer>> collections) {
        int persons = collections.size();
        Map<Integer, ArrayList<Integer>> hatMapping = new HashMap<>();
            
        hatMapping = new HashMap<>();
        for (int i = 0; i < persons; i++) {
            for (int hat: collections.get(i)) {
                if (!hatMapping.containsKey(hat)) {
                    hatMapping.put(hat, new ArrayList<>());
                }
                    
                hatMapping.get(hat).add(i);
            }
        }
            
        int allAssigned = (int) Math.pow(2, persons) - 1;
        int MODULO = 1000000007;
            
        int[][] dpTable = new int[42][allAssigned + 1];
        for (int i = 0; i < dpTable.length; i++) {
            dpTable[i][allAssigned] = 1;
        }
            
        for (int hat = 40; hat > 0; hat--) {
            for (int mask = allAssigned; mask >= 0; mask--) {
                int current = dpTable[hat + 1][mask];
                if (hatMapping.containsKey(hat)) {
                    for (int person: hatMapping.get(hat)) {
                        if ((mask & (1 << person)) == 0) {
                            current = (current + dpTable[hat + 1][mask | (1 << person)]) % MODULO;
                        }
                    }
                }
                    
                dpTable[hat][mask] = current;
            }
        }
    
        return dpTable[1][0];
    }
}

The Java solution for finding the number of unique ways to distribute hats such that each person wears a different hat involves the use of a dynamic programming approach along with bit manipulation. The input is a list of lists where each sublist represents the hats available for each person.

Here's a breakdown of the approach:

• Initialize Variables:

  • Determine the number of persons with int persons = collections.size().
  • Initialize a mapping (hatMapping) from each hat to a list of persons who can wear it.

• Build Hat to Person Mapping:

  • Loop through each person and each hat they can wear.
  • For each hat, add the current person to the hatMapping list for that hat if they can wear it.

• Set Up Dynamic Programming Table:

  • Define a 2D DP table dpTable where dimensions consider the maximum number of hats (here considered as 41) and all possible states of persons wearing hats represented by allAssigned (bitmask).
  • Initially, set the base case for each state where everyone is assigned a hat.

• Recursive Calculation with DP Table:

  • Use a nested loop starting from the last hat and decrement down. Within each hat, consider every possible state of who has hats (tracked through bitmask mask).
  • For each state, attempt to assign the current hat to each person who can wear it (and isn't already wearing a hat in that state), updating the count combining the state where the person wears the hat with the state where they don't.
  • Employ mod operations to handle very large numbers as specified by MODULO.

• Return Result:

  • The value dpTable[1][0] gives the number of ways zero persons are assigned hats starting from consideration of hat 1 (i.e., the result for no one wearing a hat initially).

This implementation leverages the power of bitmasking and dynamic programming to efficiently explore all combinations and configurations, consequentially finding the number of ways to distribute the hats respecting the problem constraints. This is done in a polynomial time complexity which is manageable given typical constraints encountered in such problems.

class Solution:
    def countWays(self, hat_preferences: List[List[int]]) -> int:
        # Mapping hats to people
        hat_assignments = defaultdict(list)
        # Populate the mapping
        for person_id, preferences in enumerate(hat_preferences):
            for preference in preferences:
                hat_assignments[preference].append(person_id)
    
        # Number of people
        total_people = len(hat_preferences)
        # Modulo value for the result
        MODULO = 10**9 + 7
        # All people have a hat
        all_hats_assigned = (1 << total_people) - 1
            
        # Dynamic programming table definitions
        dp_states = [[0] * (all_hats_assigned + 1) for _ in range(42)]
        # Set base dp states where all are satisfied
        for pos in range(len(dp_states)):
            dp_states[pos][all_hats_assigned] = 1
            
        # Fill in dp table using existing states
        for current_mask in range(all_hats_assigned, -1, -1):
            for current_hat in range(40, 0, -1):
                current_ways = dp_states[current_hat + 1][current_mask]
                for individual in hat_assignments[current_hat]:
                    if current_mask & (1 << individual) == 0:
                        current_ways = (current_ways + dp_states[current_hat + 1][current_mask | (1 << individual)]) % MODULO
    
                dp_states[current_hat][current_mask] = current_ways
            
        return dp_states[1][0]

This Python solution approaches the problem of assigning hats uniquely to each person using a dynamic programming strategy. Here’s how it works:

1. Initialize a dictionary to map hat preferences to individuals.
2. Populate this dictionary by iterating over each person's preferences, linking each hat to the list of people who have that hat as a preference.
3. Define constants:
   • total_people gets the total number of individuals from the length of hat_preferences.
   • MODULO is set to 10^9+7 which is used later to ensure results of calculations are within an acceptable range.
   • all_hats_assigned represents a bitmask where all bits are set (indicating each person gets a unique hat).
4. Create a dynamic programming table (dp_states) whereby each element is initiated to zero, except at the state where all individuals are satisfied (all hats are assigned correctly), which is set to 1.
5. The dynamic programming transition iterates over all possible states (represented by current_mask) and for each hat considered backward from the last to the first. It calculates the number of ways to distribute the current hat among the individuals based on the persons' preferences.
6. For each person with a preference for the current hat, if they haven't been assigned any hat at current_mask, update current_ways by adding the states derived from placing the hat on this individual.
7. After evaluating all options, record the computed ways in dp_states.
8. Return the value dp_states[1][0], which represents the count of ways to satisfy the initial condition where no hats have been assigned.

This solution is efficient as it reduces the complexity by only considering valid preferences and uses bitmasking for state representation, thereby leveraging the power of bit operations for efficient state transitions. The use of the modulo operation ensures that the function handles very large numbers gracefully.