Count Ways To Assign Unique Cap To Every Person
Last Updated :
13 Nov, 2024
Given n people and 100 types of caps labelled from 1 to 100, along with a 2D integer array caps where caps[i] represents the list of caps preferred by the i-th person, the task is to determine the number of ways the n people can wear different caps.
Example:
Input: caps = [[3, 4], [4, 5], [5]]
Output: 1
Explanation: First person choose cap 3, Second person choose cap 4 and last one cap 5.
Input: caps = [[3, 5, 1], [3, 5]]
Output: 4
Explanation: There are 4 ways to choose hats: (3, 5), (5, 3), (1, 3) and (1, 5)
Using Recursion
In the recursive approach, there are two cases for each cap:
- Skip the current cap, this means we move to the next cap without assigning the current cap to any person, keeping the assigned count unchanged. The recursive call will look like: dfs(assignedCount, cap+1)
- Assign the current cap to each person who prefers it: For each person who likes the current cap, if they do not already have a cap assigned, we assign this cap to them and move to the next cap with the assigned count incremented by 1. After the recursive call, we backtrack by unassigning the cap to explore other possibilities.
The recurrence relation is as follows:
Base Cases:
- If assignedCount == totalPeople, return 1, as this indicates all people have a unique cap assigned.
- If cap > 100, return 0, as this means there are no more caps left to assign but not all people have received a cap.
- dfs(assignedCount, cap) = dfs(assignedCount, cap+1) + ∑person in capToPeople[cap] dfs(assignedCount+1, cap+1) if assignedPeople[person] == false , we loop through each person who prefers the current cap, check if they have already been assigned a cap, and if not, assign it to them and recurse to explore further assignments with the updated assignedCount.
C++
// C++ Code to Assign Unique Cap To Every Person
// using Recursion
#include <bits/stdc++.h>
using namespace std;
// Recursive function to calculate the number of ways
// to assign caps to people such that each person has
// a unique cap
int dfs(int assignedCount, vector<bool>& assignedPeople,
int cap, vector<vector<int>>& capToPeople, int totalPeople) {
// Base case: if all people have a cap assigned, return 1
if (assignedCount == totalPeople) {
return 1;
}
// If we've considered all caps and not everyone
// has a cap, return 0
if (cap > 100) {
return 0;
}
// Case: skip the current cap
int ways = dfs(assignedCount, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Assign the current cap to each person who likes it
for (int person : capToPeople[cap]) {
// Check if the person already has a cap assigned
if (!assignedPeople[person]) {
// Assign current cap to the person
assignedPeople[person] = true;
// Recurse with increased assigned count
ways = ways + dfs(assignedCount + 1, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Backtrack: unassign the cap for other possibilities
assignedPeople[person] = false;
}
}
return ways;
}
// Main function to calculate the number of ways to assign caps
int numberWays(vector<vector<int>>& caps) {
int n = caps.size();
// Map each cap to the list of people who prefer it
vector<vector<int>> capToPeople(101);
for (int i = 0; i < n; ++i) {
for (int cap : caps[i]) {
capToPeople[cap].push_back(i);
}
}
// Initialize assignedPeople vector to track
// assigned caps
vector<bool> assignedPeople(n, false);
// Call the recursive function starting from the first cap
return dfs(0, assignedPeople, 1, capToPeople, n);
}
int main() {
vector<vector<int>> caps
= {{1, 2, 3}, {1, 2}, {3, 4}, {4, 5}};
cout << numberWays(caps) << endl;
return 0;
}
// Java Code to Assign Unique Cap To Every Person
// using Recursion
import java.util.ArrayList;
import java.util.List;
class GfG {
// Recursive function to calculate the number of ways
// to assign caps to people such that each person has
// a unique cap
static int dfs(int assignedCount, ArrayList<Boolean> assignedPeople,
int cap, ArrayList<ArrayList<Integer>> capToPeople, int totalPeople) {
// Base case: if all people have a cap assigned, return 1
if (assignedCount == totalPeople) {
return 1;
}
// If we've considered all caps and not everyone
// has a cap, return 0
if (cap > 100) {
return 0;
}
// Case: skip the current cap
int ways = dfs(assignedCount, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Assign the current cap to each person who likes it
for (int person : capToPeople.get(cap)) {
// Check if the person already has a cap assigned
if (!assignedPeople.get(person)) {
// Assign current cap to the person
assignedPeople.set(person, true);
// Recurse with increased assigned count
ways = ways + dfs(assignedCount + 1, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Backtrack: unassign the cap for other possibilities
assignedPeople.set(person, false);
}
}
return ways;
}
// Main function to calculate the number of ways to assign caps
static int numberWays(ArrayList<ArrayList<Integer>> caps) {
int n = caps.size();
// Map each cap to the list of people who prefer it
ArrayList<ArrayList<Integer>> capToPeople = new ArrayList<>(101);
for (int i = 0; i <= 100; i++) {
capToPeople.add(new ArrayList<>());
}
for (int i = 0; i < n; ++i) {
for (int cap : caps.get(i)) {
capToPeople.get(cap).add(i);
}
}
// Initialize assignedPeople list to track assigned caps
ArrayList<Boolean> assignedPeople = new ArrayList<>(n);
for (int i = 0; i < n; i++) {
assignedPeople.add(false);
}
// Call the recursive function starting from the first cap
return dfs(0, assignedPeople, 1, capToPeople, n);
}
public static void main(String[] args) {
ArrayList<ArrayList<Integer>> caps = new ArrayList<>();
caps.add(new ArrayList<>(List.of(1, 2, 3)));
caps.add(new ArrayList<>(List.of(1, 2)));
caps.add(new ArrayList<>(List.of(3, 4)));
caps.add(new ArrayList<>(List.of(4, 5)));
System.out.println(numberWays(caps));
}
}
# Python Code to Assign Unique Cap To Every Person
# using Recursion
def dfs(assigned_count, assigned_people, cap, cap_to_people, total_people):
# Base case: if all people have a cap assigned, return 1
if assigned_count == total_people:
return 1
# If we've considered all caps and not everyone
# has a cap, return 0
if cap > 100:
return 0
# Case: skip the current cap
ways = dfs(assigned_count, assigned_people,
cap + 1, cap_to_people, total_people)
# Assign the current cap to each person who likes it
for person in cap_to_people[cap]:
# Check if the person already has a cap assigned
if not assigned_people[person]:
# Assign current cap to the person
assigned_people[person] = True
# Recurse with increased assigned count
ways += dfs(assigned_count + 1, assigned_people,
cap + 1, cap_to_people, total_people)
# Backtrack: unassign the cap for other possibilities
assigned_people[person] = False
return ways
# Main function to calculate the number
# of ways to assign caps
def number_ways(caps):
n = len(caps)
# Map each cap to the list of people who prefer it
cap_to_people = [[] for _ in range(101)]
for i in range(n):
for cap in caps[i]:
cap_to_people[cap].append(i)
# Initialize assigned_people list to track assigned caps
assigned_people = [False] * n
# Call the recursive function starting from the first cap
return dfs(0, assigned_people, 1, cap_to_people, n)
if __name__ == "__main__":
caps = [[1, 2, 3], [1, 2], [3, 4], [4, 5]]
print(number_ways(caps))
// C# Code to Assign Unique Cap To Every Person
// using Recursion
using System;
using System.Collections.Generic;
class GfG {
// Recursive function to calculate the number of ways
// to assign caps to people such that each person has
// a unique cap
static int Dfs(int assignedCount,
bool[] assignedPeople, int cap,
List<List<int>> capToPeople, int totalPeople) {
// Base case: if all people have a cap assigned, return 1
if (assignedCount == totalPeople) {
return 1;
}
// If we've considered all caps and not everyone
// has a cap, return 0
if (cap > 100) {
return 0;
}
// Case: skip the current cap
int ways = Dfs(assignedCount, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Assign the current cap to each person who likes it
foreach (int person in capToPeople[cap]) {
// Check if the person already has a cap assigned
if (!assignedPeople[person]) {
// Assign current cap to the person
assignedPeople[person] = true;
// Recurse with increased assigned count
ways += Dfs(assignedCount + 1, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Backtrack: unassign the cap for other
// possibilities
assignedPeople[person] = false;
}
}
return ways;
}
// Main function to calculate the number of ways
// to assign caps
static int NumberWays(List<List<int>> caps) {
int n = caps.Count;
// Map each cap to the list of people who prefer it
List<List<int>> capToPeople = new List<List<int>>();
for (int i = 0; i <= 100; i++) {
capToPeople.Add(new List<int>());
}
for (int i = 0; i < n; i++) {
foreach (int cap in caps[i]) {
capToPeople[cap].Add(i);
}
}
// Initialize assignedPeople array to
// track assigned caps
bool[] assignedPeople = new bool[n];
// Call the recursive function starting
// from the first cap
return Dfs(0, assignedPeople, 1, capToPeople, n);
}
static void Main() {
List<List<int>> caps = new List<List<int>> {
new List<int> {1, 2, 3},
new List<int> {1, 2},
new List<int> {3, 4},
new List<int> {4, 5}
};
Console.WriteLine(NumberWays(caps));
}
}
// Javascript Code to Assign Unique Cap To Every Person
// using Recursion
function dfs(assignedCount, assignedPeople, cap,
capToPeople, totalPeople) {
// Base case: if all people have a cap assigned, return 1
if (assignedCount === totalPeople) {
return 1;
}
// If we've considered all caps and not everyone
// has a cap, return 0
if (cap > 100) {
return 0;
}
// Case: skip the current cap
let ways = dfs(assignedCount, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Assign the current cap to each person who likes it
for (let person of capToPeople[cap]) {
// Check if the person already has a cap assigned
if (!assignedPeople[person]) {
// Assign current cap to the person
assignedPeople[person] = true;
// Recurse with increased assigned count
ways += dfs(assignedCount + 1,
assignedPeople, cap + 1, capToPeople, totalPeople);
// Backtrack: unassign the cap for other possibilities
assignedPeople[person] = false;
}
}
return ways;
}
// Main function to calculate the number of
// ways to assign caps
function numberWays(caps) {
const n = caps.length;
// Map each cap to the list of people who prefer it
const capToPeople = Array.from({ length: 101 }, () => []);
for (let i = 0; i < n; i++) {
for (let cap of caps[i]) {
capToPeople[cap].push(i);
}
}
// Initialize assignedPeople array to track assigned caps
const assignedPeople = new Array(n).fill(false);
// Call the recursive function starting from the first cap
return dfs(0, assignedPeople, 1, capToPeople, n);
}
const caps = [[1, 2, 3], [1, 2], [3, 4], [4, 5]];
console.log(numberWays(caps));
The above solution will have a exponential time complexity.
Using Top-Down DP (Memoization) - O(n * 2^n) Time and O(2^n) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming.
1. Optimal Substructure: The problem exhibits optimal substructure, meaning that the solution to the problem can be derived from the solutions of smaller subproblems.
The recursive relation is:
- ways = dfs(allMask, assignedPeople, cap+1) (skip current cap)
- ways = ways + dfs(allMask, assignedPeople ∣ (1 << person), cap+1) (assign current cap to a person)
2. Overlapping Subproblems:
Many subproblems are computed multiple times with the same parameters (cap, assignedPeople). To avoid recomputing the same subproblems, we store the result in a memoization table memo[cap][assignedPeople], which stores the number of ways to assign caps for a given cap and assignedPeople combination.
C++
// C++ Code to Assign Unique Cap To Every Person
// using Memoization and Bitmasking
#include <bits/stdc++.h>
using namespace std;
// Recursive function to count ways to assign
// caps with memoization
int dfs(int allMask, int assignedPeople, int cap,
vector<vector<int>>& capToPeople,
vector<vector<int>>& memo) {
// Base case: if all people have hats assigned
if (assignedPeople == allMask) {
return 1;
}
// If we've considered all caps and not everyone
// has a cap, return 0
if (cap > 100) {
return 0;
}
// Return memoized result if already computed
if (memo[cap][assignedPeople] != -1) {
return memo[cap][assignedPeople];
}
// Case: skip the current cap
int ways = dfs(allMask, assignedPeople,
cap + 1, capToPeople, memo);
// Try assigning the current cap to each person
// who can wear it
for (int person : capToPeople[cap]) {
// Check if the person hasn't been assigned a cap yet
if ((assignedPeople & (1 << person)) == 0) {
// Assign current cap to the person
ways = ways + dfs(allMask,
assignedPeople | (1 << person),
cap + 1, capToPeople, memo);
}
}
// Memoize and return the result
return memo[cap][assignedPeople] = ways;
}
// Main function to calculate the number of
// ways to assign caps
int numberWays(vector<vector<int>>& caps) {
int n = caps.size();
int allMask = (1 << n) - 1;
// Create adjacency matrix for
// cap-to-people distribution
vector<vector<int>> capToPeople(101);
for (int i = 0; i < n; ++i) {
for (int cap : caps[i]) {
capToPeople[cap].push_back(i);
}
}
// Memo array to store computed results
vector<vector<int>> memo(101, vector<int>(1 << n, -1));
// Call the recursive function starting
// from the first cap
return dfs(allMask, 0, 1, capToPeople, memo);
}
int main() {
vector<vector<int>> caps
= {{1, 2, 3}, {1, 2}, {3, 4}, {4, 5}};
cout << numberWays(caps) << endl;
return 0;
}
// Java Code to Assign Unique Cap To Every Person
// using Memoization and Bitmasking
import java.util.ArrayList;
import java.util.List;
class GfG {
// Recursive function to calculate the number of ways
// to assign caps to people such that each person has
// a unique cap
static int dfs(int assignedCount, ArrayList<Boolean> assignedPeople,
int cap, ArrayList<ArrayList<Integer>> capToPeople,
int totalPeople) {
// Base case: if all people have a cap assigned, return 1
if (assignedCount == totalPeople) {
return 1;
}
// If we've considered all caps and not everyone
// has a cap, return 0
if (cap > 100) {
return 0;
}
// Case: skip the current cap
int ways = dfs(assignedCount, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Assign the current cap to each person who likes it
for (int person : capToPeople.get(cap)) {
// Check if the person already has a cap assigned
if (!assignedPeople.get(person)) {
// Assign current cap to the person
assignedPeople.set(person, true);
// Recurse with increased assigned count
ways = ways + dfs(assignedCount + 1, assignedPeople,
cap + 1, capToPeople, totalPeople);
// Backtrack: unassign the cap for other possibilities
assignedPeople.set(person, false);
}
}
return ways;
}
// Main function to calculate the number of ways to assign caps
static int numberWays(ArrayList<ArrayList<Integer>> caps) {
int n = caps.size();
// Map each cap to the list of people who prefer it
ArrayList<ArrayList<Integer>> capToPeople = new ArrayList<>(101);
for (int i = 0; i <= 100; i++) {
capToPeople.add(new ArrayList<>());
}
for (int i = 0; i < n; ++i) {
for (int cap : caps.get(i)) {
capToPeople.get(cap).add(i);
}
}
// Initialize assignedPeople list to track assigned caps
ArrayList<Boolean> assignedPeople = new ArrayList<>(n);
for (int i = 0; i < n; i++) {
assignedPeople.add(false);
}
// Call the recursive function starting from the first cap
return dfs(0, assignedPeople, 1, capToPeople, n);
}
public static void main(String[] args) {
ArrayList<ArrayList<Integer>> caps2 = new ArrayList<>();
caps2.add(new ArrayList<>(List.of(1, 2, 3)));
caps2.add(new ArrayList<>(List.of(1, 2)));
caps2.add(new ArrayList<>(List.of(3, 4)));
caps2.add(new ArrayList<>(List.of(4, 5)));
System.out.println(numberWays(caps2));
}
}
# Python Code to Assign Unique Cap To Every Person
# using Memoization and Bitmasking
# Recursive function to count ways to assign
# caps with memoization
def dfs(all_mask, assigned_people, cap, cap_to_people, memo):
# Base case: if all people have hats assigned
if assigned_people == all_mask:
return 1
# If we've considered all caps and not everyone
# has a cap, return 0
if cap > 100:
return 0
# Return memoized result if already computed
if memo[cap][assigned_people] != -1:
return memo[cap][assigned_people]
# Case: skip the current cap
ways = dfs(all_mask, assigned_people, cap + 1, cap_to_people, memo)
# Try assigning the current cap to each person
# who can wear it
for person in cap_to_people[cap]:
# Check if the person hasn't been assigned a cap yet
if (assigned_people & (1 << person)) == 0:
# Assign current cap to the person
ways += dfs(all_mask, assigned_people | (1 << person),\
cap + 1, cap_to_people, memo)
# Memoize and return the result
memo[cap][assigned_people] = ways
return ways
# Main function to calculate the number of
# ways to assign caps
def number_ways(caps):
n = len(caps)
all_mask = (1 << n) - 1
# Create adjacency matrix for
# cap-to-people distribution
cap_to_people = [[] for _ in range(101)]
for i in range(n):
for cap in caps[i]:
cap_to_people[cap].append(i)
# Memo array to store computed results
memo = [[-1] * (1 << n) for _ in range(101)]
# Call the recursive function starting
# from the first cap
return dfs(all_mask, 0, 1, cap_to_people, memo)
if __name__ == "__main__":
caps = [[1, 2, 3], [1, 2], [3, 4], [4, 5]]
print(number_ways(caps))
// C# Code to Assign Unique Cap To Every Person
// using Memoization and Bitmasking
using System;
using System.Collections.Generic;
class GfG {
// Recursive function to count ways to assign
// caps with memoization
static int dfs(int allMask, int assignedPeople, int cap,
List<List<int>> capToPeople,
List<List<int>> memo) {
// Base case: if all people have hats assigned
if (assignedPeople == allMask) {
return 1;
}
// If we've considered all caps and not everyone
// has a cap, return 0
if (cap > 100) {
return 0;
}
// Return memoized result if already computed
if (memo[cap][assignedPeople] != -1) {
return memo[cap][assignedPeople];
}
// Case: skip the current cap
int ways = dfs(allMask, assignedPeople,
cap + 1, capToPeople, memo);
// Try assigning the current cap to each person
// who can wear it
foreach (int person in capToPeople[cap]) {
// Check if the person hasn't been assigned a cap yet
if ((assignedPeople & (1 << person)) == 0) {
// Assign current cap to the person
ways += dfs(allMask,
assignedPeople | (1 << person),
cap + 1, capToPeople, memo);
}
}
// Memoize and return the result
memo[cap][assignedPeople] = ways;
return ways;
}
// Main function to calculate the number of
// ways to assign caps
static int NumberWays(List<List<int>> caps) {
int n = caps.Count;
int allMask = (1 << n) - 1;
// Create adjacency matrix for
// cap-to-people distribution
List<List<int>> capToPeople
= new List<List<int>>(new List<int>[101]);
for (int i = 0; i < 101; i++) {
capToPeople[i] = new List<int>();
}
for (int i = 0; i < n; i++) {
foreach (int cap in caps[i]) {
capToPeople[cap].Add(i);
}
}
// Memo array to store computed results
List<List<int>> memo
= new List<List<int>>(new List<int>[101]);
for (int i = 0; i < 101; i++) {
memo[i] = new List<int>(new int[1 << n]);
for (int j = 0; j < (1 << n); j++) {
memo[i][j] = -1;
}
}
// Call the recursive function starting
// from the first cap
return dfs(allMask, 0, 1, capToPeople, memo);
}
static void Main(string[] args) {
List<List<int>> caps = new List<List<int>>() {
new List<int>{1, 2, 3},
new List<int>{1, 2},
new List<int>{3, 4},
new List<int>{4, 5}
};
Console.WriteLine(NumberWays(caps));
}
}
// Javascript Code to Assign Unique Cap To Every Person
// using Memoization and Bitmasking
// Recursive function to count ways to assign
// caps with memoization
function dfs(allMask, assignedPeople, cap, capToPeople, memo) {
// Base case: if all people have hats assigned
if (assignedPeople === allMask) {
return 1;
}
// If we've considered all caps and not everyone
// has a cap, return 0
if (cap > 100) {
return 0;
}
// Return memoized result if already computed
if (memo[cap][assignedPeople] !== -1) {
return memo[cap][assignedPeople];
}
// Case: skip the current cap
let ways = dfs(allMask, assignedPeople,
cap + 1, capToPeople, memo);
// Try assigning the current cap to each person
// who can wear it
for (let person of capToPeople[cap]) {
// Check if the person hasn't been assigned a cap yet
if ((assignedPeople & (1 << person)) === 0) {
// Assign current cap to the person
ways += dfs(allMask,
assignedPeople | (1 << person),
cap + 1, capToPeople, memo);
}
}
// Memoize and return the result
memo[cap][assignedPeople] = ways;
return ways;
}
// Main function to calculate the number of
// ways to assign caps
function numberWays(caps) {
let n = caps.length;
let allMask = (1 << n) - 1;
// Create adjacency matrix for
// cap-to-people distribution
let capToPeople = Array.from({ length: 101 }, () => []);
for (let i = 0; i < n; i++) {
for (let cap of caps[i]) {
capToPeople[cap].push(i);
}
}
// Memo array to store computed results
let memo = Array.from({ length: 101 }, () => Array(1 << n).fill(-1));
// Call the recursive function starting
// from the first cap
return dfs(allMask, 0, 1, capToPeople, memo);
}
let caps = [
[1, 2, 3],
[1, 2],
[3, 4],
[4, 5]
];
console.log(numberWays(caps));
Using Bottom-Up DP (Tabulation) – O(n * 2^n) Time and O(2^n) Space
We use a 2D DP table of size (number of caps + 1) * (2^n). The state dp[cap][assignedPeople] represents the number of ways to assign caps to people considering the first cap caps and assigning caps to the people represented by the bitmask assignedPeople.
Dynamic Programming Relation:
Base Case: dp[0][0] = 1, If no caps are assigned and no people are assigned any caps, there is 1 way to do nothing.
Skip the current cap:
- dp[cap][assignedPeople] += dp[cap-1][assignedPeople]
Assign the current cap to a person who hasn't been assigned a cap:
- dp[cap][assignedPeople | (1 << person)] += dp[cap - 1][assignedPeople]
After filling the DP table, the final result will be stored in dp[100][allMask], which represents the number of ways to assign all caps to all people.
C++
// C++ code to calculate the number of ways
// to assign caps to people using Tabulation
#include <bits/stdc++.h>
using namespace std;
int numberWays(vector<vector<int>>& caps) {
int n = caps.size();
int allMask = (1 << n) - 1;
// Create adjacency matrix for cap-to-people distribution
vector<vector<int>> capToPeople(101);
for (int i = 0; i < n; ++i) {
for (int cap : caps[i]) {
capToPeople[cap].push_back(i);
}
}
// DP table: dp[cap][assignedPeople]
// stores the number of ways
// to assign caps for the first 'cap'
// caps with 'assignedPeople' bitmask
vector<vector<int>> dp(102, vector<int>(1 << n, 0));
// Base case: With 0 caps, no people assigned,
// there's 1 way (do nothing)
dp[0][0] = 1;
// Fill the DP table
for (int cap = 1; cap <= 100; ++cap) {
for (int assignedPeople = 0;
assignedPeople <= allMask; ++assignedPeople) {
// If there are no ways to assign caps for
// this state, continue
if (dp[cap - 1][assignedPeople] == 0) {
continue;
}
// Case 1: Skip the current cap
dp[cap][assignedPeople] += dp[cap - 1][assignedPeople];
// Case 2: Assign current cap to each person who can wear it
for (int person : capToPeople[cap]) {
// If the person hasn't been assigned a cap yet
if ((assignedPeople & (1 << person)) == 0) {
dp[cap][assignedPeople | (1 << person)]
+= dp[cap - 1][assignedPeople];
}
}
}
}
// The result will be in dp[100][allMask],
// as we have considered all caps and assigned all people
return dp[100][allMask];
}
int main() {
vector<vector<int>> caps
= {{1, 2, 3}, {1, 2}, {3, 4}, {4, 5}};
cout << numberWays(caps) << endl;
return 0;
}
// Java code to calculate the number of ways
// to assign caps to people using Tabulation
import java.util.*;
class GfG {
// Method to calculate the number of ways
// to assign caps using Tabulation
static int numberWays(List<List<Integer>> caps) {
int n = caps.size();
int allMask = (1 << n) - 1;
// Create adjacency matrix for cap-to-people distribution
List<Integer>[] capToPeople = new ArrayList[101];
for (int i = 0; i < 101; i++) {
capToPeople[i] = new ArrayList<>();
}
// Fill the cap-to-people adjacency list
for (int i = 0; i < n; ++i) {
for (int cap : caps.get(i)) {
capToPeople[cap].add(i);
}
}
// DP table: dp[cap][assignedPeople]
// stores the number of ways to
// assign caps for the first 'cap' caps
// with 'assignedPeople' bitmask
int[][] dp = new int[102][1 << n];
// Base case: With 0 caps, no people assigned,
// there's 1 way (do nothing)
dp[0][0] = 1;
// Fill the DP table
for (int cap = 1; cap <= 100; ++cap) {
for (int assignedPeople = 0;
assignedPeople <= allMask; ++assignedPeople) {
// If there are no ways to assign caps for
// this state, continue
if (dp[cap - 1][assignedPeople] == 0) {
continue;
}
// Case 1: Skip the current cap
dp[cap][assignedPeople] += dp[cap - 1][assignedPeople];
// Case 2: Assign current cap to each person who can wear it
for (int person : capToPeople[cap]) {
// If the person hasn't been assigned a cap yet
if ((assignedPeople & (1 << person)) == 0) {
dp[cap][assignedPeople | (1 << person)]
+= dp[cap - 1][assignedPeople];
}
}
}
}
// The result will be in dp[100][allMask],
// as we have considered all caps and assigned all people
return dp[100][allMask];
}
public static void main(String[] args) {
List<List<Integer>> caps = new ArrayList<>();
caps.add(Arrays.asList(1, 2, 3));
caps.add(Arrays.asList(1, 2));
caps.add(Arrays.asList(3, 4));
caps.add(Arrays.asList(4, 5));
System.out.println(numberWays(caps));
}
}
# Python code to calculate the number of ways
# to assign caps to people using Tabulation
def numberWays(caps):
n = len(caps)
allMask = (1 << n) - 1
# Create adjacency list for cap-to-people
# distribution
capToPeople = [[] for _ in range(101)]
# Fill the cap-to-people adjacency list
for i in range(n):
for cap in caps[i]:
capToPeople[cap].append(i)
# DP table: dp[cap][assignedPeople]
# stores the number of ways to assign
# caps for the first 'cap' caps
# with 'assignedPeople' bitmask
dp = [[0] * (1 << n) for _ in range(102)]
# Base case: With 0 caps, no people assigned,
# there's 1 way (do nothing)
dp[0][0] = 1
# Fill the DP table
for cap in range(1, 101):
for assignedPeople in range(allMask + 1):
# If there are no ways to assign caps for
# this state, continue
if dp[cap - 1][assignedPeople] == 0:
continue
# Case 1: Skip the current cap
dp[cap][assignedPeople] += dp[cap - 1][assignedPeople]
# Case 2: Assign current cap to each person who can wear it
for person in capToPeople[cap]:
# If the person hasn't been assigned a cap yet
if (assignedPeople & (1 << person)) == 0:
dp[cap][assignedPeople | (1 << person)] \
+= dp[cap - 1][assignedPeople]
# The result will be in dp[100][allMask],
# as we have considered all caps and assigned all people
return dp[100][allMask]
if __name__ == "__main__":
caps = [[1, 2, 3], [1, 2], [3, 4], [4, 5]]
print(numberWays(caps))
// C# code to calculate the number of ways
// to assign caps to people using Tabulation
using System;
using System.Collections.Generic;
class GfG {
static int NumberWays(List<List<int>> caps) {
int n = caps.Count;
int allMask = (1 << n) - 1;
// Create adjacency list for cap-to-people
// distribution
List<List<int>> capToPeople
= new List<List<int>>(new List<int>[101]);
for (int i = 0; i < 101; ++i) {
capToPeople[i] = new List<int>();
}
// Fill the cap-to-people adjacency list
for (int i = 0; i < n; ++i) {
foreach (int cap in caps[i]) {
capToPeople[cap].Add(i);
}
}
// DP table: dp[cap][assignedPeople]
// stores the number of ways to assign
// caps for the first 'cap' caps
// with 'assignedPeople' bitmask
int[,] dp = new int[102, 1 << n];
// Base case: With 0 caps, no people assigned,
// there's 1 way (do nothing)
dp[0, 0] = 1;
// Fill the DP table
for (int cap = 1; cap <= 100; ++cap) {
for (int assignedPeople = 0;
assignedPeople <= allMask; ++assignedPeople) {
// If there are no ways to assign caps
// for this state, continue
if (dp[cap - 1, assignedPeople] == 0) {
continue;
}
// Case 1: Skip the current cap
dp[cap, assignedPeople] += dp[cap - 1, assignedPeople];
// Case 2: Assign current cap to each person who can wear it
foreach (int person in capToPeople[cap]) {
// If the person hasn't been assigned a cap yet
if ((assignedPeople & (1 << person)) == 0) {
dp[cap, assignedPeople | (1 << person)]
+= dp[cap - 1, assignedPeople];
}
}
}
}
// The result will be in dp[100, allMask],
// as we have considered all caps and assigned all people
return dp[100, allMask];
}
static void Main() {
List<List<int>> caps = new List<List<int>> {
new List<int> { 1, 2, 3 },
new List<int> { 1, 2 },
new List<int> { 3, 4 },
new List<int> { 4, 5 }
};
Console.WriteLine(NumberWays(caps));
}
}
// JavaScript code to calculate the number of ways
// to assign caps to people using Tabulation
function numberWays(caps) {
const n = caps.length;
const allMask = (1 << n) - 1;
// Create adjacency list for cap-to-people distribution
const capToPeople = Array.from({ length: 101 }, () => []);
// Fill the cap-to-people adjacency list
for (let i = 0; i < n; ++i) {
for (let cap of caps[i]) {
capToPeople[cap].push(i);
}
}
// DP table: dp[cap][assignedPeople]
// stores the number of ways to assign caps
// for the first 'cap' caps with 'assignedPeople' bitmask
const dp = Array.from({ length: 102 }, () => Array(1 << n).fill(0));
// Base case: With 0 caps, no people
// assigned, there's 1 way (do nothing)
dp[0][0] = 1;
// Fill the DP table
for (let cap = 1; cap <= 100; ++cap) {
for (let assignedPeople = 0;
assignedPeople <= allMask; ++assignedPeople) {
// If there are no ways to assign caps
// for this state, continue
if (dp[cap - 1][assignedPeople] === 0) {
continue;
}
// Case 1: Skip the current cap
dp[cap][assignedPeople] += dp[cap - 1][assignedPeople];
// Case 2: Assign current cap to each person who can wear it
for (let person of capToPeople[cap]) {
// If the person hasn't been assigned a cap yet
if ((assignedPeople & (1 << person)) === 0) {
dp[cap][assignedPeople | (1 << person)]
+= dp[cap - 1][assignedPeople];
}
}
}
}
// The result will be in dp[100][allMask],
// as we have considered all caps and assigned all people
return dp[100][allMask];
}
const caps = [
[1, 2, 3],
[1, 2],
[3, 4],
[4, 5]
];
console.log(numberWays(caps));
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