What is the time complexity of Number of Ways to Wear Different Hats to Each Other?

The optimal solution has a time complexity of O(40 * 2^n * n).

What is the space complexity of Number of Ways to Wear Different Hats to Each Other?

The space complexity is O(2^n).

What data structures are used to solve Number of Ways to Wear Different Hats to Each Other?

This problem uses Array, Dynamic Programming, Bit Manipulation, Bitmask. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Ways to Wear Different Hats to Each Other?

Insight 1: Dynamic programming is well-suited since we're counting combinations and can break the problem into subproblems. We can iterate through hats and, for each hat, decide whether each person who likes that hat will wear it or not. Insight 2: A bitmask is efficient to represent which people have already been assigned a hat. Each bit in the mask corresponds to a person, and if the bit is set, it means that person has already been assigned a hat.

Which companies ask LeetCode 1434. Number of Ways to Wear Different Hats to Each Other in interviews?

This problem is frequently asked at Mindtickle.

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Number of Ways to Wear Different Hats to Each Other - LeetCode 1434 Solution

Number of Ways to Wear Different Hats to Each Other - Complete Solution Guide

Number of Ways to Wear Different Hats to Each Other is LeetCode problem 1434, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There are n people and 40 types of hats labeled from 1 to 40 . Given a 2D integer array hats , where hats[i] is a list of all hats preferred by the i th person. Return the number of ways that n people can wear different hats from each other. Since the answer may be too large, return it modulo 10 9 + 7 . Example 1: Input: hats = [[3,4],[4,5],[5]] Output: 1 Explanation: There is only one way to choose hats given the conditions. First person choose hat 3, Second person choose hat 4 and last one hat

Detailed Explanation

The problem requires us to find the number of ways `n` people can wear different hats, given a list of hats each person prefers. We're given a 2D array `hats` where `hats[i]` is a list of hats the i-th person likes. Each person must wear a *different* hat. We need to return the number of possible arrangements modulo 10^9 + 7 because the number of ways can be very large. There are `n` people and 40 different hat types, labeled 1 to 40.

Solution Approach

The solution uses dynamic programming with bitmasking. `dp[mask]` stores the number of ways to assign hats to the people represented by the set bits in `mask`. The algorithm iterates through each hat from 1 to 40. For each hat, it iterates through all possible masks, representing states where some people have already been assigned hats. Then, for each person who likes the current hat, the algorithm checks if that person has not already been assigned a hat (i.e., the corresponding bit in the mask is not set). If they haven't been assigned, the algorithm updates the current state by adding the number of ways to reach the previous state (where the person hadn't been assigned a hat).

Step-by-Step Algorithm

Step 1: Create a `hat_to_people` mapping (list of lists). For each hat `h`, `hat_to_people[h]` contains a list of people who like hat `h`.
Step 2: Initialize a DP array `dp` of size `2^n`. `dp[0] = 1` because there is one way to assign hats to no one (the initial state).
Step 3: Iterate through the hats from 1 to 40. For each hat `h`:
Step 4: Iterate through all possible masks from `2^n - 1` down to 0 (reverse order to avoid overwriting values needed in the current iteration). For each mask:
Step 5: Iterate through the people who like hat `h` (from `hat_to_people[h]`). For each person `p`:
Step 6: Check if person `p` has already been assigned a hat (i.e., the `p`-th bit in `mask` is set). If it *is* set:
Step 7: Calculate the previous mask `prev_mask` by unsetting the `p`-th bit in `mask`.
Step 8: Update `dp[mask]` by adding `dp[prev_mask]` (modulo 10^9 + 7). `dp[mask] = (dp[mask] + dp[prev_mask]) % MOD`
Step 9: After iterating through all hats and masks, `dp[(1 << n) - 1]` contains the total number of ways to assign hats to all `n` people.

Key Insights

Insight 1: Dynamic programming is well-suited since we're counting combinations and can break the problem into subproblems. We can iterate through hats and, for each hat, decide whether each person who likes that hat will wear it or not.
Insight 2: A bitmask is efficient to represent which people have already been assigned a hat. Each bit in the mask corresponds to a person, and if the bit is set, it means that person has already been assigned a hat.
Insight 3: Preprocessing the input into a `hat_to_people` mapping greatly simplifies the iteration. Instead of iterating through people for each hat, we iterate through people *who like* a particular hat.

Complexity Analysis

Time Complexity: O(40 * 2^n * n)

Space Complexity: O(2^n)

Topics

This problem involves: Array, Dynamic Programming, Bit Manipulation, Bitmask.

Companies

Asked at: Mindtickle.

Number of Ways to Wear Different Hats to Each Other | LeetCode 1434

Hard

Array Dynamic Programming Bit Manipulation Bitmask

Time: O(40 * 2^n * n)Space: O(2^n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int numberWays(List<List<Integer>> hats) {
        int n = hats.size();
        int MOD = (int)1e9 + 7;

        List<List<Integer>> hatToPeople = new ArrayList<>();
        for (int i = 0; i < 41; i++) {
            hatToPeople.add(new ArrayList<>());
        }

        for (int p = 0; p < n; p++) {
            for (int h : hats.get(p)) {
                hatToPeople.get(h).add(p);
            }
        }

        int[] dp = new int[1 << n];
        dp[0] = 1;

        for (int h = 1; h < 41; h++) {
            for (int mask = (1 << n) - 1; mask >= 0; mask--) {
                for (int p : hatToPeople.get(h)) {
                    if ((mask >> p) % 2 == 1) {
                        int prevMask = mask ^ (1 << p);
                        dp[mask] = (dp[mask] + dp[prevMask]) % MOD;
                    }
                }
            }
        }

        return dp[(1 << n) - 1];
    }
}

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Problem Info

Description

Problem 1434: Number of Ways to Wear Different Hats to Each Other - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Bit Manipulation Bitmask

Companies

Mindtickle

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1434

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Create a `hat_to_people` mapping (list of lists). For each hat `h`, `hat_to_people[h]` contains a list of people who like hat `h`.
2
Step 2: Initialize a DP array `dp` of size `2^n`. `dp[0] = 1` because there is one way to assign hats to no one (the initial state).
3
Step 3: Iterate through the hats from 1 to 40. For each hat `h`:
4
Step 4: Iterate through all possible masks from `2^n - 1` down to 0 (reverse order to avoid overwriting values needed in the current iteration). For each mask:
5
Step 5: Iterate through the people who like hat `h` (from `hat_to_people[h]`). For each person `p`:
6
Step 6: Check if person `p` has already been assigned a hat (i.e., the `p`-th bit in `mask` is set). If it *is* set:
7
Step 7: Calculate the previous mask `prev_mask` by unsetting the `p`-th bit in `mask`.
8
Step 8: Update `dp[mask]` by adding `dp[prev_mask]` (modulo 10^9 + 7). `dp[mask] = (dp[mask] + dp[prev_mask]) % MOD`
9
Step 9: After iterating through all hats and masks, `dp[(1 << n) - 1]` contains the total number of ways to assign hats to all `n` people.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(40 * 2^n * n)

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity stems from the three nested loops. The outer loop iterates through the hats (up to 40), the middle loop considers all possible assignments of hats to people using the bitmask representation (2^n possibilities), and the inner loop considers all possible people who can wear a particular hat (at most n). The space complexity is mainly due to the DP table (`dp`), which requires space proportional to the number of possible mask states (2^n). The `hat_to_people` vector takes O(40n) space, which is less than O(2^n) when n is small (<= 10).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

Solve the problem recursively. The base case would be when all people have been assigned hats. Use memoization to store the results of subproblems, preventing redundant calculations.

Time: O(40 * 2^n * n)Space: O(2^n)

Trade-offs: Recursion might be more intuitive for some, but it might lead to stack overflow errors for larger inputs. Iterative DP is generally more efficient because it avoids recursion overhead.

✨

Key Insights

1Insight 1: Dynamic programming is well-suited since we're counting combinations and can break the problem into subproblems. We can iterate through hats and, for each hat, decide whether each person who likes that hat will wear it or not.
2Insight 2: A bitmask is efficient to represent which people have already been assigned a hat. Each bit in the mask corresponds to a person, and if the bit is set, it means that person has already been assigned a hat.
3Insight 3: Preprocessing the input into a `hat_to_people` mapping greatly simplifies the iteration. Instead of iterating through people for each hat, we iterate through people *who like* a particular hat.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty hats list for a person. The algorithm correctly handles this because if `hats[p]` is empty, then `hat_to_people[h]` will not include `p` for any `h`, so person `p` will never be assigned a hat, and the result will be 0 if it's impossible. The code doesn't explicitly check for this case, but it implicitly handles it correctly.
Edge case 2: Only one person. If there is only one person, the `mask` will have only one bit. The algorithm iterates and sums up the number of hats that person likes.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo correctly. The number of ways can be very large, so it is crucial to take the modulo at each step to prevent overflow.

✗Common mistake 2: Incorrect bit manipulation. Getting the bitwise operations wrong can lead to incorrect mask states and thus incorrect results. Ensure the bit shifts and XOR operations are applied correctly to represent person assignments.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Assignment problems where each task (hat) can be performed by certain agents (people), and each agent must be assigned a different task. This comes up in scheduling and resource allocation.

🏦Real-world scenario 2: Configuration management. Finding the number of ways to configure a system where each component can have different settings and certain settings are incompatible across components. The hats represent the configuration choices, and the people represent the system components.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Number of Ways to Wear Different Hats to Each Other"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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