How do you solve Number of Ways to Wear Different Hats to Each Other on LeetCode?

Number of Ways to Wear Different Hats to Each Other (LeetCode #1434) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 40 * 2^n) time complexity and O(2^n) space complexity. Key insight: Using bitmasking allows us to efficiently track which people have been assigned hats.

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

The optimal solution for Number of Ways to Wear Different Hats to Each Other runs in O(n * 40 * 2^n) time and uses O(2^n) space. The time complexity is O(n * 40 * 2^n) because we iterate over each hat (up to 40), and for each hat, we update the DP array for all possible combinations of people (2^n).

What is the brute force solution for Number of Ways to Wear Different Hats to Each Other?

The brute force approach for Number of Ways to Wear Different Hats to Each Other is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we explore all possible combinations of hats for each person. This means we will try every possible assignment of hats to people, checking if each assignment is valid (i.e., no two people wear the same hat). The optimal Optimal Solution approach improves this to O(n * 40 * 2^n).

What are the interview tips for Number of Ways to Wear Different Hats to Each Other?

Clearly explain your thought process and how you arrived at the solution. Be prepared to discuss the trade-offs between different approaches. Practice implementing bitmasking and dynamic programming, as they are common in combinatorial problems.

What are common mistakes when solving Number of Ways to Wear Different Hats to Each Other?

Failing to consider all people when updating the DP state. Not properly handling the modulo operation, leading to overflow.

#1434

Number of Ways to Wear Different Hats to Each Other

Hard

ArrayDynamic ProgrammingBit ManipulationBitmaskDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 40 * 2^n)
Space	O(1)	O(2^n)

💡

Intuition

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

The optimal solution uses dynamic programming with bitmasking to efficiently count the number of ways to assign hats. By representing the state of which people have been assigned hats using a bitmask, we can avoid redundant calculations and significantly reduce the number of combinations we need to check.

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Algorithm

3 steps

1Step 1: Create a DP array where dp[mask] represents the number of ways to assign hats to the people represented by 'mask'.
2Step 2: Iterate through each hat type and update the DP array based on which people can wear that hat.
3Step 3: For each person that can wear the current hat, update the DP state by adding the number of ways from previous states.

solution.py16 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def countWays(hats):
5    n = len(hats)
6    dp = [0] * (1 << n)
7    dp[0] = 1
8    for hat in range(1, 41):
9        new_dp = dp[:]
10        for person in range(n):
11            if hat in hats[person]:
12                for mask in range(1 << n):
13                    if (mask & (1 << person)) == 0:
14                        new_dp[mask | (1 << person)] = (new_dp[mask | (1 << person)] + dp[mask]) % MOD
15        dp = new_dp
16    return dp[(1 << n) - 1]

ℹ

Complexity note: The time complexity is O(n * 40 * 2^n) because we iterate over each hat (up to 40), and for each hat, we update the DP array for all possible combinations of people (2^n).

1Using bitmasking allows us to efficiently track which people have been assigned hats.
2Dynamic programming helps avoid recalculating the same states multiple times.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.