Leetcode #1434: Number of Ways to Wear Different Hats to Each Other
In this guide, we solve Leetcode #1434 Number of Ways to Wear Different Hats to Each Other in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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 ith person.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Array, Dynamic Programming, Bitmask
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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 5.
Python Solution
class Solution:
def numberWays(self, hats: List[List[int]]) -> int:
g = defaultdict(list)
for i, h in enumerate(hats):
for v in h:
g[v].append(i)
mod = 10**9 + 7
n = len(hats)
m = max(max(h) for h in hats)
f = [[0] * (1 << n) for _ in range(m + 1)]
f[0][0] = 1
for i in range(1, m + 1):
for j in range(1 << n):
f[i][j] = f[i - 1][j]
for k in g[i]:
if j >> k & 1:
f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod
return f[m][-1]
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.