README.md

1434. Number of Ways to Wear Different Hats to Each Other

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.

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 109 + 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 5.

Example 2:

Input: hats = [[3,5,1],[3,5]]
Output: 4
Explanation: There are 4 ways to choose hats:
(3,5), (5,3), (1,3) and (1,5)

Example 3:

Input: hats = [[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]]
Output: 24
Explanation: Each person can choose hats labeled from 1 to 4.
Number of Permutations of (1,2,3,4) = 24.

Constraints:

• n == hats.length
• 1 <= n <= 10
• 1 <= hats[i].length <= 40
• 1 <= hats[i][j] <= 40
• hats[i] contains a list of unique integers.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn number_ways(hats: Vec<Vec<i32>>) -> i32 {
        let n = hats.len();
        let mut dp = vec![vec![0; 1 << n]; 41];
        dp[0][0] = 1;

        for i in 1..=40 {
            dp[i] = dp[i - 1].clone();

            for j in 0..n {
                if !hats[j].contains(&(i as i32)) {
                    continue;
                }

                for k in 0..(1 << n) {
                    if (k >> j) & 1 == 0 {
                        dp[i][k | (1 << j)] = (dp[i][k | (1 << j)] + dp[i - 1][k]) % 1_000_000_007;
                    }
                }
            }
        }

        dp[40][(1 << n) - 1]
    }
}