LeetCode in Kotlin

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

Hard

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 the n people wear different hats to 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.

Solution

class Solution {
    fun numberWays(hats: List<List<Int>>): Int {
        val mod = 1000000007L
        val size = hats.size
        val possible = Array(size) { BooleanArray(41) }
        for (i in 0 until size) {
            for (j in hats[i]) {
                possible[i][j] = true
            }
        }
        val dp = LongArray(1 shl size)
        dp[0] = 1
        for (i in 1..40) {
            for (j in dp.size - 1 downTo 1) {
                for (k in 0 until size) {
                    if (j shr k and 1 == 1 && possible[k][i]) {
                        dp[j] += dp[j xor (1 shl k)]
                    }
                }
                dp[j] %= mod
            }
        }
        return dp[(1 shl size) - 1].toInt()
    }
}

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