README_EN.md

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

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Description

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.

Solutions

Python3

class Solution:
    def numberWays(self, hats: List[List[int]]) -> int:
        d = defaultdict(list)
        for i, h in enumerate(hats):
            for v in h:
                d[v].append(i)
        n = len(hats)
        mx = max(max(h) for h in hats)
        dp = [[0] * (1 << n) for _ in range(mx + 1)]
        dp[0][0] = 1
        mod = int(1e9) + 7
        for i in range(1, mx + 1):
            for mask in range(1 << n):
                dp[i][mask] = dp[i - 1][mask]
                for j in d[i]:
                    if (mask >> j) & 1:
                        dp[i][mask] += dp[i - 1][mask ^ (1 << j)]
                dp[i][mask] %= mod
        return dp[mx][(1 << n) - 1]

Java

class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int numberWays(List<List<Integer>> hats) {
        List<Integer>[] d = new List[41];
        Arrays.setAll(d, k -> new ArrayList<>());
        int n = hats.size();
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            for (int h : hats.get(i)) {
                d[h].add(i);
                mx = Math.max(mx, h);
            }
        }
        long[][] dp = new long[mx + 1][1 << n];
        dp[0][0] = 1;
        for (int i = 1; i < mx + 1; ++i) {
            for (int mask = 0; mask < 1 << n; ++mask) {
                dp[i][mask] = dp[i - 1][mask];
                for (int j : d[i]) {
                    if (((mask >> j) & 1) == 1) {
                        dp[i][mask] = (dp[i][mask] + dp[i - 1][mask ^ (1 << j)]) % MOD;
                    }
                }
            }
        }
        return (int) dp[mx][(1 << n) - 1];
    }
}

C++

using ll = long long;

class Solution {
public:
    int numberWays(vector<vector<int>>& hats) {
        vector<vector<int>> d(41);
        int n = hats.size();
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            for (int& h : hats[i]) {
                d[h].push_back(i);
                mx = max(mx, h);
            }
        }
        vector<vector<ll>> dp(mx + 1, vector<ll>(1 << n));
        dp[0][0] = 1;
        int mod = 1e9 + 7;
        for (int i = 1; i <= mx; ++i) {
            for (int mask = 0; mask < 1 << n; ++mask) {
                dp[i][mask] = dp[i - 1][mask];
                for (int& j : d[i]) {
                    if ((mask >> j) & 1) {
                        dp[i][mask] = (dp[i][mask] + dp[i - 1][mask ^ (1 << j)]) % mod;
                    }
                }
            }
        }
        return dp[mx][(1 << n) - 1];
    }
};

Go

func numberWays(hats [][]int) int {
	d := make([][]int, 41)
	mx := 0
	for i, h := range hats {
		for _, v := range h {
			d[v] = append(d[v], i)
			mx = max(mx, v)
		}
	}
	dp := make([][]int, mx+1)
	n := len(hats)
	for i := range dp {
		dp[i] = make([]int, 1<<n)
	}
	dp[0][0] = 1
	mod := int(1e9) + 7
	for i := 1; i <= mx; i++ {
		for mask := 0; mask < 1<<n; mask++ {
			dp[i][mask] = dp[i-1][mask]
			for _, j := range d[i] {
				if ((mask >> j) & 1) == 1 {
					dp[i][mask] = (dp[i][mask] + dp[i-1][mask^(1<<j)]) % mod
				}
			}
		}
	}
	return dp[mx][(1<<n)-1]
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

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