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Formatted question description: https://leetcode.ca/all/1434.html

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 list of list of integers hats, where hats[i] is a list of all hats preferred by the i-th 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 10^9 + 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.

Example 4:

Input: hats = [[1,2,3],[2,3,5,6],[1,3,7,9],[1,8,9],[2,5,7]]
Output: 111

 

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.

Related Topics:
Dynamic Programming, Bit Manipulation

Solution 1.

  • class Solution {
    public int numberWays(List<List<Integer>> hats) {
        final int MODULO = 1000000007;
        int peopleCount = hats.size();
        int hatsCount = 40;
        Map<Integer, List<Integer>> map = new HashMap<Integer, List<Integer>>();
        for (int i = 0; i < peopleCount; i++) {
            List<Integer> curHats = hats.get(i);
            for (int hat : curHats) {
                List<Integer> list = map.getOrDefault(hat, new ArrayList<Integer>());
                list.add(i);
                map.put(hat, list);
            }
        }
        int[][] dp = new int[hatsCount + 1][1 << peopleCount];
        dp[0][0] = 1;
        for (int i = 1; i <= hatsCount; i++) {
            for (int j = 0; j < 1 << peopleCount; j++)
                dp[i][j] = dp[i - 1][j];
            if (map.containsKey(i)) {
                List<Integer> list = map.get(i);
                for (int j = 0; j < 1 << peopleCount; j++) {
                    for (int k : list) {
                        if ((j & (1 << k)) == 1 << k)
                            dp[i][j] = (dp[i][j] + dp[i - 1][j - (1 << k)]) % MODULO;
                    }
                }
            }
        }
        return dp[hatsCount][(1 << peopleCount) - 1];
    }
}

############

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];
    }
}

  • // OJ: https://leetcode.com/problems/number-of-ways-to-wear-different-hats-to-each-other/
// Time: O(2^N * M)
// Space: O(2^N)
class Solution {
public:
    int numberWays(vector<vector<int>>& hats) {
        vector<vector<int>> persons(40);
        int N = hats.size(), mod = 1e9+7;
        vector<int> masks(1 << N);
        masks[0] = 1;
        for (int i = 0; i < N; ++i) {
            for (int h : hats[i]) persons[h - 1].push_back(i);
        }
        for (int i = 0; i < 40; ++i) {
            for (int j = (1 << N) - 1; j >= 0; --j) {
                for (int p : persons[i]) {
                    if (j & (1 << p)) continue;
                    masks[j | (1 << p)] += masks[j];
                    masks[j | (1 << p)] %= mod;
                }
            }
        }
        return masks[(1 << N) - 1];
    }
};

  • 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]



  • 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
}

  • function numberWays(hats: number[][]): number {
    const n = hats.length;
    const m = Math.max(...hats.flat());
    const g: number[][] = Array.from({ length: m + 1 }, () => []);
    for (let i = 0; i < n; ++i) {
        for (const v of hats[i]) {
            g[v].push(i);
        }
    }
    const f: number[][] = Array.from({ length: m + 1 }, () =>
        Array.from({ length: 1 << n }, () => 0),
    );
    f[0][0] = 1;
    const mod = 1e9 + 7;
    for (let i = 1; i <= m; ++i) {
        for (let j = 0; j < 1 << n; ++j) {
            f[i][j] = f[i - 1][j];
            for (const k of g[i]) {
                if (((j >> k) & 1) === 1) {
                    f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod;
                }
            }
        }
    }
    return f[m][(1 << n) - 1];
}

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