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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.length1 <= n <= 101 <= hats[i].length <= 401 <= hats[i][j] <= 40hats[i]contains a list of unique integers.
Related Topics:
Dynamic Programming, Bit Manipulation
Solution 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 { 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 }