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
// OJ: https://leetcode.com/problems/number-of-ways-to-wear-different-hats-to-each-other/
// Author: github.com/lzl124631x
// 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];
}
};