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];
}
};
Java
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];
}
}