Number of Ways to Wear Different Hats to Each Other
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.
/**
* The first idea come to our mind is that we will do straightforward like the
* problem said: Return the number of ways that the n people wear different hats
* to each other. We will assign n people with different hats and use dp to
* calculate number of ways.
*/
// But that will result in TLE
class Solution {
int mod = 1_000_000_007;
public int numberWays(List<List<Integer>> hats) {
int n = hats.size();
// h2p[i] indicates the list of people who can wear i_th hat
List<Integer>[] h2p = new List[41];
for (int i = 1; i <= 40; i++)
h2p[i] = new ArrayList<>();
for (int i = 0; i < n; i++)
for (int hat : hats.get(i))
h2p[hat].add(i);
// max value of full mask can be 2^n -> 1024
Integer[][] dp = new Integer[41][1024];
// Using a bitmask where each bit shows whether the ith perseon has been
// assigned or not
return dfs(h2p, (1 << n) - 1, 1, 0, dp);
// we are using 2 masks here
// 1st-> (1<<n)-1 -> this shows the final mask configuration we want to reach
// 2nd-> 0 -> this shows the current mask configuration
}
// dp[i][j] -> ways to assign ith hat to mask value of j
// Remember that (i+1)th hat will be assigned after the ith hat
int dfs(List<Integer>[] h2p, int allMask, int hat, int assignedPeople, Integer[][] dp) {
if (assignedPeople == allMask)
return 1; // Check if assigned different hats to all people
if (hat > 40)
return 0; // no more hats to process
if (dp[hat][assignedPeople] != null)
return dp[hat][assignedPeople];
int ans = dfs(h2p, allMask, hat + 1, assignedPeople, dp); // Don't wear this hat
for (int p : h2p[hat]) {
// Skip if person `p` was assigned hat
if (((assignedPeople >> p) & 1) == 1)
continue;
// Wear this hat for p_th person
ans += dfs(h2p, allMask, hat + 1, assignedPeople | (1 << p), dp);
ans %= mod;
}
dp[hat][assignedPeople] = ans;
return ans;
}
}