Given a m * n matrix seats that represent seats distributions in a classroom. If a seat is broken, it is denoted by '#' character otherwise it is denoted by a '.' character.
Students can see the answers of those sitting next to the left, right, upper left and upper right, but he cannot see the answers of the student sitting directly in front or behind him. Return the maximum number of students that can take the exam together without any cheating being possible..
Students must be placed in seats in good condition.
Example 1:
Input: seats = [["#",".","#","#",".","#"], [".","#","#","#","#","."], ["#",".","#","#",".","#"]] Output: 4 Explanation: Teacher can place 4 students in available seats so they don't cheat on the exam.
Example 2:
Input: seats = [[".","#"], ["#","#"], ["#","."], ["#","#"], [".","#"]] Output: 3 Explanation: Place all students in available seats.
Example 3:
Input: seats = [["#",".",".",".","#"], [".","#",".","#","."], [".",".","#",".","."], [".","#",".","#","."], ["#",".",".",".","#"]] Output: 10 Explanation: Place students in available seats in column 1, 3 and 5.
Constraints:
seatscontains only characters'.' and'#'.m == seats.lengthn == seats[i].length1 <= m <= 81 <= n <= 8
class Solution:
def maxStudents(self, seats: List[List[str]]) -> int:
def f(seat: List[str]) -> int:
mask = 0
for i, c in enumerate(seat):
if c == '.':
mask |= 1 << i
return mask
@cache
def dfs(seat: int, i: int) -> int:
ans = 0
for mask in range(1 << n):
if (seat | mask) != seat or (mask & (mask << 1)):
continue
cnt = mask.bit_count()
if i == len(ss) - 1:
ans = max(ans, cnt)
else:
nxt = ss[i + 1]
nxt &= ~(mask << 1)
nxt &= ~(mask >> 1)
ans = max(ans, cnt + dfs(nxt, i + 1))
return ans
n = len(seats[0])
ss = [f(s) for s in seats]
return dfs(ss[0], 0)class Solution {
private Integer[][] f;
private int n;
private int[] ss;
public int maxStudents(char[][] seats) {
int m = seats.length;
n = seats[0].length;
ss = new int[m];
f = new Integer[1 << n][m];
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (seats[i][j] == '.') {
ss[i] |= 1 << j;
}
}
}
return dfs(ss[0], 0);
}
private int dfs(int seat, int i) {
if (f[seat][i] != null) {
return f[seat][i];
}
int ans = 0;
for (int mask = 0; mask < 1 << n; ++mask) {
if ((seat | mask) != seat || (mask & (mask << 1)) != 0) {
continue;
}
int cnt = Integer.bitCount(mask);
if (i == ss.length - 1) {
ans = Math.max(ans, cnt);
} else {
int nxt = ss[i + 1];
nxt &= ~(mask << 1);
nxt &= ~(mask >> 1);
ans = Math.max(ans, cnt + dfs(nxt, i + 1));
}
}
return f[seat][i] = ans;
}
}class Solution {
public:
int maxStudents(vector<vector<char>>& seats) {
int m = seats.size();
int n = seats[0].size();
vector<int> ss(m);
vector<vector<int>> f(1 << n, vector<int>(m, -1));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (seats[i][j] == '.') {
ss[i] |= 1 << j;
}
}
}
function<int(int, int)> dfs = [&](int seat, int i) -> int {
if (f[seat][i] != -1) {
return f[seat][i];
}
int ans = 0;
for (int mask = 0; mask < 1 << n; ++mask) {
if ((seat | mask) != seat || (mask & (mask << 1)) != 0) {
continue;
}
int cnt = __builtin_popcount(mask);
if (i == m - 1) {
ans = max(ans, cnt);
} else {
int nxt = ss[i + 1];
nxt &= ~(mask >> 1);
nxt &= ~(mask << 1);
ans = max(ans, cnt + dfs(nxt, i + 1));
}
}
return f[seat][i] = ans;
};
return dfs(ss[0], 0);
}
};func maxStudents(seats [][]byte) int {
m, n := len(seats), len(seats[0])
ss := make([]int, m)
f := make([][]int, 1<<n)
for i, seat := range seats {
for j, c := range seat {
if c == '.' {
ss[i] |= 1 << j
}
}
}
for i := range f {
f[i] = make([]int, m)
for j := range f[i] {
f[i][j] = -1
}
}
var dfs func(int, int) int
dfs = func(seat, i int) int {
if f[seat][i] != -1 {
return f[seat][i]
}
ans := 0
for mask := 0; mask < 1<<n; mask++ {
if (seat|mask) != seat || (mask&(mask<<1)) != 0 {
continue
}
cnt := bits.OnesCount(uint(mask))
if i == m-1 {
ans = max(ans, cnt)
} else {
nxt := ss[i+1] & ^(mask >> 1) & ^(mask << 1)
ans = max(ans, cnt+dfs(nxt, i+1))
}
}
f[seat][i] = ans
return ans
}
return dfs(ss[0], 0)
}
func max(a, b int) int {
if a > b {
return a
}
return b
}