A cinema has n rows of seats, numbered from 1 to n and there are ten seats in each row, labelled from 1 to 10 as shown in the figure above.
Given the array reservedSeats containing the numbers of seats already reserved, for example, reservedSeats[i] = [3,8] means the seat located in row 3 and labelled with 8 is already reserved.
Return the maximum number of four-person groups you can assign on the cinema seats. A four-person group occupies four adjacent seats in one single row. Seats across an aisle (such as [3,3] and [3,4]) are not considered to be adjacent, but there is an exceptional case on which an aisle split a four-person group, in that case, the aisle split a four-person group in the middle, which means to have two people on each side.
Example 1:
Input: n = 3, reservedSeats = [[1,2],[1,3],[1,8],[2,6],[3,1],[3,10]]
Output: 4
Explanation: The figure above shows the optimal allocation for four groups, where seats mark with blue are already reserved and contiguous seats mark with orange are for one group.
Example 2:
Input: n = 2, reservedSeats = [[2,1],[1,8],[2,6]]
Output: 2
Example 3:
Input: n = 4, reservedSeats = [[4,3],[1,4],[4,6],[1,7]]
Output: 4
Constraints:
1 <= n <= 10^9
1 <= reservedSeats.length <= min(10*n, 10^4)
reservedSeats[i].length == 2
1 <= reservedSeats[i][0] <= n
1 <= reservedSeats[i][1] <= 10
All reservedSeats[i] are distinct.
class Solution:
    def maxNumberOfFamilies(self, n: int, r: List[List[int]]) -> int:
        r=sorted(r)
        l=[0 for i in range(10)]
        ans=0
        n_occ=0
        r.append([0,0])
        for i in range(0,len(r)):
            if i==0:
                l[r[0][1]-1]=1
            else:
                if r[i][0]==r[i-1][0]:
                    l[r[i][1]-1]=1
                else:
                    n_occ+=1
                    if (l[1]==0)&(l[2]==0)&(l[3]==0)&(l[4]==0):
                        ans+=1
                        if (l[5]==0)&(l[6]==0)&(l[7]==0)&(l[8]==0):
                            ans+=1
                    elif (l[5]==0)&(l[6]==0)&(l[7]==0)&(l[8]==0):
                        ans+=1
                    elif (l[3]==0)&(l[4]==0)&(l[5]==0)&(l[6]==0):
                        ans+=1
                    #print(r[i],l)
                    l=[0 for i in range(10)]   
                    l[r[i][1]-1]=1
        ans+=(n-n_occ)*2
        return ans
class Solution:
    def maxNumberOfFamilies(self, n: int, r: List[List[int]]) -> int:
        r=sorted(r)
        l=[0 for i in range(10)]
        ans=0
        n_occ=0
        r.append([0,0])
        for i in range(0,len(r)):
            if i==0:
                l[r[0][1]-1]=1
            else:
                if r[i][0]==r[i-1][0]:
                    l[r[i][1]-1]=1
                else:
                    n_occ+=1
                    if (l[1]==0)&(l[2]==0)&(l[3]==0)&(l[4]==0):
                        ans+=1
                        if (l[5]==0)&(l[6]==0)&(l[7]==0)&(l[8]==0):
                            ans+=1
                    elif (l[5]==0)&(l[6]==0)&(l[7]==0)&(l[8]==0):
                        ans+=1
                    elif (l[3]==0)&(l[4]==0)&(l[5]==0)&(l[6]==0):
                        ans+=1
                    #print(r[i],l)
                    l=[0 for i in range(10)]   
                    l[r[i][1]-1]=1
        ans+=(n-n_occ)*2
        return ans