refactor 1386

fishercoder1534 · fishercoder1534 · commit 871a67830691 · 2020-11-12T06:37:40.000-08:00
diff --git a/src/main/java/com/fishercoder/solutions/_1386.java b/src/main/java/com/fishercoder/solutions/_1386.java
@@ -5,35 +5,6 @@
 import java.util.Map;
 import java.util.Set;
 
-/**
- * 1386. Cinema Seat Allocation
- *
- * 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 families you can allocate on the cinema seats. A four-person family occupies fours seats in one row, that are next to each other.
- * Seats across an aisle (such as [3,3] and [3,4]) are not considered to be next to each other, however, It is permissible for the four-person family to be separated by an aisle, but in that case, exactly two people have to sit on each side of the aisle.
- *
- * 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 families, where seats mark with blue are already reserved and contiguous seats mark with orange are for one family.
- *
- * 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.
- * */
 public class _1386 {
     public static class Solution1 {
         public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
