[アルゴリズム]LeetCode-Cinema Seat Allocation
LeetCode - Cinema Seat Allocation
問題の説明
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.
I/O例
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
せいげんじょうけん
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.
Solution
[戦略]
1.配列を行、列の順に並べる、
2.各保留駅の間に4人が座れるかどうかを判断するimport java.util.*;
class Solution {
public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
Arrays.sort(reservedSeats, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
int rowNum = 1;
int lastOccupied = 0;
int count = 0;
for (int i = 0; i < reservedSeats.length; i++) {
if (reservedSeats[i][0] > rowNum) { // row 바뀜
int diffRow = reservedSeats[i][0] - rowNum;
// 한줄이 전체 비었을때 최대 2 그룹이 앉을 수 있음
if (diffRow > 1) {
count += (diffRow - 1) * 2;
}
// 이전 줄의 마지막 자리가 어디냐에 따라 달라짐
if (lastOccupied < 2) {
count += 2;
} else if (lastOccupied < 6) {
count++;
}
rowNum = reservedSeats[i][0];
lastOccupied = 0;
}
if (reservedSeats[i][1] >= 6 && reservedSeats[i][1] <= 7) {
if (lastOccupied < 2) {
count++;
}
} else if (reservedSeats[i][1] >= 8 && reservedSeats[i][1] <= 9) {
if (lastOccupied < 4) {
count++;
}
} else if (reservedSeats[i][1] == 10) {
if (lastOccupied < 2) {
count += 2;
} else if (lastOccupied < 6) {
count++;
}
}
lastOccupied = reservedSeats[i][1];
}
if (lastOccupied < 2) {
count += 2;
} else if (lastOccupied < 6) {
count++;
}
if (rowNum < n) {
count += (n - rowNum) * 2;
}
return count;
}
}
Reference
この問題について([アルゴリズム]LeetCode-Cinema Seat Allocation), 我々は、より多くの情報をここで見つけました
https://velog.io/@jerry92/알고리즘-LeetCode-Cinema-Seat-Allocation
テキストは自由に共有またはコピーできます。ただし、このドキュメントのURLは参考URLとして残しておいてください。
Collection and Share based on the CC Protocol
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.
import java.util.*;
class Solution {
public int maxNumberOfFamilies(int n, int[][] reservedSeats) {
Arrays.sort(reservedSeats, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
int rowNum = 1;
int lastOccupied = 0;
int count = 0;
for (int i = 0; i < reservedSeats.length; i++) {
if (reservedSeats[i][0] > rowNum) { // row 바뀜
int diffRow = reservedSeats[i][0] - rowNum;
// 한줄이 전체 비었을때 최대 2 그룹이 앉을 수 있음
if (diffRow > 1) {
count += (diffRow - 1) * 2;
}
// 이전 줄의 마지막 자리가 어디냐에 따라 달라짐
if (lastOccupied < 2) {
count += 2;
} else if (lastOccupied < 6) {
count++;
}
rowNum = reservedSeats[i][0];
lastOccupied = 0;
}
if (reservedSeats[i][1] >= 6 && reservedSeats[i][1] <= 7) {
if (lastOccupied < 2) {
count++;
}
} else if (reservedSeats[i][1] >= 8 && reservedSeats[i][1] <= 9) {
if (lastOccupied < 4) {
count++;
}
} else if (reservedSeats[i][1] == 10) {
if (lastOccupied < 2) {
count += 2;
} else if (lastOccupied < 6) {
count++;
}
}
lastOccupied = reservedSeats[i][1];
}
if (lastOccupied < 2) {
count += 2;
} else if (lastOccupied < 6) {
count++;
}
if (rowNum < n) {
count += (n - rowNum) * 2;
}
return count;
}
}
Reference
この問題について([アルゴリズム]LeetCode-Cinema Seat Allocation), 我々は、より多くの情報をここで見つけました https://velog.io/@jerry92/알고리즘-LeetCode-Cinema-Seat-Allocationテキストは自由に共有またはコピーできます。ただし、このドキュメントのURLは参考URLとして残しておいてください。
Collection and Share based on the CC Protocol