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Huahua's Tech Road

花花酱 LeetCode 1394. Find Lucky Integer in an Array

By zxi on March 28, 2020

Given an array of integers arr, a lucky integer is an integer which has a frequency in the array equal to its value.

Return a lucky integer in the array. If there are multiple lucky integers return the largest of them. If there is no lucky integer return -1.

Example 1:

Input: arr = [2,2,3,4]
Output: 2
Explanation: The only lucky number in the array is 2 because frequency[2] == 2.

Example 2:

Input: arr = [1,2,2,3,3,3]
Output: 3
Explanation: 1, 2 and 3 are all lucky numbers, return the largest of them.

Example 3:

Input: arr = [2,2,2,3,3]
Output: -1
Explanation: There are no lucky numbers in the array.

Example 4:

Input: arr = [5]
Output: -1

Example 5:

Input: arr = [7,7,7,7,7,7,7]
Output: 7

Constraints:

  • 1 <= arr.length <= 500
  • 1 <= arr[i] <= 500

Solution: Hashtable

Time complexity: O(n)
Space complexity: O(n)

C++

花花酱 LeetCode 1396. Design Underground System

By zxi on March 28, 2020

Implement the class UndergroundSystem that supports three methods:

1. checkIn(int id, string stationName, int t)

  • A customer with id card equal to id, gets in the station stationName at time t.
  • A customer can only be checked into one place at a time.

2. checkOut(int id, string stationName, int t)

  • A customer with id card equal to id, gets out from the station stationName at time t.

3. getAverageTime(string startStation, string endStation) 

  • Returns the average time to travel between the startStation and the endStation.
  • The average time is computed from all the previous traveling from startStation to endStation that happened directly.
  • Call to getAverageTime is always valid.

You can assume all calls to checkIn and checkOut methods are consistent. That is, if a customer gets in at time t1 at some station, then it gets out at time t2 with t2 > t1. All events happen in chronological order.

Example 1:

Input
["UndergroundSystem","checkIn","checkIn","checkIn","checkOut","checkOut","checkOut","getAverageTime","getAverageTime","checkIn","getAverageTime","checkOut","getAverageTime"]
[[],[45,"Leyton",3],[32,"Paradise",8],[27,"Leyton",10],[45,"Waterloo",15],[27,"Waterloo",20],[32,"Cambridge",22],["Paradise","Cambridge"],["Leyton","Waterloo"],[10,"Leyton",24],["Leyton","Waterloo"],[10,"Waterloo",38],["Leyton","Waterloo"]]

Output
[null,null,null,null,null,null,null,14.0,11.0,null,11.0,null,12.0]

Explanation
UndergroundSystem undergroundSystem = new UndergroundSystem();
undergroundSystem.checkIn(45, "Leyton", 3);
undergroundSystem.checkIn(32, "Paradise", 8);
undergroundSystem.checkIn(27, "Leyton", 10);
undergroundSystem.checkOut(45, "Waterloo", 15);
undergroundSystem.checkOut(27, "Waterloo", 20);
undergroundSystem.checkOut(32, "Cambridge", 22);
undergroundSystem.getAverageTime("Paradise", "Cambridge");       // return 14.0. There was only one travel from "Paradise" (at time 8) to "Cambridge" (at time 22)
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.0. There were two travels from "Leyton" to "Waterloo", a customer with id=45 from time=3 to time=15 and a customer with id=27 from time=10 to time=20. So the average time is ( (15-3) + (20-10) ) / 2 = 11.0
undergroundSystem.checkIn(10, "Leyton", 24);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 11.0
undergroundSystem.checkOut(10, "Waterloo", 38);
undergroundSystem.getAverageTime("Leyton", "Waterloo");          // return 12.0

Constraints:

  • There will be at most 20000 operations.
  • 1 <= id, t <= 10^6
  • All strings consist of uppercase, lowercase English letters and digits.
  • 1 <= stationName.length <= 10
  • Answers within 10^-5 of the actual value will be accepted as correct.

Solution: Hashtable

For each user, store the checkin station and time.
For each trip (startStation + “_” + endStation), store the total time and counts.

Time complexity: O(n)
Space complexity: O(n)

C++

花花酱 LeetCode 1390. Four Divisors

By zxi on March 22, 2020

Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors.

If there is no such integer in the array, return 0.

Example 1:

Input: nums = [21,4,7]
Output: 32
Explanation:
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.

Constraints:

  • 1 <= nums.length <= 10^4
  • 1 <= nums[i] <= 10^5

Solution: Math

If a number is a perfect square (e.g. 9 = 3 * 3), it will have odd number of divisors. (9: 1, 3, 9).

Time complexity: O(sum(sqrt(num_i))
Space complexity: O(1)

C++

花花酱 LeetCode 1389. Create Target Array in the Given Order

By zxi on March 22, 2020

Given two arrays of integers nums and index. Your task is to create target array under the following rules:

  • Initially target array is empty.
  • From left to right read nums[i] and index[i], insert at index index[i] the value nums[i] in target array.
  • Repeat the previous step until there are no elements to read in nums and index.

Return the target array.

It is guaranteed that the insertion operations will be valid.

Example 1:

Input: nums = [0,1,2,3,4], index = [0,1,2,2,1]
Output: [0,4,1,3,2]
Explanation:
nums       index     target
0            0        [0]
1            1        [0,1]
2            2        [0,1,2]
3            2        [0,1,3,2]
4            1        [0,4,1,3,2]

Example 2:

Input: nums = [1,2,3,4,0], index = [0,1,2,3,0]
Output: [0,1,2,3,4]
Explanation:
nums       index     target
1            0        [1]
2            1        [1,2]
3            2        [1,2,3]
4            3        [1,2,3,4]
0            0        [0,1,2,3,4]

Example 3:

Input: nums = [1], index = [0]
Output: [1]

Constraints:

  • 1 <= nums.length, index.length <= 100
  • nums.length == index.length
  • 0 <= nums[i] <= 100
  • 0 <= index[i] <= i

Solution: Simulation

Time complexity: O(n) ~ O(n^2)
Space complexity: O(n)

C++

花花酱 LeetCode 1386. Cinema Seat Allocation

By zxi on March 22, 2020

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.

Solution: HashTable + Greedy

if both seat[2~5] seat[6~9] are empty, seat two groups.
if any of seat[2~5] seat[4~7] seat[6~9] is empty seat one group.
if there is no one sit in a row, seat two groups.

Time complexity: O(|reservedSeats|)
Space complexity: O(|rows|)

C++