Q1. Design Parking System
Design a parking system for a parking lot. The parking lot has three kinds of parking spaces: big, medium, and small, with a fixed number of slots for each size.
Implement the ParkingSystem class:
ParkingSystem(int big, int medium, int small)Initializes object of theParkingSystemclass. The number of slots for each parking space are given as part of the constructor.bool addCar(int carType)Checks whether there is a parking space ofcarTypefor the car that wants to get into the parking lot.carTypecan be of three kinds: big, medium, or small, which are represented by1,2, and3respectively. A car can only park in a parking space of itscarType. If there is no space available, returnfalse, else park the car in that size space and returntrue.
Example 1:
Input
["ParkingSystem", "addCar", "addCar", "addCar", "addCar"]
[[1, 1, 0], [1], [2], [3], [1]]
Output
[null, true, true, false, false]Explanation
ParkingSystem parkingSystem = new ParkingSystem(1, 1, 0);
parkingSystem.addCar(1); // return true because there is 1 available slot for a big car
parkingSystem.addCar(2); // return true because there is 1 available slot for a medium car
parkingSystem.addCar(3); // return false because there is no available slot for a small car
parkingSystem.addCar(1); // return false because there is no available slot for a big car. It is already occupied.
Solution:
The question is straightforward, in ParkingSystem constructor, we store values into separate integer variables. Then, in addCar method, we determine each integer variable corresponding to carType is available or not, if there are still available spots, then return true, else false.
Q2. Alert Using Same Key-Card Three or More Times in a One Hour Period
Leetcode company workers use key-cards to unlock office doors. Each time a worker uses their key-card, the security system saves the worker’s name and the time when it was used. The system emits an alert if any worker uses the key-card three or more times in a one-hour period.
You are given a list of strings keyName and keyTime where [keyName[i], keyTime[i]] corresponds to a person's name and the time when their key-card was used in a single day.
Access times are given in the 24-hour time format “HH:MM”, such as "23:51" and "09:49".
Return a list of unique worker names who received an alert for frequent keycard use. Sort the names in ascending order alphabetically.
Notice that "10:00" - "11:00" is considered to be within a one-hour period, while "23:51" - "00:10" is not considered to be within a one-hour period.
Example 1:
Input: keyName = ["daniel","daniel","daniel","luis","luis","luis","luis"], keyTime = ["10:00","10:40","11:00","09:00","11:00","13:00","15:00"]
Output: ["daniel"]
Explanation: "daniel" used the keycard 3 times in a one-hour period ("10:00","10:40", "11:00").Example 2:
Input: keyName = ["alice","alice","alice","bob","bob","bob","bob"], keyTime = ["12:01","12:00","18:00","21:00","21:20","21:30","23:00"]
Output: ["bob"]
Explanation: "bob" used the keycard 3 times in a one-hour period ("21:00","21:20", "21:30").Example 3:
Input: keyName = ["john","john","john"], keyTime = ["23:58","23:59","00:01"]
Output: []Example 4:
Input: keyName = ["leslie","leslie","leslie","clare","clare","clare","clare"], keyTime = ["13:00","13:20","14:00","18:00","18:51","19:30","19:49"]
Output: ["clare","leslie"]Solution:
Q3. Find Valid Matrix Given Row and Column Sums
You are given two arrays rowSum and colSum of non-negative integers where rowSum[i] is the sum of the elements in the ith row and colSum[j] is the sum of the elements of the jth column of a 2D matrix. In other words, you do not know the elements of the matrix, but you do know the sums of each row and column.
Find any matrix of non-negative integers of size rowSum.length x colSum.length that satisfies the rowSum and colSum requirements.
Return a 2D array representing any matrix that fulfills the requirements. It’s guaranteed that at least one matrix that fulfills the requirements exists.
Example 1:
Input: rowSum = [3,8], colSum = [4,7]
Output: [[3,0],
[1,7]]
Explanation:
0th row: 3 + 0 = 0 == rowSum[0]
1st row: 1 + 7 = 8 == rowSum[1]
0th column: 3 + 1 = 4 == colSum[0]
1st column: 0 + 7 = 7 == colSum[1]
The row and column sums match, and all matrix elements are non-negative.
Another possible matrix is: [[1,2],
[3,5]]Example 2:
Input: rowSum = [5,7,10], colSum = [8,6,8]
Output: [[0,5,0],
[6,1,0],
[2,0,8]]Example 3:
Input: rowSum = [14,9], colSum = [6,9,8]
Output: [[0,9,5],
[6,0,3]]Example 4:
Input: rowSum = [1,0], colSum = [1]
Output: [[1],
[0]]Example 5:
Input: rowSum = [0], colSum = [0]
Output: [[0]]Solution:
Time complexity: O(rowSum.length * colSum.length)
Space complexity: O (rowSum.length * colSum.length)
Implement greedy search by Math.min(rowSum[i], colSum[j])
Update both rowSum & colSum arrays by subtracting value of arr indexes.
Based on the question provided, it concludes that
rowSum == colSum == 0