Posts tagged as “hashtable”

花花酱 LeetCode 1636. Sort Array by Increasing Frequency

By zxi on October 31, 2020

Given an array of integers nums, sort the array in increasing order based on the frequency of the values. If multiple values have the same frequency, sort them in decreasing order.

Return the sorted array.

Example 1:

Input: nums = [1,1,2,2,2,3]
Output: [3,1,1,2,2,2]
Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 3.

Example 2:

Input: nums = [2,3,1,3,2]
Output: [1,3,3,2,2]
Explanation: '2' and '3' both have a frequency of 2, so they are sorted in decreasing order.

Example 3:

Input: nums = [-1,1,-6,4,5,-6,1,4,1]
Output: [5,-1,4,4,-6,-6,1,1,1]

Constraints:

1 <= nums.length <= 100
-100 <= nums[i] <= 100

Solution: Hashtable + Sorting

Use a hashtable to track the frequency of each number.

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

C++

class Solution {

public:

vector<int> frequencySort(vector<int>& nums) {

unordered_map<int, int> freq;

for (int x : nums) ++freq[x];

sort(begin(nums), end(nums), [&](int a, int b) {

if (freq[a] != freq[b]) return freq[a] < freq[b];

return a > b;

});

return nums;

}

};

花花酱 LeetCode 1624. Largest Substring Between Two Equal Characters

By zxi on October 18, 2020

Given a string s, return the length of the longest substring between two equal characters, excluding the two characters. If there is no such substring return -1.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "aa"
Output: 0
Explanation: The optimal substring here is an empty substring between the two 'a's.

Example 2:

Input: s = "abca"
Output: 2
Explanation: The optimal substring here is "bc".

Example 3:

Input: s = "cbzxy"
Output: -1
Explanation: There are no characters that appear twice in s.

Example 4:

Input: s = "cabbac"
Output: 4
Explanation: The optimal substring here is "abba". Other non-optimal substrings include "bb" and "".

Constraints:

1 <= s.length <= 300
s contains only lowercase English letters.

Solution: Hashtable

Remember the first position each letter occurs.

Time complexity: O(n)
Space complexity: O(26)

C++

class Solution {

public:

int maxLengthBetweenEqualCharacters(string s) {

vector<int> first(26, -1);

int ans = -1;

for (int i = 0; i < s.length(); ++i) {

int& p = first[s[i] - 'a'];

if (p != -1) {

ans = max(ans, i - p - 1);

} else {

p = i;

}

return ans;

}

};

花花酱 LeetCode 1604. Alert Using Same Key-Card Three or More Times in a One Hour Period

By zxi on October 3, 2020

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"]

Constraints:

1 <= keyName.length, keyTime.length <= 10⁵
keyName.length == keyTime.length
keyTime are in the format “HH:MM”.
[keyName[i], keyTime[i]] is unique.
1 <= keyName[i].length <= 10
keyName[i] contains only lowercase English letters.

Solution: Hashtable + sorting

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

C++

class Solution {

public:

vector<string> alertNames(vector<string>& keyName, vector<string>& keyTime) {

unordered_map<string, vector<int>> t; // {name -> time}

for (size_t i = 0; i < keyName.size(); ++i) {

const int h = stoi(keyTime[i].substr(0, 2));

const int m = stoi(keyTime[i].substr(3));

t[keyName[i]].push_back(h * 60 + m);

}

vector<string> ans;

for (auto& [name, times] : t) {

sort(begin(times), end(times));

for (size_t i = 2; i < times.size(); ++i)

if (times[i] - times[i - 2] <= 60) {

ans.push_back(name);

break;

}

sort(begin(ans), end(ans));

return ans;

}

};

花花酱 LeetCode 1603. Design Parking System

By zxi on October 3, 2020

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 the ParkingSystem class. 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 of carType for the car that wants to get into the parking lot. carType can be of three kinds: big, medium, or small, which are represented by 1, 2, and 3 respectively. A car can only park in a parking space of its carType. If there is no space available, return false, else park the car in that size space and return true.

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.

Constraints:

0 <= big, medium, small <= 1000
carType is 1, 2, or 3
At most 1000 calls will be made to addCar

Solution: Simulation

Time complexity: O(1) per addCar call
Space complexity: O(1)

C++

// Author: Huahua

class ParkingSystem {

public:

ParkingSystem(int big, int medium, int small):

slots_{{big, medium, small}} {}

bool addCar(int carType) {

return slots_[carType - 1]-- > 0;

}

private:

vector<int> slots_;

};

Python3

# Author: Huahua

class ParkingSystem:

def __init__(self, big: int, medium: int, small: int):

self.slots = {1: big, 2: medium, 3: small}

def addCar(self, carType: int) -> bool:

if self.slots[carType] == 0: return False

self.slots[carType] -= 1

return True

花花酱 LeetCode 1600. Throne Inheritance

By zxi on September 27, 2020

A kingdom consists of a king, his children, his grandchildren, and so on. Every once in a while, someone in the family dies or a child is born.

The kingdom has a well-defined order of inheritance that consists of the king as the first member. Let’s define the recursive function Successor(x, curOrder), which given a person x and the inheritance order so far, returns who should be the next person after x in the order of inheritance.

Successor(x, curOrder):

if x has no children or all of x's children are in curOrder:

if x is the king return null

else return Successor(x's parent, curOrder)

else return x's oldest child who's not in curOrder

For example, assume we have a kingdom that consists of the king, his children Alice and Bob (Alice is older than Bob), and finally Alice’s son Jack.

In the beginning, curOrder will be ["king"].
Calling Successor(king, curOrder) will return Alice, so we append to curOrder to get ["king", "Alice"].
Calling Successor(Alice, curOrder) will return Jack, so we append to curOrder to get ["king", "Alice", "Jack"].
Calling Successor(Jack, curOrder) will return Bob, so we append to curOrder to get ["king", "Alice", "Jack", "Bob"].
Calling Successor(Bob, curOrder) will return null. Thus the order of inheritance will be ["king", "Alice", "Jack", "Bob"].

Using the above function, we can always obtain a unique order of inheritance.

Implement the ThroneInheritance class:

ThroneInheritance(string kingName) Initializes an object of the ThroneInheritance class. The name of the king is given as part of the constructor.
void birth(string parentName, string childName) Indicates that parentName gave birth to childName.
void death(string name) Indicates the death of name. The death of the person doesn’t affect the Successor function nor the current inheritance order. You can treat it as just marking the person as dead.
string[] getInheritanceOrder() Returns a list representing the current order of inheritance excluding dead people.

Example 1:

Input
["ThroneInheritance", "birth", "birth", "birth", "birth", "birth", "birth", "getInheritanceOrder", "death", "getInheritanceOrder"]
[["king"], ["king", "andy"], ["king", "bob"], ["king", "catherine"], ["andy", "matthew"], ["bob", "alex"], ["bob", "asha"], [null], ["bob"], [null]]
Output
[null, null, null, null, null, null, null, ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"], null, ["king", "andy", "matthew", "alex", "asha", "catherine"]]
Explanation
ThroneInheritance t= new ThroneInheritance("king"); // order: king
t.birth("king", "andy"); // order: king > andy
t.birth("king", "bob"); // order: king > andy > bob
t.birth("king", "catherine"); // order: king > andy > bob > catherine
t.birth("andy", "matthew"); // order: king > andy > matthew > bob > catherine
t.birth("bob", "alex"); // order: king > andy > matthew > bob > alex > catherine
t.birth("bob", "asha"); // order: king > andy > matthew > bob > alex > asha > catherine
t.getInheritanceOrder(); // return ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"]
t.death("bob"); // order: king > andy > matthew > bob > alex > asha > catherine
t.getInheritanceOrder(); // return ["king", "andy", "matthew", "alex", "asha", "catherine"]

Constraints:

1 <= kingName.length, parentName.length, childName.length, name.length <= 15
kingName, parentName, childName, and name consist of lowercase English letters only.
All arguments childName and kingName are distinct.
All name arguments of death will be passed to either the constructor or as childName to birth first.
For each call to birth(parentName, childName), it is guaranteed that parentName is alive.
At most 10⁵ calls will be made to birth and death.
At most 10 calls will be made to getInheritanceOrder.

Solution: HashTable + DFS

Record :
1. mapping from parent to children (ordered)
2. who has dead

Time complexity: getInheritanceOrder O(n), other O(1)
Space complexity: O(n)

C++

// Author: Huahua

class ThroneInheritance {

public:

ThroneInheritance(string kingName): king_(kingName) {

m_[kingName] = {};

}

void birth(string parentName, string childName) {

m_[parentName].push_back(childName);

}

void death(string name) {

dead_.insert(name);

}

vector<string> getInheritanceOrder() {

vector<string> ans;

function<void(const string&)> dfs = [&](const string& name) {

if (!dead_.count(name)) ans.push_back(name);

for (const string& child: m_[name]) dfs(child);

};

dfs(king_);

return ans;

}

private:

string king_;

unordered_map<string, vector<string>> m_; // parent -> list[children]

unordered_set<string> dead_;

};

Python3

# Author: Huahua

class ThroneInheritance:

def __init__(self, kingName: str):

self.kingName = kingName

self.family = defaultdict(list)

self.dead = set()

def birth(self, parentName: str, childName: str) -> None:

self.family[parentName].append(childName)

def death(self, name: str) -> None:

self.dead.add(name)

def getInheritanceOrder(self) -> List[str]:

order = []

def dfs(name: str):

if name not in self.dead: order.append(name)

for child in self.family[name]: dfs(child)

dfs(self.kingName)

return order