Posts tagged as “hashtable”

花花酱 LeetCode 1797. Design Authentication Manager

By zxi on March 20, 2021

There is an authentication system that works with authentication tokens. For each session, the user will receive a new authentication token that will expire timeToLive seconds after the currentTime. If the token is renewed, the expiry time will be extended to expire timeToLive seconds after the (potentially different) currentTime.

Implement the AuthenticationManager class:

AuthenticationManager(int timeToLive) constructs the AuthenticationManager and sets the timeToLive.
generate(string tokenId, int currentTime) generates a new token with the given tokenId at the given currentTime in seconds.
renew(string tokenId, int currentTime) renews the unexpired token with the given tokenId at the given currentTime in seconds. If there are no unexpired tokens with the given tokenId, the request is ignored, and nothing happens.
countUnexpiredTokens(int currentTime) returns the number of unexpired tokens at the given currentTime.

Note that if a token expires at time t, and another action happens on time t (renew or countUnexpiredTokens), the expiration takes place before the other actions.

Example 1:

Input
["AuthenticationManager", "renew", "generate", "countUnexpiredTokens", "generate", "renew", "renew", "countUnexpiredTokens"]
[[5], ["aaa", 1], ["aaa", 2], [6], ["bbb", 7], ["aaa", 8], ["bbb", 10], [15]]
Output
[null, null, null, 1, null, null, null, 0]

Explanation AuthenticationManager authenticationManager = new AuthenticationManager(5); // Constructs the AuthenticationManager with timeToLive = 5 seconds. authenticationManager.renew(“aaa”, 1); // No token exists with tokenId “aaa” at time 1, so nothing happens. authenticationManager.generate(“aaa”, 2); // Generates a new token with tokenId “aaa” at time 2. authenticationManager.countUnexpiredTokens(6); // The token with tokenId “aaa” is the only unexpired one at time 6, so return 1. authenticationManager.generate(“bbb”, 7); // Generates a new token with tokenId “bbb” at time 7. authenticationManager.renew(“aaa”, 8); // The token with tokenId “aaa” expired at time 7, and 8 >= 7, so at time 8 the renew request is ignored, and nothing happens. authenticationManager.renew(“bbb”, 10); // The token with tokenId “bbb” is unexpired at time 10, so the renew request is fulfilled and now the token will expire at time 15. authenticationManager.countUnexpiredTokens(15); // The token with tokenId “bbb” expires at time 15, and the token with tokenId “aaa” expired at time 7, so currently no token is unexpired, so return 0.

Constraints:

1 <= timeToLive <= 10⁸
1 <= currentTime <= 10⁸
1 <= tokenId.length <= 5
tokenId consists only of lowercase letters.
All calls to generate will contain unique values of tokenId.
The values of currentTime across all the function calls will be strictly increasing.
At most 2000 calls will be made to all functions combined.

Solution: Hashtable

Use a hashtable to store the token and its expiration time.

Time complexity: at most O(n) per operation
Space complexity: O(n)

C++

// Author: Huahua, 160 ms, 30.1 MB

class AuthenticationManager {

public:

AuthenticationManager(int timeToLive) : ttl_(timeToLive) {}

void generate(string tokenId, int currentTime) {

clear(currentTime);

tokens_[tokenId] = currentTime + ttl_;

}

void renew(string tokenId, int currentTime) {

clear(currentTime);

if (!tokens_.count(tokenId)) return;

tokens_[tokenId] = currentTime + ttl_;

}

int countUnexpiredTokens(int currentTime) {

clear(currentTime);

return tokens_.size();

}

private:

void clear(int currentTime) {

vector<string> ids;

for (const auto& [id, t] : tokens_)

if (t <= currentTime) ids.push_back(id);

for (const string& id : ids)

tokens_.erase(id);

}

unordered_map<string, int> tokens_;

int ttl_;

};

/**

* Your AuthenticationManager object will be instantiated and called as such:

* AuthenticationManager* obj = new AuthenticationManager(timeToLive);

* obj->generate(tokenId,currentTime);

* obj->renew(tokenId,currentTime);

* int param_3 = obj->countUnexpiredTokens(currentTime);

花花酱 LeetCode 1796. Second Largest Digit in a String

By zxi on March 20, 2021

Given an alphanumeric string s, return the second largest numerical digit that appears in s, or -1 if it does not exist.

An alphanumericstring is a string consisting of lowercase English letters and digits.

Example 1:

Input: s = "dfa12321afd"
Output: 2
Explanation: The digits that appear in s are [1, 2, 3]. The second largest digit is 2.

Example 2:

Input: s = "abc1111"
Output: -1
Explanation: The digits that appear in s are [1]. There is no second largest digit.

Constraints:

1 <= s.length <= 500
s consists of only lowercase English letters and/or digits.

Solution: Hashtable

Time complexity: O(n)
Space complexity: O(10)

C++

// Author: Huahua

class Solution {

public:

int secondHighest(string s) {

vector<int> d(10);

for (char c : s)

if (c >= '0' && c <= '9')

d[c - '0'] = 1;

int order = 0;

for (int i = 9; i >= 0; --i)

if (d[i] && ++order == 2)

return i;

return -1;

}

};

花花酱 LeetCode 1791. Find Center of Star Graph

By zxi on March 13, 2021

There is an undirected star graph consisting of n nodes labeled from 1 to n. A star graph is a graph where there is one center node and exactly n - 1 edges that connect the center node with every other node.

You are given a 2D integer array edges where each edges[i] = [u_i, v_i] indicates that there is an edge between the nodes u_i and v_i. Return the center of the given star graph.

Example 1:

Input: edges = [[1,2],[2,3],[4,2]]
Output: 2
Explanation: As shown in the figure above, node 2 is connected to every other node, so 2 is the center.

Example 2:

Input: edges = [[1,2],[5,1],[1,3],[1,4]]
Output: 1

Constraints:

3 <= n <= 10⁵
edges.length == n - 1
edges[i].length == 2
1 <= u_i, v_i <= n
u_i != v_i
The given edges represent a valid star graph.

Solution: Graph / Hashtable

Count the degree of each node, return the one with n-1 degrees.

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

C++

// Author: Huahua

class Solution {

public:

int findCenter(vector<vector<int>>& edges) {

unordered_map<int, int> degrees;

for (const auto& e : edges) {

++degrees[e[0]];

++degrees[e[1]];

}

const int n = degrees.size();

for (const auto& [id, d] : degrees)

if (d == n - 1) return id;

return -1;

}

};

Since the center node must appear in each edge, we just need to find the mode of edges[0] + edges[1]

Time complexity: O(1)
Space complexity: O(1)

Python

# Author: Huahua

class Solution:

def findCenter(self, e):

return mode(e[0] + e[1])

花花酱 LeetCode 1748. Sum of Unique Elements

By zxi on February 6, 2021

You are given an integer array nums. The unique elements of an array are the elements that appear exactly once in the array.

Return the sum of all the unique elements of nums.

Example 1:

Input: nums = [1,2,3,2]
Output: 4
Explanation: The unique elements are [1,3], and the sum is 4.

Example 2:

Input: nums = [1,1,1,1,1]
Output: 0
Explanation: There are no unique elements, and the sum is 0.

Example 3:

Input: nums = [1,2,3,4,5]
Output: 15
Explanation: The unique elements are [1,2,3,4,5], and the sum is 15.

Constraints:

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

Solution: Hashtable

Time complexity: O(n)
Space complexity: O(100)

C++

// Author: Huahua

class Solution {

public:

int sumOfUnique(vector<int>& nums) {

vector<int> seen(101);

int ans = 0;

for (int x : nums)

++seen[x];

for (int x : nums)

if (seen[x] == 1) ans += x;

return ans;

}

};

花花酱 LeetCode 1743. Restore the Array From Adjacent Pairs

By zxi on January 30, 2021

There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums.

You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i] = [u_i, v_i] indicates that the elements u_i and v_i are adjacent in nums.

It is guaranteed that every adjacent pair of elements nums[i] and nums[i+1] will exist in adjacentPairs, either as [nums[i], nums[i+1]] or [nums[i+1], nums[i]]. The pairs can appear in any order.

Return the original array nums. If there are multiple solutions, return any of them.

Example 1:

Input: adjacentPairs = [[2,1],[3,4],[3,2]]
Output: [1,2,3,4]
Explanation: This array has all its adjacent pairs in adjacentPairs.
Notice that adjacentPairs[i] may not be in left-to-right order.

Example 2:

Input: adjacentPairs = [[4,-2],[1,4],[-3,1]]
Output: [-2,4,1,-3]
Explanation: There can be negative numbers.
Another solution is [-3,1,4,-2], which would also be accepted.

Example 3:

Input: adjacentPairs = [[100000,-100000]]
Output: [100000,-100000]

Constraints:

nums.length == n
adjacentPairs.length == n - 1
adjacentPairs[i].length == 2
2 <= n <= 10⁵
-10⁵ <= nums[i], u_i, v_i <= 10⁵
There exists some nums that has adjacentPairs as its pairs.

Solution: Hashtable

Reverse thinking! For a given input array, e.g.
[1, 2, 3, 4, 5]
it’s adjacent pairs are [1,2] , [2,3], [3,4], [4,5]
all numbers appeared exactly twice except 1 and 5, since they are on the boundary.
We just need to find the head or tail of the input array, and construct the rest of the array in order.

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

C++

// Author: Huahua

class Solution {

public:

vector<int> restoreArray(vector<vector<int>>& adjacentPairs) {

const int n = adjacentPairs.size() + 1;

unordered_map<int, vector<int>> g;

for (const auto& p : adjacentPairs) {

g[p[0]].push_back(p[1]);

g[p[1]].push_back(p[0]);

}

vector<int> ans(n);

for (const auto& [u, vs] : g)

if (vs.size() == 1) {

ans[0] = u;

ans[1] = vs[0];

break;

}

for (int i = 2; i < n; ++i) {

const auto& vs = g[ans[i - 1]];

ans[i] = vs[0] == ans[i - 2] ? vs[1] : vs[0];

}

return ans;

}

};