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花花酱 LeetCode 721. Accounts Merge

Given a list accounts, each element accounts[i] is a list of strings, where the first element accounts[i][0] is a name, and the rest of the elements are emails representing emails of the account.

Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email that is common to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.

After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails in sorted order. The accounts themselves can be returned in any order.

Example 1:

Input: 
accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
Explanation: 
The first and third John's are the same person as they have the common email "johnsmith@mail.com".
The second John and Mary are different people as none of their email addresses are used by other accounts.
We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.

Note:The length of accounts will be in the range [1, 1000].The length of accounts[i] will be in the range [1, 10].The length of accounts[i][j] will be in the range [1, 30].

Solution: Union-Find

C++

花花酱 LeetCode 1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance

There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities fromi and toi, and given the integer distanceThreshold.

Return the city with the smallest numberof cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number.

Notice that the distance of a path connecting cities i and j is equal to the sum of the edges’ weights along that path.

Example 1:

Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
Output: 3
Explanation: The figure above describes the graph. 
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2] 
City 1 -> [City 0, City 2, City 3] 
City 2 -> [City 0, City 1, City 3] 
City 3 -> [City 1, City 2] 
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.

Example 2:

Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
Output: 0
Explanation: The figure above describes the graph. 
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1] 
City 1 -> [City 0, City 4] 
City 2 -> [City 3, City 4] 
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3] 
The city 0 has 1 neighboring city at a distanceThreshold = 2.

Constraints:

  • 2 <= n <= 100
  • 1 <= edges.length <= n * (n - 1) / 2
  • edges[i].length == 3
  • 0 <= fromi < toi < n
  • 1 <= weighti, distanceThreshold <= 10^4
  • All pairs (fromi, toi) are distinct.

Solution1: Floyd-Warshall

All pair shortest path

Time complexity: O(n^3)
Space complexity: O(n^2)

C++

Solution 2: Dijkstra’s Algorithm

Time complexity: O(V * ElogV) / worst O(n^3*logn), best O(n^2*logn)
Space complexity: O(V + E)

C++

花花酱 LeetCode 1319. Number of Operations to Make Network Connected

There are n computers numbered from 0 to n-1 connected by ethernet cables connections forming a network where connections[i] = [a, b] represents a connection between computers a and b. Any computer can reach any other computer directly or indirectly through the network.

Given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected. Return the minimum number of times you need to do this in order to make all the computers connected. If it’s not possible, return -1. 

Example 1:

Input: n = 4, connections = [[0,1],[0,2],[1,2]]
Output: 1
Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.

Example 2:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
Output: 2

Example 3:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
Output: -1
Explanation: There are not enough cables.

Example 4:

Input: n = 5, connections = [[0,1],[0,2],[3,4],[2,3]]
Output: 0

Constraints:

  • 1 <= n <= 10^5
  • 1 <= connections.length <= min(n*(n-1)/2, 10^5)
  • connections[i].length == 2
  • 0 <= connections[i][0], connections[i][1] < n
  • connections[i][0] != connections[i][1]
  • There are no repeated connections.
  • No two computers are connected by more than one cable.

Solution 1: Union-Find

Time complexity: O(V+E)
Space complexity: O(V)

C++

Solution 2: DFS

Time complexity: O(V+E)
Space complexity: O(V+E)

C++

花花酱 LeetCode 1202. Smallest String With Swaps

You are given a string s, and an array of pairs of indices in the string pairs where pairs[i] = [a, b] indicates 2 indices(0-indexed) of the string.

You can swap the characters at any pair of indices in the given pairs any number of times.

Return the lexicographically smallest string that s can be changed to after using the swaps.

Example 1:

Input: s = "dcab", pairs = [[0,3],[1,2]]
Output: "bacd"
Explaination: 
Swap s[0] and s[3], s = "bcad"
Swap s[1] and s[2], s = "bacd"

Example 2:

Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
Output: "abcd"
Explaination: 
Swap s[0] and s[3], s = "bcad"
Swap s[0] and s[2], s = "acbd"
Swap s[1] and s[2], s = "abcd"

Example 3:

Input: s = "cba", pairs = [[0,1],[1,2]]
Output: "abc"
Explaination: 
Swap s[0] and s[1], s = "bca"
Swap s[1] and s[2], s = "bac"
Swap s[0] and s[1], s = "abc"

Constraints:

  • 1 <= s.length <= 10^5
  • 0 <= pairs.length <= 10^5
  • 0 <= pairs[i][0], pairs[i][1] < s.length
  • s only contains lower case English letters.

Solution: Connected Components

Use DFS / Union-Find to find all the connected components of swapable indices. For each connected components (index group), extract the subsequence of corresponding chars as a string, sort it and put it back to the original string in the same location.

e.g. s = “dcab”, pairs = [[0,3],[1,2]]
There are two connected components: {0,3}, {1,2}
subsequences:
1. 0,3 “db”, sorted: “bd”
2. 1,2 “ca”, sorted: “ac”
0 => b
1 => a
2 => c
3 => d
final = “bacd”

Time complexity: DFS: O(nlogn + k*(V+E)), Union-Find: O(nlogn + V+E)
Space complexity: O(n)

C++/DFS

C++/Union-Find

花花酱 LeetCode 1192. Critical Connections in a Network

There are n servers numbered from 0 to n-1 connected by undirected server-to-server connections forming a network where connections[i] = [a, b] represents a connection between servers a and b. Any server can reach any other server directly or indirectly through the network.

critical connection is a connection that, if removed, will make some server unable to reach some other server.

Return all critical connections in the network in any order.

Example 1:

Input: n = 4, connections = [[0,1],[1,2],[2,0],[1,3]]
Output: [[1,3]]
Explanation: [[3,1]] is also accepted.

Constraints:

  • 1 <= n <= 10^5
  • n-1 <= connections.length <= 10^5
  • connections[i][0] != connections[i][1]
  • There are no repeated connections.

Solution: Tarjan

Time complexity: O(v+e)
Space complexity: O(v+e)

C++