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Posts tagged as “graph”

花花酱 LeetCode 1042. Flower Planting With No Adjacent

You have N gardens, labelled 1 to N.  In each garden, you want to plant one of 4 types of flowers.

paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y.

Also, there is no garden that has more than 3 paths coming into or leaving it.

Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers.

Return any such a choice as an array answer, where answer[i] is the type of flower planted in the (i+1)-th garden.  The flower types are denoted 1, 2, 3, or 4.  It is guaranteed an answer exists.

Example 1:

Input: N = 3, paths = [[1,2],[2,3],[3,1]]
Output: [1,2,3]

Example 2:

Input: N = 4, paths = [[1,2],[3,4]]
Output: [1,2,1,2]

Example 3:

Input: N = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]]
Output: [1,2,3,4]

Note:

  • 1 <= N <= 10000
  • 0 <= paths.size <= 20000
  • No garden has 4 or more paths coming into or leaving it.
  • It is guaranteed an answer exists.

Solution: Graph coloring, choose any available color

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

C++

花花酱 LeetCode 851. Loud and Rich

In a group of N people (labelled 0, 1, 2, ..., N-1), each person has different amounts of money, and different levels of quietness.

For convenience, we’ll call the person with label x, simply “person x“.

We’ll say that richer[i] = [x, y] if person x definitely has more money than person y.  Note that richer may only be a subset of valid observations.

Also, we’ll say quiet[x] = q if person x has quietness q.

Now, return answer, where answer[x] = y if y is the least quiet person (that is, the person y with the smallest value of quiet[y]), among all people who definitely have equal to or more money than person x.

Example 1:

Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
Output: [5,5,2,5,4,5,6,7]
Explanation: 
answer[0] = 5.
Person 5 has more money than 3, which has more money than 1, which has more money than 0.
The only person who is quieter (has lower quiet[x]) is person 7, but
it isn't clear if they have more money than person 0.

answer[7] = 7.
Among all people that definitely have equal to or more money than person 7
(which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x])
is person 7.

The other answers can be filled out with similar reasoning.

Note:

  1. 1 <= quiet.length = N <= 500
  2. 0 <= quiet[i] < N, all quiet[i] are different.
  3. 0 <= richer.length <= N * (N-1) / 2
  4. 0 <= richer[i][j] < N
  5. richer[i][0] != richer[i][1]
  6. richer[i]‘s are all different.
  7. The observations in richer are all logically consistent.

Solution: DFS + Memoization

For person i , remember the quietest person who is richer than person i.

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

C++

花花酱 LeetCode 997. Find the Town Judge

In a town, there are N people labelled from 1 to N.  There is a rumor that one of these people is secretly the town judge.

If the town judge exists, then:

  1. The town judge trusts nobody.
  2. Everybody (except for the town judge) trusts the town judge.
  3. There is exactly one person that satisfies properties 1 and 2.

You are given trust, an array of pairs trust[i] = [a, b] representing that the person labelled a trusts the person labelled b.

If the town judge exists and can be identified, return the label of the town judge.  Otherwise, return -1.

Example 1:

Input: N = 2, trust = [[1,2]]
Output: 2

Example 2:

Input: N = 3, trust = [[1,3],[2,3]]
Output: 3

Example 3:

Input: N = 3, trust = [[1,3],[2,3],[3,1]]
Output: -1

Example 4:

Input: N = 3, trust = [[1,2],[2,3]]
Output: -1

Example 5:

Input: N = 4, trust = [[1,3],[1,4],[2,3],[2,4],[4,3]]
Output: 3

Note:

  1. 1 <= N <= 1000
  2. trust.length <= 10000
  3. trust[i] are all different
  4. trust[i][0] != trust[i][1]
  5. 1 <= trust[i][0], trust[i][1] <= N

Solution: Degree

node with degree (in_degree – out_degree) N – 1 is the judge.

Time complexity: O(N+T)
Space complexity: O(N)

C++

花花酱 LeetCode 133. Clone Graph

Given the head of a graph, return a deep copy (clone) of the graph. Each node in the graph contains a label (int) and a list (List[UndirectedGraphNode]) of its neighbors. There is an edge between the given node and each of the nodes in its neighbors.
OJ’s undirected graph serialization (so you can understand error output):

Nodes are labeled uniquely.We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.

As an example, consider the serialized graph {0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by #.

  1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
  2. Second node is labeled as 1. Connect node 1 to node 2.
  3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.

Visually, the graph looks like the following:

       1
      / \
     /   \
    0 --- 2
         / \
         \_/

Note: The information about the tree serialization is only meant so that you can understand error output if you get a wrong answer. You don’t need to understand the serialization to solve the problem.

Solution: Queue + Hashtable

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

C++

花花酱 LeetCode 990. Satisfiability of Equality Equations

Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily different) that represent one-letter variable names.

Return true if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.

Example 1:

Input: ["a==b","b!=a"]
Output: false
Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.  There is no way to assign the variables to satisfy both equations.

Example 2:

Input: ["b==a","a==b"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.

Example 3:

Input: ["a==b","b==c","a==c"]
Output: true

Example 4:

Input: ["a==b","b!=c","c==a"]
Output: false

Example 5:

Input: ["c==c","b==d","x!=z"]
Output: true

Note:

  1. 1 <= equations.length <= 500
  2. equations[i].length == 4
  3. equations[i][0] and equations[i][3] are lowercase letters
  4. equations[i][1] is either '=' or '!'
  5. equations[i][2] is '='

Solution: Union Find

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

C++