Posts tagged as “union find”

花花酱 LeetCode Disjoint set / Union Find Forest SP1

By zxi on November 29, 2017

Disjoint-set/Union-find Forest

Find(x): find the root/cluster-id of x

Union(x, y): merge two clusters

Check whether two elements are in the same set or not in O(1)*.

Find: O(ɑ(n))* ≈ O(1)

Union: O(ɑ(n))* ≈ O(1)

Space: O(n)

Without optimization: Find: O(n)

Two key optimizations:

Path compression: make tree flat
Union by rank: merge low rank tree to high rank one

*: amortized

ɑ(.): inverse Ackermann function

Implementations:

C++

// Author: Huahua

class UnionFindSet {

public:

UnionFindSet(int n) {

ranks_ = vector<int>(n + 1, 0);

parents_ = vector<int>(n + 1, 0);

for (int i = 0; i < parents_.size(); ++i)

parents_[i] = i;

}

// Merge sets that contains u and v.

// Return true if merged, false if u and v are already in one set.

bool Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

// Meger low rank tree into high rank tree

if (ranks_[pv] < ranks_[pu])

parents_[pv] = pu;

else if (ranks_[pu] < ranks_[pv])

parents_[pu] = pv;

else {

parents_[pv] = pu;

ranks_[pu] += 1;

}

return true;

}

// Get the root of u.

int Find(int u) {

// Compress the path during traversal

if (u != parents_[u])

parents_[u] = Find(parents_[u]);

return parents_[u];

}

private:

vector<int> parents_;

vector<int> ranks_;

};

Java

// Author: Huahua

class UnionFindSet {

private int[] parents_;

private int[] ranks_;

public UnionFindSet(int n) {

parents_ = new int[n + 1];

ranks_ = new int[n + 1];

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

parents_[i] = i;

ranks_[i] = 1;

}

public boolean Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

if (ranks_[pv] > ranks_[pu])

parents_[pu] = pv;

else if (ranks_[pu] > ranks_[pv])

parents_[pv] = pu;

else {

parents_[pv] = pu;

ranks_[pu] += 1;

}

return true;

}

public int Find(int u) {

while (parents_[u] != u) {

parents_[u] = parents_[parents_[u]];

u = parents_[u];

}

return u;

}

Python

"""

Author: Huahua

"""

class UnionFindSet:

def __init__(self, n):

self._parents = [i for i in range(n + 1)]

self._ranks = [1 for i in range(n + 1)]

def find(self, u):

while u != self._parents[u]:

self._parents[u] = self._parents[self._parents[u]]

u = self._parents[u]

return u

def union(self, u, v):

pu, pv = self.find(u), self.find(v)

if pu == pv: return False

if self._ranks[pu] < self._ranks[pv]:

self._parents[pu] = pv

elif self._ranks[pu] > self._ranks[pv]:

self._parents[pv] = pu

else:

self._parents[pv] = pu

self._ranks[pu] += 1

return True

Union-Find Problems

References

花花酱 LeetCode 737. Sentence Similarity II

By zxi on November 28, 2017

Problem:

Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar.

For example, words1 = ["great", "acting", "skills"] and words2 = ["fine", "drama", "talent"] are similar, if the similar word pairs are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]].

Note that the similarity relation is transitive. For example, if “great” and “good” are similar, and “fine” and “good” are similar, then “great” and “fine” are similar.

Similarity is also symmetric. For example, “great” and “fine” being similar is the same as “fine” and “great” being similar.

Also, a word is always similar with itself. For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though there are no specified similar word pairs.

Finally, sentences can only be similar if they have the same number of words. So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].

Note:

The length of words1 and words2 will not exceed 1000.
The length of pairs will not exceed 2000.
The length of each pairs[i] will be 2.
The length of each words[i] and pairs[i][j] will be in the range [1, 20].

题目大意：

给你两个句子（由单词数组表示）和一些近义词对，问你这两个句子是否相似，即每组相对应的单词都要相似。

注意相似性可以传递，比如只给你”great”和”fine”相似、”fine”和”good”相似，能推断”great”和”good”也相似。

Idea:

DFS / Union Find

Solution1:

Time complexity: O(|Pairs| * |words1|)

Space complexity: O(|Pairs|)

C++ / DFS

// Author: Huahua

// Runtime: 346 ms

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

g_.clear();

for (const auto& p : pairs) {

g_[p.first].insert(p.second);

g_[p.second].insert(p.first);

}

unordered_set<string> visited;

for (int i = 0; i < words1.size(); ++i) {

visited.clear();

if (!dfs(words1[i], words2[i], visited)) return false;

}

return true;

}

private:

bool dfs(const string& src, const string& dst, unordered_set<string>& visited) {

if (src == dst) return true;

visited.insert(src);

for (const auto& next : g_[src]) {

if (visited.count(next)) continue;

if (dfs(next, dst, visited)) return true;

}

return false;

}

unordered_map<string, unordered_set<string>> g_;

};

Time complexity: O(|Pairs| + |words1|)

Space complexity: O(|Pairs|)

C++ / DFS Optimized

// Author: Huahua

// Runtime 229 ms

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

g_.clear();

ids_.clear();

for (const auto& p : pairs) {

g_[p.first].insert(p.second);

g_[p.second].insert(p.first);

}

int id = 0;

for (const auto& p : pairs) {

if(!ids_.count(p.first)) dfs(p.first, ++id);

if(!ids_.count(p.second)) dfs(p.second, ++id);

}

for (int i = 0; i < words1.size(); ++i) {

if (words1[i] == words2[i]) continue;

auto it1 = ids_.find(words1[i]);

auto it2 = ids_.find(words2[i]);

if (it1 == ids_.end() || it2 == ids_.end()) return false;

if (it1->second != it2->second) return false;

}

return true;

}

private:

bool dfs(const string& curr, int id) {

ids_[curr] = id;

for (const auto& next : g_[curr]) {

if (ids_.count(next)) continue;

if (dfs(next, id)) return true;

}

return false;

}

unordered_map<string, int> ids_;

unordered_map<string, unordered_set<string>> g_;

};

Solution2:

Time complexity: O(|Pairs| + |words1|)

Space complexity: O(|Pairs|)

C++ / Union Find

// Author: Huahua

// Runtime: 219 ms

class UnionFindSet {

public:

bool Union(const string& word1, const string& word2) {

const string& p1 = Find(word1, true);

const string& p2 = Find(word2, true);

if (p1 == p2) return false;

parents_[p1] = p2;

return true;

}

const string& Find(const string& word, bool create = false) {

if (!parents_.count(word)) {

if (!create) return word;

return parents_[word] = word;

}

string w = word;

while (w != parents_[w]) {

parents_[w] = parents_[parents_[w]];

w = parents_[w];

}

return parents_[w];

}

private:

unordered_map<string, string> parents_;

};

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

UnionFindSet s;

for (const auto& pair : pairs)

s.Union(pair.first, pair.second);

for (int i = 0; i < words1.size(); ++i)

if (s.Find(words1[i]) != s.Find(words2[i])) return false;

return true;

}

};

C++ / Union Find, Optimized

// Author: Huahua

// Runtime: 175 ms (< 98.45%)

class UnionFindSet {

public:

UnionFindSet(int n) {

parents_ = vector<int>(n + 1, 0);

ranks_ = vector<int>(n + 1, 0);

for (int i = 0; i < parents_.size(); ++i)

parents_[i] = i;

}

bool Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

if (ranks_[pu] > ranks_[pv]) {

parents_[pv] = pu;

} else if (ranks_[pv] > ranks_[pu]) {

parents_[pu] = pv;

} else {

parents_[pu] = pv;

++ranks_[pv];

}

return true;

}

int Find(int id) {

if (id != parents_[id])

parents_[id] = Find(parents_[id]);

return parents_[id];

}

private:

vector<int> parents_;

vector<int> ranks_;

};

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

UnionFindSet s(pairs.size() * 2);

unordered_map<string, int> indies; // word to index

for (const auto& pair : pairs) {

int u = getIndex(pair.first, indies, true);

int v = getIndex(pair.second, indies, true);

s.Union(u, v);

}

for (int i = 0; i < words1.size(); ++i) {

if (words1[i] == words2[i]) continue;

int u = getIndex(words1[i], indies);

int v = getIndex(words2[i], indies);

if (u < 0 || v < 0) return false;

if (s.Find(u) != s.Find(v)) return false;

}

return true;

}

private:

int getIndex(const string& word, unordered_map<string, int>& indies, bool create = false) {

auto it = indies.find(word);

if (it == indies.end()) {

if (!create) return INT_MIN;

int index = indies.size();

indies.emplace(word, index);

return index;

}

return it->second;

}

};

Related Problems:

花花酱 LeetCode 685. Redundant Connection II

By zxi on October 7, 2017

Problem:

In this problem, a rooted tree is a directed graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.

The given input is a directed graph that started as a rooted tree with N nodes (with distinct values 1, 2, …, N), with one additional directed edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.

The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] that represents a directed edge connecting nodes u and v, where u is a parent of child v.

Return an edge that can be removed so that the resulting graph is a rooted tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.

Example 1:

Input: [[1,2], [1,3], [2,3]]

Output: [2,3]

Explanation: The given directed graph will be like this:

/ \

v v

2-->3

Example 2:

Input: [[1,2], [2,3], [3,4], [4,1], [1,5]]

Output: [4,1]

Explanation: The given directed graph will be like this:

5 <- 1 -> 2

^ |

| v

4 <- 3

Note:

The size of the input 2D-array will be between 3 and 1000.
Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

Idea:

Union Find

Time complexity: O(nlog*n) ~ O(n)

Space complexity: O(n)

Solution:

C++

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

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

vector<int> parents(edges.size() + 1, 0);

vector<int> roots(edges.size() + 1, 0);

vector<int> sizes(edges.size() + 1, 1);

vector<int> ans1;

vector<int> ans2;

for(auto& edge: edges) {

int u = edge[0];

int v = edge[1];

// A node has two parents

if (parents[v] > 0) {

ans1 = {parents[v], v};

ans2 = edge;

// Delete the later edge

edge[0] = edge[1] = -1;

}

parents[v] = u;

}

for(const auto& edge: edges) {

int u = edge[0];

int v = edge[1];

// Invalid edge (we deleted in step 1)

if (u < 0 || v < 0) continue;

if (!roots[u]) roots[u] = u;

if (!roots[v]) roots[v] = v;

int pu = find(u, roots);

int pv = find(v, roots);

// Both u and v are already in the tree

if (pu == pv)

return ans1.empty() ? edge : ans1;

// Unoin, always merge smaller set (pv) to larger set (pu)

if (sizes[pv] > sizes[pu])

swap(pu, pv);

roots[pv] = pu;

sizes[pu] += sizes[pv];

}

return ans2;

}

private:

int find(int node, vector<int>& roots) {

while (roots[node] != node) {

roots[node] = roots[roots[node]];

node = roots[node];

}

return node;

}

};

C++ / without using Union find

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

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

vector<int> parents(edges.size() + 1, 0);

vector<int> ans1;

vector<int> ans2;

bool dup_parents = false;

for(auto& edge: edges) {

int u = edge[0];

int v = edge[1];

// A node has two parents

if (parents[v] > 0) {

ans1 = {parents[v], v};

ans2 = edge;

dup_parents = true;

// Delete the later edge

edge[0] = edge[1] = -1;

} else {

parents[v] = u;

}

// Reset parents

parents = vector<int>(edges.size() + 1, 0);

for(const auto& edge: edges) {

int u = edge[0];

int v = edge[1];

// Invalid edge (we deleted in step 1)

if (u < 0 || v < 0) continue;

parents[v] = u;

if (cycle(v, parents))

return dup_parents ? ans1 : edge;

}

return ans2;

}

private:

bool cycle(int v, const vector<int>& parents) {

int u = parents[v];

while (u) {

if (u == v) return true;

u = parents[u];

}

return false;

}

};

Python

"""

Author: Huahua

Runtime: 62 ms

"""

class Solution:

def findRedundantDirectedConnection(self, edges):

def cycle(v, p):

u = p[v]

while u != 0:

if u == v: return True

u = p[u]

return False

n = len(edges)

p = [0] * (n + 1)

ans1 = []

ans2 = []

dup_p = False

for e in edges:

u, v = e

if p[v] > 0:

ans1 = [p[v], v]

ans2 = [u, v]

dup_p = True

e[0] = e[1] = -1

else:

p[v] = u

p = [0] * (n + 1)

for u, v in edges:

if u < 0: continue

p[v] = u

if cycle(v, p):

return ans1 if dup_p else [u, v]

return ans2

花花酱 LeetCode 684. Redundant Connection

By zxi on September 28, 2017

https://leetcode.com/problems/redundant-connection/description/

Problem:

In this problem, a tree is an undirected graph that is connected and has no cycles.

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, …, N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.

The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.

Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

Input: [[1,2], [1,3], [2,3]]

Output: [2,3]

Explanation: The given undirected graph will be like this:

/ \

2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]

Output: [1,4]

Explanation: The given undirected graph will be like this:

5 - 1 - 2

| |

4 - 3

Note:

The size of the input 2D-array will be between 3 and 1000.
Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

Idea:
DFS / Union-Find

Solutions:

C++ / DFS

// Author: Huahua

// Time Complexity: O(n^2)

// Running Time: 22 ms

class Solution {

public:

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

unordered_map<int, vector<int>> graph;

for (const auto& edge : edges) {

int u = edge[0];

int v = edge[1];

unordered_set<int> visited;

if (hasPath(u, v, graph, visited))

return edge;

graph[u].push_back(v);

graph[v].push_back(u);

}

return {};

}

private:

bool hasPath(int curr,

int goal,

const unordered_map<int, vector<int>>& graph,

unordered_set<int>& visited) {

if (curr == goal) return true;

visited.insert(curr);

if (!graph.count(curr) || !graph.count(goal)) return false;

for (int next : graph.at(curr)) {

if (visited.count(next)) continue;

if (hasPath(next, goal, graph, visited)) return true;

}

return false;

}

};

C++ / Union Find

// Author: Huahua

// Running time: 6 ms

class UnionFindSet {

public:

UnionFindSet(int n) {

ranks_ = vector<int>(n + 1, 0);

parents_ = vector<int>(n + 1, 0);

for (int i = 0; i < parents_.size(); ++i)

parents_[i] = i;

}

// Merge sets that contains u and v.

// Return true if merged, false if u and v are already in one set.

bool Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

// Meger low rank tree into high rank tree

if (ranks_[pv] > ranks_[pu])

parents_[pu] = pv;

else if (ranks_[pu] > ranks_[pv])

parents_[pv] = pu;

else {

parents_[pv] = pu;

ranks_[pv] += 1;

}

return true;

}

// Get the root of u.

int Find(int u) {

// Compress the path during traversal

if (u != parents_[u])

parents_[u] = Find(parents_[u]);

return parents_[u];

}

private:

vector<int> parents_;

vector<int> ranks_;

};

class Solution {

public:

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

UnionFindSet s(edges.size());

for(const auto& edge: edges)

if (!s.Union(edge[0], edge[1]))

return edge;

return {};

}

};

Java / Union Find

// Author: Huahua

// Runtime: 6 ms

class Solution {

class UnionFindSet {

private int[] parents_;

private int[] ranks_;

public UnionFindSet(int n) {

parents_ = new int[n + 1];

ranks_ = new int[n + 1];

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

parents_[i] = i;

ranks_[i] = 1;

}

public boolean Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

if (ranks_[pv] > ranks_[pu])

parents_[pu] = pv;

else if (ranks_[pu] > ranks_[pv])

parents_[pv] = pu;

else {

parents_[pv] = pu;

ranks_[pu] += 1;

}

return true;

}

public int Find(int u) {

while (parents_[u] != u) {

parents_[u] = parents_[parents_[u]];

u = parents_[u];

}

return u;

}

public int[] findRedundantConnection(int[][] edges) {

UnionFindSet s = new UnionFindSet(edges.length);

for (int[] edge : edges)

if (!s.Union(edge[0], edge[1]))

return edge;

return null;

}

Python: Union Find

"""

Author: Huahua

Running Time: 66 ms

"""

class Solution:

def findRedundantConnection(self, edges):

p = [0]*(len(edges) + 1)

s = [1]*(len(edges) + 1)

def find(u):

while p[u] != u:

p[u] = p[p[u]]

u = p[u]

return u

for u, v in edges:

if p[u] == 0: p[u] = u

if p[v] == 0: p[v] = v

pu, pv = find(u), find(v)

if pu == pv: return [u, v]

if s[pv] > s[pu]: u, v = v, u

p[pv] = pu

s[pu] += s[pv]

return []

Python / Union Find V2

"""

Author: Huahua

Runtime: 59 ms (<92.42%)

"""

class UnionFindSet:

def __init__(self, n):

self._parents = [i for i in range(n + 1)]

self._ranks = [1 for i in range(n + 1)]

def find(self, u):

while u != self._parents[u]:

self._parents[u] = self._parents[self._parents[u]]

u = self._parents[u]

return u

def union(self, u, v):

pu, pv = self.find(u), self.find(v)

if pu == pv: return False

if self._ranks[pu] < self._ranks[pv]:

self._parents[pu] = pv

elif self._ranks[pu] > self._ranks[pv]:

self._parents[pv] = pu

else:

self._parents[pv] = pu

self._ranks[pu] += 1

return True

class Solution:

def findRedundantConnection(self, edges):

s = UnionFindSet(len(edges))

for edge in edges:

if not s.union(edge[0], edge[1]): return edge

return None

ndSet(len(edges))

for edge in edges:

if not s.union(edge[0], edge[1]): return edge

return None

花花酱 LeetCode 547. Friend Circles

By zxi on September 21, 2017

Problem:

There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.

Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the i_thand j_th students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.

Example 1:

Input:

[[1,1,0],

[1,1,0],

[0,0,1]]

Output: 2

Example 2:

Input:

[[1,1,0],

[1,1,1],

[0,1,1]]

Output: 1

N is in range [1,200].
M[i][i] = 1 for all students.
If M[i][j] = 1, then M[j][i] = 1.

Idea:

Find all connected components using DFS

Solution: DFS

C++

// Author: Huahua

// Time complexity: O(n^2)

// Space complexity: O(n)

// Running Time: 25 ms

class Solution {

public:

int findCircleNum(vector<vector<int>>& M) {

if (M.empty()) return 0;

int n = M.size();

int ans = 0;

for (int i = 0; i < n; ++i) {

if (!M[i][i]) continue;

++ans;

dfs(M, i, n);

}

return ans;

}

private:

void dfs(vector<vector<int>>& M, int curr, int n) {

// Visit all friends (neighbors)

for (int i = 0; i < n; ++i) {

if (!M[curr][i]) continue;

M[curr][i] = M[i][curr] = 0;

dfs(M, i, n);

}

};

Java

// Author: Huahua

// Time complexity: O(n^2)

// Space complexity: O(n)

// Running Time: 20 ms

class Solution {

public int findCircleNum(int[][] M) {

int n = M.length;

if (n == 0) return 0;

int ans = 0;

for (int i = 0; i < n; ++i) {

if (M[i][i] == 0) continue;

++ans;

dfs(M, i, n);

}

return ans;

}

private void dfs(int[][] M, int curr, int n) {

for (int i = 0; i < n; ++i) {

if (M[curr][i] == 0) continue;

M[curr][i] = M[i][curr] = 0;

dfs(M, i, n);

}

Python

"""

Author: Huahua

Time complexity: O(n^2)

Space complexity: O(n)

Running Time: 162 ms

"""

class Solution(object):

def findCircleNum(self, M):

"""

:type M: List[List[int]]

:rtype: int

"""

def dfs(M, curr, n):

for i in xrange(n):

if M[curr][i] == 1:

M[curr][i] = M[i][curr] = 0

dfs(M, i, n)

n = len(M)

ans = 0

for i in xrange(n):

if M[i][i] == 1:

ans += 1

dfs(M, i, n)

return ans

Solution 2: Union Find

C++

// Author: Huahua

// Runtime: 19 ms

class UnionFindSet {

public:

UnionFindSet(int n) {

parents_ = vector<int>(n + 1, 0);

ranks_ = vector<int>(n + 1, 0);

for (int i = 0; i < parents_.size(); ++i)

parents_[i] = i;

}

bool Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

if (ranks_[pu] > ranks_[pv]) {

parents_[pv] = pu;

} else if (ranks_[pv] > ranks_[pu]) {

parents_[pu] = pv;

} else {

parents_[pu] = pv;

++ranks_[pv];

}

return true;

}

int Find(int id) {

if (id != parents_[id])

parents_[id] = Find(parents_[id]);

return parents_[id];

}

private:

vector<int> parents_;

vector<int> ranks_;

};

class Solution {

public:

int findCircleNum(vector<vector<int>>& M) {

int n = M.size();

UnionFindSet s(n);

for (int i = 0; i < n; ++i)

for (int j = i + 1; j < n; ++j)

if (M[i][j] == 1) s.Union(i, j);

unordered_set<int> circles;

for (int i = 0; i < n; ++i)

circles.insert(s.Find(i));

return circles.size();

}

};