Posts tagged as “graph”

花花酱 LeetCode 952. Largest Component Size by Common Factor

By zxi on December 1, 2018

Problem

Given a non-empty array of unique positive integers A, consider the following graph:

There are A.length nodes, labelled A[0] to A[A.length - 1];
There is an edge between A[i] and A[j] if and only if A[i] and A[j] share a common factor greater than 1.

Return the size of the largest connected component in the graph.

Example 1:

Input: [4,6,15,35]
Output: 4

Example 2:

Input: [20,50,9,63]
Output: 2

Example 3:

Input: [2,3,6,7,4,12,21,39]
Output: 8

Note:

1 <= A.length <= 20000
1 <= A[i] <= 100000

Solution: Union Find

For each number, union itself with all its factors.

E.g. 6, union(6,2), union(6,3)

Time complexity: \( O(\Sigma{sqrt(A[i])}) \)

Space complexity: \( O(max(A)) \)

C++

// Author: Huahua, running time: 148 ms

class DSU {

public:

DSU(int n): p_(n) {

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

p_[i] = i;

}

void Union(int x, int y) {

p_[Find(x)] = p_[Find(y)];

}

int Find(int x) {

if (p_[x] != x) p_[x] = Find(p_[x]);

return p_[x];

}

private:

vector<int> p_;

};

class Solution {

public:

int largestComponentSize(vector<int>& A) {

int n = *max_element(begin(A), end(A));

DSU dsu(n + 1);

for (int a : A) {

int t = sqrt(a);

for (int k = 2; k <= t; ++k)

if (a % k == 0) {

dsu.Union(a, k);

dsu.Union(a, a / k);

}

unordered_map<int, int> c;

int ans = 1;

for (int a : A)

ans = max(ans, ++c[dsu.Find(a)]);

return ans;

}

};

Python3

# Author: Huahua, running time: 3432 ms

class Solution:

def largestComponentSize(self, A):

p = list(range(max(A) + 1))

def find(x):

while p[x] != x:

p[x] = p[p[x]]

x = p[x]

return x

def union(x, y):

p[find(x)] = p[find(y)]

for a in A:

for k in range(2, int(math.sqrt(a) + 1)):

if a % k == 0:

union(a, k)

union(a, a // k)

return collections.Counter([find(a) for a in A]).most_common(1)[0][1]

花花酱 LeetCode 924. Minimize Malware Spread

By zxi on October 13, 2018

Problem

In a network of nodes, each node i is directly connected to another node j if and only if graph[i][j] = 1.

Some nodes initial are initially infected by malware. Whenever two nodes are directly connected and at least one of those two nodes is infected by malware, both nodes will be infected by malware. This spread of malware will continue until no more nodes can be infected in this manner.

Suppose M(initial) is the final number of nodes infected with malware in the entire network, after the spread of malware stops.

We will remove one node from the initial list. Return the node that if removed, would minimize M(initial). If multiple nodes could be removed to minimize M(initial), return such a node with the smallest index.

Note that if a node was removed from the initial list of infected nodes, it may still be infected later as a result of the malware spread.

Example 1:

Input: graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
Output: 0

Example 2:

Input: graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2]
Output: 0

Example 3:

Input: graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2]
Output: 1

Note:

1 < graph.length = graph[0].length <= 300
0 <= graph[i][j] == graph[j][i] <= 1
graph[i][i] = 1
1 <= initial.length < graph.length
0 <= initial[i] < graph.length

Solution: BFS

Time complexity: O(n^3)

Space complexity: O(n^2)

C++

// Author: Huahua

class Solution {

public:

int minMalwareSpread(vector<vector<int>>& graph, vector<int>& initial) {

int min_affected = INT_MAX;

int min_node = -1;

sort(begin(initial), end(initial));

for (int t : initial) {

vector<int> bad(graph.size(), 0);

queue<int> q;

for (int n : initial)

if (n != t) {

bad[n] = 1;

q.push(n);

}

int affected = initial.size() - 1;

while (!q.empty()) {

int size = q.size();

while (size--) {

int n = q.front(); q.pop();

for (int i = 0; i < graph[n].size(); ++i) {

if (graph[n][i] == 0 || bad[i]) continue;

++affected;

bad[i] = 1;

q.push(i);

}

if (affected < min_affected) {

min_affected = affected;

min_node = t;

}

return min_node;

}

};

花花酱 LeetCode 886. Possible Bipartition

By zxi on August 12, 2018

Problem

Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.

Each person may dislike some other people, and they should not go into the same group.

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only if it is possible to split everyone into two groups in this way.

Example 1:

Input: N = 4, dislikes = [[1,2],[1,3],[2,4]]
Output: true
Explanation: group1 [1,4], group2 [2,3]

Example 2:

Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false

Example 3:

Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false

Note:

1 <= N <= 2000
0 <= dislikes.length <= 10000
1 <= dislikes[i][j] <= N
dislikes[i][0] < dislikes[i][1]
There does not exist i != j for which dislikes[i] == dislikes[j].

Solution: Graph Coloring

Color a node with one color, and color all it’s disliked nodes with another color, if can not finish return false.

Time complexity: O(V+E)

Space complexity: O(V+E)

C++ / DFS

// Author: Huahua

// Running time: 96 ms

class Solution {

public:

bool possibleBipartition(int N, vector<vector<int>>& dislikes) {

g_ = vector<vector<int>>(N);

for (const auto& d : dislikes) {

g_[d[0] - 1].push_back(d[1] - 1);

g_[d[1] - 1].push_back(d[0] - 1);

}

colors_ = vector<int>(N, 0); // 0: unknown, 1: red, -1: blue

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

if (colors_[i] == 0 && !dfs(i, 1)) return false;

return true;

}

private:

vector<vector<int>> g_;

vector<int> colors_;

bool dfs(int cur, int color) {

colors_[cur] = color;

for (int nxt : g_[cur]) {

if (colors_[nxt] == color) return false;

if (colors_[nxt] == 0 && !dfs(nxt, -color)) return false;

}

return true;

}

};

C++ / BFS

// Author: Huahua

// Running time: 100 ms

class Solution {

public:

bool possibleBipartition(int N, vector<vector<int>>& dislikes) {

vector<vector<int>> g(N);

for (const auto& d : dislikes) {

g[d[0] - 1].push_back(d[1] - 1);

g[d[1] - 1].push_back(d[0] - 1);

}

queue<int> q;

vector<int> colors(N, 0); // 0: unknown, 1: red, -1: blue

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

if (colors[i] != 0) continue;

q.push(i);

colors[i] = 1;

while (!q.empty()) {

int cur = q.front(); q.pop();

for (int nxt : g[cur]) {

if (colors[nxt] == colors[cur]) return false;

if (colors[nxt] != 0) continue;

colors[nxt] = -colors[cur];

q.push(nxt);

}

return true;

}

};

Problem

Starting with an undirected graph (the “original graph”) with nodes from 0 to N-1, subdivisions are made to some of the edges.

The graph is given as follows: edges[k] is a list of integer pairs (i, j, n) such that (i, j) is an edge of the original graph,

and n is the total number of new nodes on that edge.

Then, the edge (i, j) is deleted from the original graph, n new nodes (x_1, x_2, ..., x_n) are added to the original graph,

and n+1 new edges (i, x_1), (x_1, x_2), (x_2, x_3), ..., (x_{n-1}, x_n), (x_n, j) are added to the original graph.

Now, you start at node 0 from the original graph, and in each move, you travel along one edge.

Return how many nodes you can reach in at most M moves.

Example 1:

Input: edge = [[0,1,10],[0,2,1],[1,2,2]], M = 6, N = 3 
Output: 13 
Explanation:  The nodes that are reachable in the final graph after M = 6 moves are indicated below.

Example 2:

Input: edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], M = 10, N = 4 
Output: 23

Note:

0 <= edges.length <= 10000
0 <= edges[i][0] < edges[i][1] < N
There does not exist any i != j for which edges[i][0] == edges[j][0] and edges[i][1] == edges[j][1].
The original graph has no parallel edges.
0 <= edges[i][2] <= 10000
0 <= M <= 10^9
1 <= N <= 3000

Solution: Dijkstra Shortest Path

Compute the shortest from 0 to rest of the nodes. Use HP to mark the maximum moves left to reach each node.

HP[u] = a, HP[v] = b, new_nodes[u][v] = c

nodes covered between a<->b = min(c, a + b)

Time complexity: O(ElogE)

Space complexity: O(E)

C++

// Author: Huahua

// Running time: 88 ms

class Solution {

public:

int reachableNodes(vector<vector<int>>& edges, int M, int N) {

unordered_map<int, unordered_map<int, int>> g;

for (const auto& e : edges)

g[e[0]][e[1]] = g[e[1]][e[0]] = e[2];

priority_queue<pair<int, int>> q; // {hp, node}, sort by HP desc

unordered_map<int, int> HP; // node -> max HP left

q.push({M, 0});

while (!q.empty()) {

int hp = q.top().first;

int cur = q.top().second;

q.pop();

if (HP.count(cur)) continue;

HP[cur] = hp;

for (const auto& pair : g[cur]) {

int nxt = pair.first;

int nxt_hp = hp - pair.second - 1;

if (HP.count(nxt) || nxt_hp < 0) continue;

q.push({nxt_hp, nxt});

}

int ans = HP.size(); // Original nodes covered.

for (const auto& e : edges) {

int uv = HP.count(e[0]) ? HP[e[0]] : 0;

int vu = HP.count(e[1]) ? HP[e[1]] : 0;

ans += min(e[2], uv + vu);

}

return ans;

}

};

Optimized Dijkstra (replace hashmap with vector)

// Author: Huahua

// Running time: 56 ms (beats 88%)

class Solution {

public:

int reachableNodes(vector<vector<int>>& edges, int M, int N) {

vector<vector<pair<int, int>>> g(N);

for (const auto& e : edges) {

g[e[0]].emplace_back(e[1], e[2]);

g[e[1]].emplace_back(e[0], e[2]);

}

set<pair<int, int>> q; // {hp, node}, sort by HP asc

vector<int> HP(N, INT_MIN); // node -> HP left

q.insert({M, 0});

int ans = 0;

while (!q.empty()) {

auto last_it = prev(end(q));

int hp = last_it->first;

int cur = last_it->second;

q.erase(last_it);

if (HP[cur] != INT_MIN) continue;

HP[cur] = hp;

++ans;

for (const auto& pair : g[cur]) {

int nxt = pair.first;

int nxt_hp = hp - pair.second - 1;

if (nxt_hp < 0 || HP[nxt] != INT_MIN) continue;

q.insert({nxt_hp, nxt});

}

for (const auto& e : edges) {

const int uv = HP[e[0]] == INT_MIN ? 0 : HP[e[0]];

const int vu = HP[e[1]] == INT_MIN ? 0 : HP[e[1]];

ans += min(e[2], uv + vu);

}

return ans;

}

};

Using SPFA

// Author: Huahua

// Running time: 56 ms (beats 88%)

class Solution {

public:

int reachableNodes(vector<vector<int>>& edges, int M, int N) {

vector<vector<pair<int, int>>> g(N);

for (const auto& e : edges) {

g[e[0]].emplace_back(e[1], e[2]);

g[e[1]].emplace_back(e[0], e[2]);

}

queue<int> q;

vector<int> HP(N, INT_MIN);

vector<bool> in_q(N);

HP[0] = M;

q.push(0);

in_q[0] = true;

// SPFA (Shortest Path Faster Algorithm).

while (!q.empty()) {

int u = q.front(); q.pop();

in_q[u] = false;

for (const auto& pair : g[u]) {

int v = pair.first;

int new_hp = HP[u] - pair.second - 1;

if (new_hp < 0 || new_hp <= HP[v]) continue;

HP[v] = new_hp;

if (in_q[v]) continue;

q.push(v);

in_q[v] = true;

}

int ans = 0;

for (int i = 0; i < N; ++i) ans += HP[i] >= 0;

for (const auto& e : edges) {

const int uv = HP[e[0]] == INT_MIN ? 0 : HP[e[0]];

const int vu = HP[e[1]] == INT_MIN ? 0 : HP[e[1]];

ans += min(e[2], uv + vu);

}

return ans;

}

};

BFS

// Author: Huahua

// Running time: 108 ms

class Solution {

public:

int reachableNodes(vector<vector<int>>& edges, int M, int N) {

unordered_map<int, unordered_map<int, int>> g;

for (const auto& e : edges)

g[e[0]][e[1]] = g[e[1]][e[0]] = e[2];

queue<pair<int, int>> q; // {hp, node}

unordered_map<int, int> HP; // node -> max HP left

q.push({M, 0});

while (!q.empty()) {

int hp = q.front().first;

int cur = q.front().second;

q.pop();

if (HP.count(cur) && HP[cur] > hp) continue;

HP[cur] = hp;

for (const auto& pair : g[cur]) {

int nxt = pair.first;

int nxt_hp = hp - pair.second - 1;

if (nxt_hp < 0 || (HP.count(nxt) && HP[nxt] > nxt_hp)) continue;

q.push({nxt_hp, nxt});

}

int ans = HP.size(); // Original nodes covered.

for (const auto& e : edges) {

int uv = HP.count(e[0]) ? HP[e[0]] : 0;

int vu = HP.count(e[1]) ? HP[e[1]] : 0;

ans += min(e[2], uv + vu);

}

return ans;

}

};

花花酱 LeetCode 785. Is Graph Bipartite?

By zxi on July 24, 2018

Video is for 花花酱 LeetCode 886. Possible Bipartition, but the algorithm is exact the same.

Problem

https://leetcode.com/problems/is-graph-bipartite/

Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it’s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn’t contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation: 
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.

Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation: 
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.

Note:

graph will have length in range [1, 100].
graph[i] will contain integers in range [0, graph.length - 1].
graph[i] will not contain i or duplicate values.
The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

Solution: Graph Coloring

For each node

If has not been colored, color it to RED(1).
Color its neighbors with a different color RED(1) to BLUE(-1) or BLUE(-1) to RED(-1).

If we can finish the coloring then the graph is bipartite. All red nodes on the left no connections between them and all blues nodes on the right, again no connections between them. red and blue nodes are neighbors.

Time complexity: O(V+E)

Space complexity: O(V)

C++ / DFS

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

bool isBipartite(vector<vector<int>>& graph) {

const int n = graph.size();

vector<int> colors(n);

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

if (!colors[i] && !coloring(graph, colors, 1, i))

return false;

return true;

}

private:

bool coloring(const vector<vector<int>>& graph, vector<int>& colors, int color, int node) {

if (colors[node]) return colors[node] == color;

colors[node] = color;

for (int nxt : graph[node])

if (!coloring(graph, colors, -color, nxt)) return false;

return true;

}

};

Posts tagged as “graph”

花花酱 LeetCode 952. Largest Component Size by Common Factor

Problem

Solution: Union Find

C++

Python3

花花酱 LeetCode 924. Minimize Malware Spread

Problem

Solution: BFS

C++

花花酱 LeetCode 886. Possible Bipartition

Problem

Solution: Graph Coloring

C++ / DFS

C++ / BFS

Related Problem

花花酱 LeetCode 882. Reachable Nodes In Subdivided Graph

Problem

Solution: Dijkstra Shortest Path

花花酱 LeetCode 785. Is Graph Bipartite?

Problem

Solution: Graph Coloring

Related Problem