Posts published in “Graph”

花花酱 LeetCode 947. Most Stones Removed with Same Row or Column

By zxi on November 29, 2018

Problem

On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at most one stone.

Now, a move consists of removing a stone that shares a column or row with another stone on the grid.

What is the largest possible number of moves we can make?

Example 1:

Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5

Example 2:

Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3

Example 3:

Input: stones = [[0,0]]
Output: 0

Note:

1 <= stones.length <= 1000
0 <= stones[i][j] < 10000

Solution 2: Union Find

Find all connected components (islands)

Ans = # of stones – # of islands

C++

// Author: Huahua, running time: 16 ms

class Solution {

public:

int removeStones(vector<vector<int>>& stones) {

const int kSize = 10000;

DSU dsu(kSize * 2);

for (const auto& stone : stones)

dsu.Union(stone[0], stone[1] + kSize);

unordered_set<int> seen;

for (const auto& stone : stones)

seen.insert(dsu.Find(stone[0]));

return stones.size() - seen.size();

}

private:

class DSU {

public:

DSU(int n): p_(n) {

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

p_[i] = i;

}

void Union(int i, int j) {

p_[Find(i)] = Find(j);

}

int Find(int i) {

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

return p_[i];

}

private:

vector<int> p_;

};

Problem

https://leetcode.com/problems/shortest-bridge/description/

In a given 2D binary array A, there are two islands. (An island is a 4-directionally connected group of 1s not connected to any other 1s.)

Now, we may change 0s to 1s so as to connect the two islands together to form 1 island.

Return the smallest number of 0s that must be flipped. (It is guaranteed that the answer is at least 1.)

Example 1:

Input: [[0,1],[1,0]]
Output: 1

Example 2:

Input: [[0,1,0],[0,0,0],[0,0,1]]
Output: 2

Example 3:

Input: [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]
Output: 1

Note:

1 <= A.length = A[0].length <= 100
A[i][j] == 0 or A[i][j] == 1

Solution: DFS + BFS

Use DFS to find one island and color all the nodes as 2 (BLUE).
Use BFS to find the shortest path from any nodes with color 2 (BLUE) to any nodes with color 1 (RED).

Time complexity: O(mn)

Space complexity: O(mn)

C++

class Solution {

public:

int shortestBridge(vector<vector<int>>& A) {

queue<pair<int, int>> q;

bool found = false;

for (int i = 0; i < A.size() && !found; ++i)

for (int j = 0; j < A[0].size() && !found; ++j)

if (A[i][j]) {

dfs(A, j, i, q);

found = true;

}

int steps = 0;

vector<int> dirs{0, 1, 0, -1, 0};

while (!q.empty()) {

size_t size = q.size();

while (size--) {

int x = q.front().first;

int y = q.front().second;

q.pop();

for (int i = 0; i < 4; ++i) {

int tx = x + dirs[i];

int ty = y + dirs[i + 1];

if (tx < 0 || ty < 0 || tx >= A[0].size() || ty >= A.size() || A[ty][tx] == 2) continue;

if (A[ty][tx] == 1) return steps;

A[ty][tx] = 2;

q.emplace(tx, ty);

}

++steps;

}

return -1;

}

private:

void dfs(vector<vector<int>>& A, int x, int y, queue<pair<int, int>>& q) {

if (x < 0 || y < 0 || x >= A[0].size() || y >= A.size() || A[y][x] != 1) return;

A[y][x] = 2;

q.emplace(x, y);

dfs(A, x - 1, y, q);

dfs(A, x, y - 1, q);

dfs(A, x + 1, y, q);

dfs(A, x, y + 1, q);

}

};

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;

}

};

Posts published in “Graph”

花花酱 LeetCode 947. Most Stones Removed with Same Row or Column

Problem

Solution 2: Union Find

C++

Related Problems

花花酱 LeetCode 934. Shortest Bridge

Problem

Solution: DFS + BFS

C++

Related Problems

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