Posts tagged as “BFS”

花花酱 LeetCode 934. Shortest Bridge

By zxi on November 4, 2018

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 909. Snakes and Ladders

By zxi on September 25, 2018

Problem

On an N x N board, the numbers from 1 to N*N are written boustrophedonically starting from the bottom left of the board, and alternating direction each row. For example, for a 6 x 6 board, the numbers are written as follows:

You start on square 1 of the board (which is always in the last row and first column). Each move, starting from square x, consists of the following:

You choose a destination square S with number x+1, x+2, x+3, x+4, x+5, or x+6, provided this number is <= N*N.
- (This choice simulates the result of a standard 6-sided die roll: ie., there are always at most 6 destinations.)
If S has a snake or ladder, you move to the destination of that snake or ladder. Otherwise, you move to S.

A board square on row r and column c has a “snake or ladder” if board[r][c] != -1. The destination of that snake or ladder is board[r][c].

Note that you only take a snake or ladder at most once per move: if the destination to a snake or ladder is the start of another snake or ladder, you do not continue moving. (For example, if the board is [[4,-1],[-1,3]], and on the first move your destination square is 2, then you finish your first move at 3, because you do notcontinue moving to 4.)

Return the least number of moves required to reach square N*N. If it is not possible, return -1.

Example 1:

Input: [
[-1,-1,-1,-1,-1,-1],
[-1,-1,-1,-1,-1,-1],
[-1,-1,-1,-1,-1,-1],
[-1,35,-1,-1,13,-1],
[-1,-1,-1,-1,-1,-1],
[-1,15,-1,-1,-1,-1]]
Output: 4
Explanation: 
At the beginning, you start at square 1 [at row 5, column 0].
You decide to move to square 2, and must take the ladder to square 15.
You then decide to move to square 17 (row 3, column 5), and must take the snake to square 13.
You then decide to move to square 14, and must take the ladder to square 35.
You then decide to move to square 36, ending the game.
It can be shown that you need at least 4 moves to reach the N*N-th square, so the answer is 4.

Note:

2 <= board.length = board[0].length <= 20
board[i][j] is between 1 and N*N or is equal to -1.
The board square with number 1 has no snake or ladder.
The board square with number N*N has no snake or ladder.

Solution: BFS

Time complexity: O(n*n)

Space complexity: O(n*n)

C++

// Author: Huahua, 8 ms (beats 100%)

class Solution {

public:

int snakesAndLadders(vector<vector<int>>& board) {

const int N = board.size();

int steps = 0;

vector<int> seen(N * N + 1, 0);

queue<int> q;

q.push(1);

seen[1] = 1;

while (!q.empty()) {

int size = q.size();

++steps;

while (size--) {

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

for (int x = s + 1; x <= min(s + 6, N * N); ++x) {

int r, c;

getRC(x, N, &r, &c);

int nx = board[r][c] == -1 ? x : board[r][c];

if (seen[nx]) continue;

if (nx == N * N) return steps;

q.push(nx);

seen[nx] = 1;

}

return -1;

}

private:

void getRC(int s, int N, int* r, int* c) {

int y = (s - 1) / N;

int x = (s - 1) % N;

*r = N - 1 - y;

*c = (y % 2) ? N - 1 - x : x;

}

};

花花酱 LeetCode 882. Reachable Nodes In Subdivided Graph

By zxi on August 4, 2018

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 542. 01 Matrix

By zxi on July 14, 2018

Problem

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1:
Input:

0 0 0
0 1 0
0 0 0

Output:

0 0 0
0 1 0
0 0 0

Example 2:
Input:

0 0 0
0 1 0
1 1 1

Output:

0 0 0
0 1 0
1 2 1

Note:

The number of elements of the given matrix will not exceed 10,000.
There are at least one 0 in the given matrix.
The cells are adjacent in only four directions: up, down, left and right.

Solution 1: DP

Two passes:

down, right
up, left

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Running time: 132 ms

class Solution {

public:

vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {

int m = matrix.size();

int n = matrix[0].size();

vector<vector<int>> ans(m, vector<int>(n, INT_MAX - m * n));

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

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

if (matrix[i][j]) {

if (i > 0) ans[i][j] = min(ans[i][j], ans[i - 1][j] + 1);

if (j > 0) ans[i][j] = min(ans[i][j], ans[i][j - 1] + 1);

} else {

ans[i][j] = 0;

}

for (int i = m - 1; i >= 0; --i)

for (int j = n - 1; j >= 0; --j) {

if (i < m - 1) ans[i][j] = min(ans[i][j], ans[i + 1][j] + 1);

if (j < n - 1) ans[i][j] = min(ans[i][j], ans[i][j + 1] + 1);

}

return ans;

}

};

Solution 2: BFS

Start from all 0 cells and find shortest paths to rest of the cells.

Time complexity: O(mn)

Space complexity: O(mn)

// Author: Huahua

// Running time: 136 ms

class Solution {

public:

vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {

const int m = matrix.size();

const int n = matrix[0].size();

queue<pair<int, int>> q;

vector<vector<int>> seen(m, vector<int>(n, 0));

vector<vector<int>> ans(m, vector<int>(n, INT_MAX));

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

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

if (matrix[i][j] == 0) {

q.push({i, j});

seen[i][j] = 1;

}

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

int steps = 0;

while (!q.empty()) {

int size = q.size();

while (size--) {

auto pair = q.front(); q.pop();

int i = pair.first;

int j = pair.second;

ans[i][j] = steps;

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

int x = j + dirs[k];

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

if (x < 0 || x >= n || y < 0 || y >= m || seen[y][x]) continue;

seen[y][x] = 1;

q.push({y, x});

}

++steps;

}

return ans;

}

};

Posts tagged as “BFS”

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 909. Snakes and Ladders

Problem

Solution: BFS

C++

花花酱 LeetCode 882. Reachable Nodes In Subdivided Graph

Problem

Solution: Dijkstra Shortest Path

花花酱 LeetCode 542. 01 Matrix

Problem

Solution 1: DP

Solution 2: BFS