## Algorithm¶

BFS search a graph by first explore all the vertices that are adjacent to the source vertex s, then explore the adjacent vertices of those already examined in layer-by-layer fashion until all the vertices are visited.

BFS (graph G, start vertex s)
[ all nodes initially unexplored ]
-- mark s as explored
-- let Q = queue data structure, initialized with s
-- while Q not empty:
-- remove the first node of Q, call it v
-- for each edge(v,w) :
-- if w unexplored
-- mark w as explored
-- add w to Q (at the end)


## BFS properties¶

1. Runtime complexity $\Theta(|V| + |E|)$.
2. It calculate shortest path from the source $s$ to the vertex $v$.
3. It build a breadth-first tree, CLRS called predecessor subgraph.
4. If edge $(u, v)$ in the BFS tree, the simple path from source vertext $s$ to $u$ is equal to the simple path from $s$ to $v$, or differ by 1.
5. Simple path from a source vertex $s$ to a destination vertex $v$ is equal to the later’s level label.
6. If $v_1, v_2, \cdots, v_n$ is the vertex enqueued in a certain point, $v_1$ is the head and $v_n$ is the back, then the level label of $v_1$ is either equal to level label of $v_n$ or less by 1. Also the level label of all the vertices in the queue is non-decreasing from head to tail. In other words, at a certain time the enqueued vertices at most comes from two different levels.
1. unweighted edges
2. Bi-directional search, (Speed up Dijkastra of 6.006)
3. At all time, the level of nodes in queue at most differ by 1
4. The relationship between BFS and DFS (The Maze II)

BFS tips for board and grid problems

1. When push a pair of coordinate to a BFS queue, one way of doing it is using pair<int, int>. Another way is to convert to a number by p = i * n + j, and retrieve the coordinate by i = p / n and j = p % n.
2. The BFS starting point and "target" are important, give a second thought about where should the search start and how it terminate.
3. visitted[m][n] is not necessary sometimes. See whether you can modify the original borad/matrix/grid/maze to reflect whether the element has already been accessed or not.
4. In a borad or maze like input, searching neighbor cells requires you to manipulate indexes for accessing its neighbors. We do this by defining two "offset" coordinates arrays. int x = {-1, 0, 1, 0}; int y = {0, 1, 0, -1};
5. The BFS queue doesn't necessarily start with one element pushed. It could be "multi-end" BFS. Like in problme Pacific Atlantic Water Flow and Longest Increasing Path in a Matrix, one could init the queue with all possible starting node then in use a while to visit all the node towards finding the optimal path.
6. The "multi-end" BFS not only can make the BFS level by level (each round of for loop visited one level of nodes), but also can make use some optimal feature of BFS, like the BFS solution of Walls and Gates.
7. The BFS while loop could include another while loop to interate throught total q.size() nodes. This is different from the "oracle" BFS solution from CRLS, but sometime could be useful. For example: Longest Increasing Path in a Matrix.

### The Maze¶

class Solution {
public:
bool hasPath(vector<vector<int>>&maze, vector<int>& start, vector<int>& destination) {
int m = maze.size();
int n = m ? maze.size() : 0;
vector<vector<bool>> visited(m, vector<bool>(n, false));

return dfs_helper(maze, start, destination, visited);
}

void dfs_helper(vector<vector<int>>& maze, vector<int>& start,
vector<int>& destination, vector<vector<bool>>& visited) {

if (start == destination && start == destination) {
return true;
}

int x = {0, -1, 0, 1};
int y = {-1, 0, 1, 0};

bool res = false;
for (int k = 0; k < 4; k++) {
int a = start + x[k];
int b = start + y[k];

// walk untill hit the wall
while (a >= 0 && b >= 0 && a < maze.size() && b < maze.size90 && maze[a][b] == 0) {
a += x[k];
b += y[k];
}

if (!visited[a - x[k]][b - y[k]]) {
visited[a - x[k]][b - y[k]] = true;
vector<int> new_start({a - x[k], b - y[k]});
res != dfs_helper(maze, new_start, destination, visited);
}
}

return res;
}
};


=== "BFS"

c++
class Solution {
public:
bool hasPath(vector<vector<int>>&maze, vector<int>& start, vector<int>& destination) {
int m = maze.size();
int n = m ? maze.size() : 0;

vector<vector<bool>> visited(m, vector<bool>(n, false));
queue<vector<int>> q;
q.push(start);

int x = {0, -1, 0, 1};
int y = {-1, 0, 1, 0};

while (!q.empty()) {
vector<int> s = q.front(); q.pop();

if (s == destination && s == destination) {
return true;
}

for (int k = 0; k < 4; k++) {
int a = s + x[k];
int b = s + y[k];

while (a >= 0 && b >= 0 && a < m && b < n && maze[a][b] == 0) {
a += x[k];
b += y[k];
}

if (!visited[a - x[k]][b - y[k]]) {
q.push(vector<int>({a - x[k], b - y[k]}));
visited[a - x[k]][b - y[k]] = true;
}
}
}

return false;
}
};



### The Maze II¶

• BFS can computer shortest path, we need additional memory to keep the optimal values.
• we update the corresponding destination cell in distance vector when we found a smaller distane can reach it from any direction. the final result will be in distance[destination][destination].
• If we don't want to use so much memory, we can use a flag to mark when the BFS switch levels.
• This solution is essentially to use BFS to solve the weighted graph problem as Victor mentioned in 6006 R15.
• Compare this solution to the Dijkastra solution. The update distance matrix is essentially the relaxiation step.
class Solution {
public:
int shortestDistance(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
int m = maze.size();
int n = m ? maze.size() : 0;
if (m == 0) return 0;

vector<vector<int>> distance(m, vector<int>(n, INT_MAX));
queue<vector<int>> q;
q.push(start);
distance[start][start] = 0;

int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};

while (!q.empty()) {
vector<int> s = q.front(); q.pop();

for (int k = 0; k < 4; k++) {
int a = s + x[k];
int b = s + y[k];
int count = 0;
while (a >= 0 && b >= 0 && a < m && b < n && maze[a][b] == 0) {
a += x[k];
b += y[k];
count++;
}

// notice here is a optimize
if (distance[s][s] + count < distance[a - x[k]][b - y[k]]) {
distance[a - x[k]][b - y[k]] = distance[s][s] + count;
q.push(vector<int>({a - x[k], b - y[k]})); /* could use emplace({a - x[k], b - y[k]}) */
}
}
}

return distance[destination][destination] == INT_MAX ? -1 : distance[destination][destination];
}
};


Note

• Based on the BFS solution, we can use a priority queue. Then the problem is very similar to the Dijkastra algorithm.
• The discussion about the improvement in leetcode forum is very interesting.

### The Maze III¶

• This problem is aksing for output the shortest path to reach the destination.
• The idea is to use a distance[i][i] and result[i][j]. we should keep update the distance[i][j] and result[i][j] with smaller distance and lexicographical order. The hardest part is that there is two constrains. While using BFS, How could we ensure the two constrains are met, and also we are not missing any possible path. (i.e. missed to push a path to the BFS queue.)
• A trick we can make use of is we searching in the lexicographically order. such as down, left, right, up. However, we face a problem: the smallest lexicon order result may not be the shortest distance. Think this through.
• We need to used additional memory to record that piece of information beside the memory used to keep the minimum distance.
class Solution {
public:
string findShortestWay(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
int m = maze.size();
int n = m ? maze.size() : 0;

if (m == 0) return 0;

vector<vector<int>> distance(m, vector<int>(n, INT_MAX));
vector<vector<string>> result(m, vector<string>(n, ""));
queue<vector<int>> q;
q.push(start);
distance[start][start] = 0;

char path = {'d', 'l', 'r', 'u'};
int x = {1, 0, 0, -1};
int y = {0, -1, 1, 0};

while (!q.empty()) {
vector<int> s = q.front(); q.pop();

for (int k = 0; k < 4; k++) {
int a = s + x[k];
int b = s + y[k];
int count = 0;
/* continue if no wall */
while (a >= 0 && b >= 0 && a < m && b < n && maze[a][b] == 0) {
/* run over the destination */
if (a == destination && b == destination){
if (distance[s][s] + count < distance[a][b]) {
distance[a][b] = distance[s][s] + count;
result[a][b] = result[s][s] + path[k];
} else if (distance[s][s] + count == distance[a][b]) {
string tmp = result[s][s] + path[k];
/* same distance, smaller lexicon */
if (tmp.compare(result[a][b]) < 0) {
result[a][b] = tmp;
}
}
}

a += x[k];
b += y[k];
count++;
}

/* hit a wall, have to retreat "0", current a, b is in wall or off maze */
if (distance[s][s] + count < distance[a - x[k]][b - y[k]]) {
distance[a - x[k]][b - y[k]] = distance[s][s] + count;
result[a - x[k]][b - y[k]] = result[s][s] + path[k];
q.push(vector<int>({a - x[k], b - y[k]}));
} else if (distance[s][s] + count == distance[a - x[k]][b - y[k]]) {
string tmp = result[s][s] + path[k];
/* same distance, smaller lexicon */
if (tmp.compare(result[a - x[k]][b - y[k]]) < 0) {
result[a - x[k]][b - y[k]] = tmp;
/* possible path, need to search it again */
q.push(vector<int>({a - x[k], b - y[k]}));
}
}
}
}

return distance[destination][destination] == INT_MAX ? "impossible" : result[destination][destination];
}
};

class Solution {
public:
string findShortestWay(vector<vector<int>>& maze, vector<int>& ball, vector<int>& hole) {
return roll(maze, ball, ball, hole, 0, 0, 0, "", pair<string, int>() = {"impossible", INT_MAX});
}

string roll(vector<vector<int>>& maze, int rowBall, int colBall, const vector<int>& hole,
int distRow, int distCol, int steps, const string& path, pair<string, int>& res)
{
if (steps < res.second) {
if (distRow != 0 || distCol != 0) {
while ((rowBall + distRow) >= 0 && (rowBall + distRow) < maze.size() &&
(colBall + distCol) >= 0 && (colBall + distCol) < maze.size() &&
maze[rowBall + distRow][colBall + distCol] != 1) {

rowBall += distRow;
colBall += distCol;
++steps;
if (rowBall == hole && colBall == hole && steps < res.second) {
res = {path, steps};
}
}
}

if (maze[rowBall][colBall] == 0 || steps + 2 < maze[rowBall][colBall]) {
maze[rowBall][colBall] = steps + 2;
if (distRow == 0) roll(maze, rowBall, colBall, hole, 1, 0, steps, path + "d", res);
if (distCol == 0) roll(maze, rowBall, colBall, hole, 0, -1, steps, path + "l", res);
if (distCol == 0) roll(maze, rowBall, colBall, hole, 0, 1, steps, path + "r", res);
if (distRow == 0) roll(maze, rowBall, colBall, hole, -1, 0, steps, path + "u", res);
}
}

return res.first;
}
};

class Vertex
{
public:
int x, y, dist;
std::string path;
Vertex(int x, int y, int dist, std::string path)
{
this->x = x;
this->y = y;
this->dist = dist;
this->path = path;
}
Vertex(int x, int y)
{
this->x = x;
this->y = y;
this->dist = INT_MAX;
this->path = "";
}
bool equals(Vertex v)
{
return this->x == v.x && this->y == v.y;
}
bool equals(int x, int y)
{
return this->x == x && this->y == y;
}
};

class Solution {
private:
// coordinates and path strings for the four directions.
std::vector<int> directions = {-1, 0, 1, 0, -1};
std::vector<std::string> paths = {"u", "r", "d", "l"};
// can't roll if it's boundry, wall or hole.
bool canRoll(vector<vector<int>> &maze, Vertex hole, int x, int y) {
int m = maze.size(), n = maze.size();
if (x >= 0 && x < m && y >= 0 && y < n)
return maze[x][y] != 1 && !hole.equals(x, y);
else
return false;
}

Vertex roll(vector<vector<int>> &maze, Vertex curr, Vertex hole, int dx,
int dy, std::string path) {
int x = curr.x, y = curr.y, steps = curr.dist;
while (canRoll(maze, hole, x + dx, y + dy))
{
x += dx;
y += dy;
steps++;
}
// if couldn't roll due to hole, then return vertex corresponding to the hole.
if (hole.equals(x + dx, y + dy))
{
x += dx;
y += dy;
}
return Vertex(x, y, steps, curr.path + path);
}
public:
/*
* to form a pq based on the increasing order of distances, and if same
* distance then sorting based on lexicographic order.
*/
struct Compare
{
bool operator()(const Vertex &a, const Vertex &b)
{
return a.dist > b.dist || a.dist == b.dist && a.path > b.path;
}
};
/*
* Applying Dijkstra's algorithm here.
* We treat hole as a gate here. The trick is look for holes while rolling
* the back in a particular direction. So, the pq here contains the vertices
* before gates as well as the holes. So, using dijkstra, when we pop a hole
* from pq, we know its that.
*/
string findShortestWay(vector<vector<int>>& maze, vector<int>& ball, vector<int>& hole) {
int m = maze.size(), n = maze.size();
std::priority_queue<Vertex, std::vector<Vertex>, Compare> pq;
Vertex startVertex(ball, ball, 0, ""), holeVertex(hole, hole);
std::unordered_set<int> visited;
pq.push(startVertex);
while (!pq.empty())
{
Vertex curr = pq.top();
visited.insert(curr.x * n + curr.y);
pq.pop();
if (curr.equals(holeVertex))
{
return curr.path;
}
for (int i = 0; i < directions.size() - 1; i++)
{
Vertex newStart = roll(maze, curr, holeVertex, directions[i], directions[i + 1], paths[i]);
if (visited.find(newStart.x * n + newStart.y) == visited.end())
pq.push(newStart);
}
}
return "impossible";
}
};


### Pacific Atlantic Water Flow¶

• Using DFS and two matrices to record the states and then to loop through the result check for both possible flow cases.
• The key is how to start the DFS? should we initiate the DFS in each (i, j)? Here we can see it is more efficient if that we start from the edges. Notice we mark visited[i][j] = true immediately enter the helper function.
class Solution {
public:
vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = m ? matrix.size() : 0;

vector<vector<int>> p_visited(m, vector<int>(n, 0));
vector<vector<int>> a_visited(m, vector<int>(n, 0));
vector<pair<int, int>> res;

for (int i = 0; i < m; i++) {
helper(matrix, i, 0, p_visited);
helper(matrix, i, n - 1, a_visited);
}

for (int j = 0; j < n; j++) {
helper(matrix, 0, j, p_visited);
helper(matrix, m - 1, j, a_visited);
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (p_visited[i][j] && a_visited[i][j]) {
//res.push_back(pair<int, int>(i, j));
res.push_back({i, j});
}
}
}

return res;
}

void helper (vector<vector<int>>& mat, int i, int j, vector<vector<int>>& visited) {
int m = mat.size();
int n = m ? mat.size() : 0;
int x[] = {0, -1, 0, 1};
int y[] = {-1, 0, 1, 0};

visited[i][j] = true; /* visite mat[i][j] first */
/* and then explore its neighbors */
for (int k = 0; k < 4; k++) {
int p = i + x[k];
int q = j + y[k];
if (p >= 0 && p < m && q >= 0 && q < n && !visited[p][q] && mat[i][j] <= mat[p][q]) {
helper(mat, p, q, visited);
}
}
}
};

class Solution {
public:
vector<pair<int, int>> pacificAtlantic(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = m ? matrix.size() : 0;

vector<vector<bool>> vPacific(m, vector<bool>(n, false));
vector<vector<bool>> vAtlantic(m, vector<bool>(n, false));
queue<pair<int, int>> qPacific;
queue<pair<int, int>> qAtlantic;

for (int i = 0; i < m; ++i) {
qPacific.push({i, 0});
vPacific[i] = true;
qAtlantic.push({i, n - 1});
vAtlantic[i][n - 1] = true;
}

for (int i = 0; i < n; ++i) {
qPacific.push({0, i});
vPacific[i] = true;
qAtlantic.push({m - 1, i});
vAtlantic[m - 1][i] = true;
}

bfs_helper(matrix, qPacific, vPacific);
bfs_helper(matrix, qAtlantic, vAtlantic);

vector<pair<int, int>> res;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (vPacific[i][j] && vAtlantic[i][j]) {
res.push_back({i, j});
}
}
}

return res;
}

void bfs_helper(vector<vector<int>>& matrix, queue<pair<int, int>>& q, vector<vector<bool>>& visited) {
int m = matrix.size();
int n = m ? matrix.size() : 0;
int x = {0, -1, 0, 1};
int y = {-1, 0, 1, 0};

while (!q.empty()) {
int a = q.front().first;
int b = q.front().second;
q.pop();
for (int k = 0; k < 4; ++k) {
int u = a + x[k];
int v = b + y[k];

if (u < 0 || u >= m || v < 0 || v >= n || visited[u][v] || matrix[a][b] > matrix[u][v]) continue;
visited[u][v] = true;
q.push({u, v});
}
}
}
};


Warning

Notice your terminate condition matrix[a][b] > matrix[u][v] in the DFS solution indicates your search direction cannot not go from high to low. The search direction is from low to high (or equal). This is to optimize the search since we don't have to revisite a node later.

### Longest Increasing Path in a Matrix¶

BFS solution

• Each cell should possibly be the starting cell of the longest increasing path. So we have to search starting from each one of the cells. It seems the running time is quadratic. How to optimize?
• use markers matrix to mark whether the matrix[i][j] have greater neighbors, left, top, right, bottom -> 1, 2, 4, 8.
• The do BFS only looking for the directions that have greater neighbors. The idea is very similar to the "branch cutting" in DFS.
class Solution {
public:
int longestIncreasingPath(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = m ? matrix.size() : 0;

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

int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};
int res = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < 4; k++) {
int p = i + x[k];
int q = j + y[k];
if (p < 0 || p >= m || q < 0 || q >= m) continue;
if (matrix[p][q] > matrix[i][j])
markers[i][j] ^= (1 << k);
}
}
}

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int level = 0;
if (markers[i][j] > 0) {
level = 0;
queue<int> q;
q.push(i * n + j);
while (!q.empty()) {
int a = q.front() / n;
int b = q.front() % n;
q.pop();
bool flag = true;
for (int k = 0; k < 4; k++) {
int u = a + x[k];
int v = b + y[k];
if (u < 0 || u >= m || v < 0 || v >= m) continue;
if (has_greater_neighbor(markers[a][b], k)) {
q.push(u * n + v);
if (flag) {
level++;
flag = false;
}
}
}
}
}
res = max(res, level);
}
}

return res;
}

bool has_greater_neighbor(int flags, int k) {
return (flags >> k) & 1;
}
};


BFS solution II

• BFS can find the shortest path. How can you use BFS to find the longest path? In this problem, we are not use BFS's shortest path properpy, we make use of BFS's property of finding the connected components. We can imagine each increasing path is a connected component, we need to find out the largest one. Notice how the solution construct the search structure (for loop inside the while loop).
• By construct the search structure in this way, the len is actually the level of nodes from all the peaks identified in the first nested for loops.
class Solution {
public int longestIncreasingPath(int[][] matrix) {
private final int[][] dirs = {{1, 0}, {-1, 0},{0, 1}, {0, -1}};
private boolean ispeak(int[][] matrix, boolean[][] marked, int i, int j) {
if (i > 0 && !marked[i-1][j] && matrix[i-1][j] > matrix[i][j]) return false;
if (i < matrix.length-1 && !marked[i+1][j] && matrix[i+1][j] > matrix[i][j]) return false;
if (j > 0 && !marked[i][j-1] && matrix[i][j-1] > matrix[i][j]) return false;
if (j < matrix.length-1 && !marked[i][j+1] && matrix[i][j+1] > matrix[i][j]) return false;
return true;
}
public int longestIncreasingPath(int[][] matrix) {
if (matrix.length == 0 || matrix.length == 0) return 0;
int len = 0;
boolean[][] marked = new boolean[matrix.length][matrix.length];
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix.length; j++) {
if (ispeak(matrix, marked, i, j)) queue.add(new int[]{i, j});
}
}
while (!queue.isEmpty()) {
len++;
int size = queue.size();
for (int i = 0; i < size; i++) {
int[] p = queue.poll();
marked[p][p] = true;
for (int j = 0; j < 4; j++) {
int r = p+dirs[j], c = p+dirs[j];
if (r >= 0 && r < matrix.length && c >= 0 && c < matrix.length && !marked[r][c] && ispeak(matrix, marked, r, c)) {
if (matrix[r][c] != matrix[p][p]) queue.add(new int[]{r, c});
}
}
}
}
return len;
}
}
}


Note

This BFS is not the same as the classic BFS routine from CLRS. It is a generalized BFS approach, which can be transformed to a problem to find the connected components.

DFS solution

• we can also use DFS for this problem, it turns out DFS is the simpliest solution.
class Solution {
public:
int longestIncreasingPath(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = m ? matrix.size() : 0;

if (n == 0) return 0;
vector<vector<int>> cache(m, vector<int>(n, 0));
int res = 1;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
int len = dfs_helper(matrix, i, j, m, n, cache);
res = max(res, len);
}
}

return res;
}

int dfs_helper(vector<vector<int>>& matrix, int i, int j, int m, int n, vector<vector<int>>& cache) {
if (cache[i][j]) return cache[i][j];
int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};
int res = 1;
for (int k = 0; k < 4; k++) {
int p = i + x[k];
int q = j + y[k];
if (p < 0 || p >= m || q < 0 || q >= n || matrix[p][q] <= matrix[i][j]) continue;
int len = 1 + dfs_helper(matrix, p, q, m, n, cache);
res = max(res, len);
}
cache[i][j] = res;
return res;
}
};


### Walls and Gates¶

BFS solution

• We push all the gate to the queue at once, then BFS to update the room. This makes use of the optimal character of BFS and bring us some optimization.
• Proof correctness: If init the queue with all gates, BFS alternate between gates, once BFS explore one level of rooms for a particular gate, it switches to explore the same level or next level rooms for another gate. The level is the distance from room to gate. Suppose room R1 is visited at the first time from G1, then if R1 is visited again via G2, the distance from G2 to R1 is possible to be the same as G1 to R1, or greater than G1 to R1 by 1. The assigned nearest distance from room to gate never be able to reduce further in future. Then the first assigned distance value for a room will be the nearest one.

DFS solution

• I initially start with each INF cell and search for 0. the issue is we have to pass the original i, j, in case we reach the base case to update the INF cell.
• Another trick is how to handle the repeated searching the same empty room. Extra space like visited[m][n] will work. but a better way to handle it is what we did in the below solution. We check if d > ma[i][j], if d > 0, the d > mat[i][j] also make sure we update mat[i][j] with a smaller value because the initial value of INF cell is INT_MAX. This is different from DFS because DFS can ensure the shortest path.
• The time complexity is $O(m^2n^2)$. The space complexity is from the stack. It is definitely not constant.
class Solution {
public:
void wallsAndGates(vector<vector<int>>& rooms) {
int m = rooms.size();
int n = m ? rooms.size() : 0;

int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};

queue<pair<int, int>> q;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (rooms[i][j] == 0) { // start from each gate
q.push({i, j});
}
}
}

while (!q.empty()) {
int a = q.front().first;
int b = q.front().second;
q.pop();
for (int k = 0; k < 4; k++) {
int u = a + x[k];
int v = b + y[k];
if (u < 0 || u >= m || v < 0 || v >= n || rooms[u][v] != INT_MAX) {
continue;
}

rooms[u][v] = rooms[a][b] + 1;
q.push({u, v});
}
}
}
};

class Solution {
public:
void wallsAndGates(vector<vector<int>>& rooms) {
int m = rooms.size();
int n = m ? rooms.size() : 0;

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (rooms[i][j] == 0) { // start from each gate, not room
helper(rooms, i, j, 0);
}
}
}
}
/* i, j are the starting point. */
void helper(vector<vector<int>>& mat, int i, int j, int d) {
int m = mat.size();
int n = m ? mat.size() : 0;

int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};

if (i < 0 || i >= m || j < 0 || j >= n || mat[i][j] < d) {
return;
}
mat[i][j] = d;
for (int k = 0; k < 4; k++) {
int a = i + x[k];
int b = j + y[k];
helper(mat, a, b, d + 1);
}
}
};


multi-end BFS

Notice this implementation of BFS is different from triditional BFS. It start with a queue that have more than one element. Some people call this multi-end BFS. It is perform better than the original BFS, The time complexity is $O(mn)$

### Surrounded Regions¶

• We first mark the boundary O, then start BFS from each O. If it cannot reach the special marks, we capture the cell.
• This solution is starting from each of inner O and search the mark at the boundary. The solution is TLE due to more inner O's than boundary O's.
class Solution {
public:
void solve(vector<vector<char>>& board) {
int m = board.size();
int n = m ? board.size() : 0;
int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0 || i == m - 1 || j == n - 1) {
board[i][j] = board[i][j] == 'O' ? '-' : 'X';
}
}
}

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == 'O') {
bool flag = true;
queue<pair<int, int>> q;
q.push({i, j});
while (!q.empty()) {
int a = q.front().first, b = q.front().second;
q.pop();
for (int k = 0; k < 4; k++) {
int u = a + x[k];
int v = b + y[k];
if (u < 0 || u >= m || v < 0 || v >= n || board[u][v] == 'X') {
continue;
}

if (board[u][v] == 'O') {
q.push({u, v});
}

if (board[u][v] == '-') {
flag = false;
break;
}
}
if (!flag)
break;
}
if (flag)
board[i][j] = 'X';
}
}
}

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0 || i == m - 1 || j == n - 1) {
board[i][j] = board[i][j] == '-' ? 'O' : 'X';
}
}
}
}
};

• Instead searching whether a 'O' have an outlet. We can discover all 'O's that lead to edge 'O's and mark them, then all the rest 'O's are captured.
• Specifically, starting from every 'O' at the edge, using BFS to search all the 'O's outside the 'X's, all these 'O's that cannot be captured. Compare to solution 1, this ideas is thinking “out of the box”, instead of thinking inside 'O's, we take care outside 'O's first.
• Similar problems: Walls and Gates, Shortest Distance from All Buildings.
class Solution {
public:
void solve(vector<vector<char>>& board) {
int m = board.size();
int n = m ? board.size() : 0;

// bfs search from boundary 'O's
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (i == 0 || j == 0 || i == m - 1 || j == n - 1) {
if (board[i][j] == 'O')
bfs_helper(board, i, j);
}
}
}

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == 'O') {
board[i][j] = 'X'; // capture all inner 'O's
} else if (board[i][j] == '-') {
board[i][j] = 'O'; // flip back the marks
}
}
}
}

// helper function to mark "outside" 'O' as '-'
void bfs_helper(vector<vector<char>>& board, int i, int j) {
int m = board.size();
int n = m ? board.size() : 0;
int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};

board[i][j] = '-';
queue<pair<int, int>> q;
q.push({i, j});
while (!q.empty()) {
int a = q.front().first, b = q.front().second;
q.pop();
for (int k = 0; k < 4; k++) {
int u = a + x[k];
int v = b + y[k];
if (u >= 0 && u < m && v >= 0 && v < n && board[u][v] == 'O') {
q.push({u, v});
board[u][v] = '-';
}
}
}
}
};


Tips

### Best Meeting Point¶

• At the first attempt, I tried to use BFS to do search level by level, which can exhaust all the possible distances from 1 to 0. Unfortunately, it is TLE in the OJ.
• Following from the idea in problem Minimum Moves to Equal Array Elements II, we can use math method to calculate the median of the x and y coordinates.
• The worst case time complexity is $O(mn\log mn)$, since there are at most $m\times n$ "1" in the grid.
class Solution {
public:
int minTotalDistance(vector<vector<int>>& grid) {
int m = grid.size();
int n = m ? grid.size() : 0;

vector<int> rows, cols;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
rows.push_back(i);
cols.push_back(j);
}
}
}

return minDistance(rows) + minDistance(cols);
}

int minDistance(vector<int> vec) {
int n = vec.size();
sort(vec.begin(), vec.end());
int l = 0, r = n - 1;
int res = 0;
while (l < r) res += vec[r--] - vec[l++];

return res;
}
};

class Solution {
public:
int minTotalDistance(vector<vector<int>>& grid) {
int m = grid.size();
int n = m ? grid.size() : 0;

vector<int> rows, cols;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
rows.push_back(i);
cols.push_back(j);
}
}
}

sort(cols.begin(), cols.end());
int l = 0, r = rows.size() - 1;
int res = 0;
while (l < r)
res += rows[r] - rows[l] + cols[r--] - cols[l++];

return res;
}
};

class Solution {
public:
int minTotalDistance(vector<vector<int>>& grid) {
int m = grid.size();
int n = m ? grid.size() : 0;
int res = INT_MAX;

vector<vector<int>> dist(m, vector<int>(n, 0));
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
bfs_helper(grid, i, j, dist);
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (res > dist[i][j]) {
res = dist[i][j];
}
}
}

return res;
}

void bfs_helper(vector<vector<int>>& grid, int i, int j, vector<vector<int>> &dist) {
int m = grid.size();
int n = m ? grid.size() : 0;

int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};

vector<vector<bool>> visited(m, vector<bool>(n, false));

int level = 1;
queue<pair<int, int>> q;
q.push({i, j});
visited[i][j] = true;
while (!q.empty()) {
int len = q.size();
for (int l = 0; l < len; l++) {
int a = q.front().first;
int b = q.front().second;
q.pop();
for (int k = 0; k < 4; k++) {
int u = a + x[k];
int v = b + y[k];
if (u >= 0 && u < m && v >= 0 && v < n && !visited[u][v]) {
dist[u][v] += level;
visited[u][v] = true;
q.push({u, v});
}
}
}
level++;
}
}
};


### 399. Evaluate Division¶

class Solution {
public:
vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
unordered_map<string, vector<pair<string, double>>> graph;
for (int i = 0; i < equations.size(); i++) {

graph[equations[i]].push_back({equations[i], values[i]});
graph[equations[i]].push_back({equations[i], 1.0});
graph[equations[i]].push_back({equations[i], 1.0 / values[i]});
graph[equations[i]].push_back({equations[i], 1.0});

}

vector<double> res;

for (auto& qr: queries ) {
if (graph.find(qr) == graph.end() || graph.find(qr) == graph.end()) {
res.push_back(-1);
continue;
}

queue<pair<string, double>> q;
unordered_set<string> visited;
q.push({qr, 1.0});
visited.insert(qr);
bool find = false;
while (!q.empty() && !find) {
auto t = q.front(); q.pop();
if (t.first == qr) {
res.push_back(t.second);
find = true;
break;
}

for (auto a: graph[t.first]) {
if (visited.find(a.first) == visited.end()) {
a.second *= t.second;
q.push(a);
visited.insert(a.first);
}
}
}
if (!find) res.push_back(-1.0);
}

return res;
}
};

class Solution {
public:
vector<double> calcEquation(vector<pair<string, string>> equations,
vector<double>& values, vector<pair<string, string>> queries) {

int m = equations.size();
int n = queries.size();
vector<double> res(n, -1);
set<string> s;
for (auto equation : equations) {
s.insert(equation.first);
s.insert(equation.second);
}

for (int i = 0; i < n; i++) {
vector<string> query({queries[i].first, queries[i].second});
if (s.count(query) && s.count(query)) {
vector<bool> visited(m, 0);
res[i] = helper(equations, values, query, visited);
}
}

return res;
}

// Parameter visited here is to help the search.
double helper (vector<pair<string, string>> equations, vector<double>& values, vector<string> query, vector<bool>visited) {
int m = equations.size();

//base case, writing in seperate loop is more efficient O(n)
for (int i = 0; i < m; i++) {
if (equations[i].first == query && equations[i].second == query) {
return values[i];
}

if (equations[i].first == query && equations[i].second == query) {
return 1 / values[i];
}
}

for (int i = 0; i < m; i++) {
if (equations[i].first == query && !visited[i]) {
visited[i] = true;
double t = values[i] * helper(equations, values, {equations[i].second, query}, visited);
if (t > 0) return t;
visited[i] = false;
}

if (equations[i].second == query && !visited[i]) {
visited[i] = true;
double t = (1 / values[i]) * helper(equations, values, {equations[i].first, query}, visited);
if (t > 0) return t;
visited[i] = false;
}
}

return -1.0;
}
};




## Compute shortest path for undirected graphs¶

### Shortest Distance from All Buildings¶

• Similar to Walls and Gates, We should search from the "dst" (buildings) to "src" (empty land), while the problem is asking search from empty land to buildings.
• We need a 2D array to record the distance obtained by searching. The result is obtained at last after we finish the searching. Attention: we have also need to check whether the new building can reach all buildings. So use another 2D array and a counter num_buildings to check it can reach all.
class Solution {
public:
int shortestDistance(vector<vector<int>>& grid) {
int m = grid.size();
int n = m ? grid.size() : 0;

int num_buildings = 0;
vector<vector<int>> dist1(m, vector<int>(n, 0));
vector<vector<int>> reach(m, vector<int>(n, 0));

int x = {-1, 0, 1, 0};
int y = {0, 1, 0, -1};

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
num_buildings++;
queue<int> q;
vector<vector<bool>> visited(m, vector<bool>(n, false));

q.push(i * n + j);
int level = 1; // grid[a][b] is level 0, grid[u][v] is level 1
while (!q.empty()) {
int l = q.size();
for (int s = 0; s < l; s++) { // BFS level by level
int a = q.front() / n;
int b = q.front() % n;
q.pop();
for (int k = 0; k < 4; k++) {
int u = a + x[k];
int v = b + y[k];
if (u >= 0 && u < m && v >= 0 && v < n && grid[u][v] == 0 && !visited[u][v]) {
dist1[u][v] += level;
reach[u][v]++; // use to count num of 1s that reach this 0.
visited[u][v] = true;
q.push(u * n + v);
}
}
}
level++;
}
}
}
}

int res = INT_MAX;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 0 && reach[i][j] == num_buildings) {
res = min(res, dist1[i][j]);
}
}
}

return res == INT_MAX ? -1 : res;
}
};

• We can modify the original 2D array to record which cell is visited and which isn't so as to optimize the space complexity. Each time we start BFS from a building grid[i][j]=1, the land will be decreased by 1. So after the first BFS, all grid[i][j]=0 becomes grid[i][j]=-1. The decreasing of grid[i][j] is equivalent to use the visited variables.
• similar to Solution 1, we use a 2D array dist[i][j] to record the distances. This distance is dist[i][j] used to record each BFS started from a building (grid[i][j]=1), not the global distances. To accumulate all the distance from 1s, we have to use another 2D array sum to record it.
class Solution {
public:
int shortestDistance(vector<vector<int>>& grid) {
int m = grid.size();
int n = m ? grid.size() : 0;

int res = INT_MAX;

int counter = 0;
vector<vector<int>> sum = grid;

int x = {0, -1, 0, 1};
int y = {-1, 0, 1, 0};

for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
vector<vector<int>> dist = grid;
queue<int> q;
q.push(i * n + j);
res = INT_MAX;
while (!q.empty()) {
int a = q.front() / n;
int b = q.front() % n;
q.pop();
for (int k = 0; k < 4; ++k) {
int u = a + x[k];
int v = b + y[k];
if (u < m && u >= 0 && v < n && v >= 0 && grid[u][v] == counter) {
grid[u][v]--; // Mark this potential building position is visited in prev round
dist[u][v] = dist[a][b] + 1;
sum[u][v] += dist[u][v] - 1;
res = min(res, sum[u][v]);
q.push(u * n + v);
}
}
}
counter--; // mark as visited
}
}
}

return res == INT_MAX ? -1 : res;
}
};


## Compute connected components for undirected graphs¶

### Number of Connected Components in an Undirected Graph¶

• Build the undirected graph using adjacency list from the edge list representation.
• Call BFS in a for loop and use a visited vector to mark the status of the vertex explorations.
• Direct transform the above BFS solution to a DFS solution.
class Solution {
public:
int countComponents(int n, vector<pair<int, int>>& edges) {
vector<vector<int>> graph(n, vector<int>(0));
vector<bool> visited(n, false);

for (auto edge : edges) {
graph[edge.first].push_back(edge.second);
graph[edge.second].push_back(edge.first);
}

queue<int> q;
int count = 0;
for (int i = 0; i < n; ++i) {
if (!visited[i]) {
q.push(i);
visited[i] = true;
while (!q.empty()) {
int t = q.front(); q.pop();
for (auto a : graph[t]) {
if (!visited[a]) {
visited[a] = true;
q.push(a);
}
}
}

count++;
}
}

return count;
}
};

class Solution {
public:
int countComponents(int n, vector<pair<int, int>>& edges) {
vector<vector<int>> graph(n, vector<int>(0));
vector<bool> visited(n, false);
int count = 0;

for (auto edge : edges) {
graph[edge.first].push_back(edge.second);
graph[edge.second].push_back(edge.first);
}

for (int i = 0; i < n; ++i) {
if (!visited[i]) {
dfs_helper(graph, visited, i);
count++;
}
}

return count;
}

void dfs_helper(vector<vector<int>>& graph, vector<bool>& visited, int i) {
visited[i] = true;
for (auto a : graph[i]) {
if (!visited[a]) {
dfs_helper(graph, visited, a);
}
}
}
};

class Solution {
vector<int> parent;
public:
int countComponents(int n, vector<pair<int, int>>& edges) {
parent = vector<int>(n, -1);

for (int i = 0; i < n; ++i) parent[i] = i;

for (auto edge : edges) {
int p = root(edge.first);
int q = root(edge.second);
if (p != q) parent[q] = p;
}

int count = 0;
for (int i = 0; i < n; ++i) {
if (parent[i] == i) {
count++;
}
}

return count;
}

int root(int i) {
while (i != parent[i]) i = parent[i];
return i;
}
};