Walled Off π½ - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional integer matrix containing 0s and 1s where 0 represents empty space and 1 represents a wall.
Return the minimum number cells that need to become walls such that there's no path from the top left cell to the bottom right cell. You cannot put walls on the top left cell and the bottom right cell. You are only allowed to travel adjacently (no diagonal moves allowed), and you can't leave the matrix.
Constraints
2 β€ n, m β€ 250 where n and m are the number of rows and columns in matrix
Example 1
Input
matrix = [
[0, 1, 0, 0],
[0, 1, 0, 0],
[0, 0, 0, 0]
]
Output
1
Explanation
We can put a wall on either matrix[2][0], matrix[2][1], matrix[1][0] or matrix[2][2].
Solution :
Solution in C++ :
struct Edge {
int u, v;
int cap, flow;
Edge() {
}
Edge(int u, int v, int cap) : u(u), v(v), cap(cap), flow(0) {
}
};
struct Dinic {
int N;
vector<Edge> E;
vector<vector<int>> g;
vector<int> d, pt;
Dinic(int N) : N(N), E(0), g(N), d(N), pt(N) {
}
void AddEdge(int u, int v, int cap) {
if (u != v) {
E.emplace_back(u, v, cap);
g[u].emplace_back(E.size() - 1);
E.emplace_back(v, u, 0);
g[v].emplace_back(E.size() - 1);
}
}
bool BFS(int S, int T) {
queue<int> q({S});
fill(d.begin(), d.end(), N + 1);
d[S] = 0;
while (!q.empty()) {
int u = q.front();
q.pop();
if (u == T) break;
for (int k : g[u]) {
Edge &e = E[k];
if (e.flow < e.cap && d[e.v] > d[e.u] + 1) {
d[e.v] = d[e.u] + 1;
q.emplace(e.v);
}
}
}
return d[T] != N + 1;
}
int DFS(int u, int T, int flow = -1) {
if (u == T || flow == 0) return flow;
for (int &i = pt[u]; i < g[u].size(); ++i) {
Edge &e = E[g[u][i]];
Edge &oe = E[g[u][i] ^ 1];
if (d[e.v] == d[e.u] + 1) {
int amt = e.cap - e.flow;
if (flow != -1 && amt > flow) amt = flow;
if (int pushed = DFS(e.v, T, amt)) {
e.flow += pushed;
oe.flow -= pushed;
return pushed;
}
}
}
return 0;
}
int MaxFlow(int S, int T) {
int total = 0;
while (BFS(S, T)) {
fill(pt.begin(), pt.end(), 0);
while (int flow = DFS(S, T)) total += flow;
}
return total;
}
};
Dinic *dinic;
int solve(vector<vector<int>> &g) {
int r = g.size();
int c = g[0].size();
dinic = new Dinic(2 * r * c);
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
if (g[i][j]) continue;
dinic->AddEdge(i * c + j, i * c + j + r * c, 1);
int dx[4]{-1, 0, 1, 0};
int dy[4]{0, -1, 0, 1};
for (int k = 0; k < 4; k++) {
int nx = i + dx[k];
int ny = j + dy[k];
if (nx >= 0 && nx < r && ny >= 0 && ny < c && g[nx][ny] == 0) {
dinic->AddEdge(i * c + j + r * c, nx * c + ny, 1);
}
}
}
}
int ret = dinic->MaxFlow(r * c, r * c - 1);
delete dinic;
return ret;
}
Solution in Java :
import java.util.*;
class Solution {
int[][] dir = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
public int solve(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
if (!dfs(matrix, 0, 0, m, n))
return 0;
if (!dfs(matrix, 0, 0, m, n))
return 1;
return 2;
}
private boolean dfs(int[][] g, int x, int y, int m, int n) {
if (x == m - 1 && y == n - 1)
return true;
g[x][y] = 1;
for (int[] d : dir) {
int a = x + d[0], b = y + d[1];
if (a >= 0 && a < m && b >= 0 && b < n && g[a][b] == 0) {
if (dfs(g, a, b, m, n))
return true;
}
}
return false;
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
R = len(matrix)
C = len(matrix[0])
def get_neighbors(i, j):
for ii, jj in ((i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1)):
if 0 <= ii < R and 0 <= jj < C and matrix[ii][jj] == 0:
yield ii, jj
visited = set()
tin = {}
low = {}
timer = 0
articulation_points = set()
prev = {}
source = (0, 0)
target = (R - 1, C - 1)
def dfs(v, parent):
nonlocal timer
visited.add(v)
prev[v] = parent
tin[v] = timer
low[v] = timer
timer += 1
children = 0
for to in get_neighbors(*v):
if to == parent:
continue
if to in visited:
low[v] = min(low[v], tin[to])
else:
dfs(to, v)
low[v] = min(low[v], low[to])
if low[to] >= tin[v] and parent is not None:
articulation_points.add(v)
children += 1
if parent is None and children > 1:
articulation_points.add(v)
def bfs(root):
Q = deque([root])
visited = set([root])
while Q:
v = Q.pop()
if v == target:
return True
for w in get_neighbors(*v):
if w not in visited:
visited.add(w)
Q.appendleft(w)
return False
dfs(source, None)
if target not in prev:
return 0
for i, j in articulation_points:
matrix[i][j] = 1
if bfs(source):
return 2
return 1
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