The Meeting Place Sequel - Amazon Top Interview Questions
Problem Statement :
You are given a two dimensional integer matrix consisting of:
0 which represents an empty cell.
1 which represents a wall.
2 which represents a person.
A person can walk up, down, left, right or stay in one time unit. Find a walkable cell such that it minimizes the time it would take for everyone to meet and return the time.
Note: two people can walk through the same empty cell and you can assume there is always some path between any two people.
Constraints
The number of cells in the matrix is at most 100,000
Any two people are at most 30 travel distance away
There are at most 30 people in the matrix
Example 1
Input
matrix = [
[2, 0, 1, 0],
[1, 0, 1, 2],
[0, 0, 2, 2]
]
Output
3
Explanation
If we set the meeting place to matrix[2][1] then everyone can get there in at most 3 steps.
Solution :
Solution in C++ :
bool isSafe(vector<vector<int>> &matrix, int r, int c) {
int m = matrix.size();
int n = matrix[0].size();
return (r >= 0 && c >= 0 && r < m && c < n && matrix[r][c] != 1);
}
void bfs(vector<vector<int>> &matrix, vector<vector<int>> &answer, int row, int col) {
int m = matrix.size();
int n = matrix[0].size();
int hor[4] = {-1, 1, 0, 0};
int ver[4] = {0, 0, 1, -1};
vector<vector<bool>> visited(m, vector<bool>(n, false));
queue<vector<int>> nodes;
nodes.push({row, col, 0});
visited[row][col] = true;
while (!nodes.empty()) {
int curRow = nodes.front()[0];
int curCol = nodes.front()[1];
int level = nodes.front()[2];
nodes.pop();
answer[curRow][curCol] = max(answer[curRow][curCol], level);
for (int i = 0; i < 4; i++) {
int newRow = curRow + hor[i];
int newCol = curCol + ver[i];
if (isSafe(matrix, newRow, newCol) && !visited[newRow][newCol]) {
visited[newRow][newCol] = true;
nodes.push({newRow, newCol, level + 1});
}
}
}
return;
}
int solve(vector<vector<int>> &matrix) {
int m = matrix.size();
int n = matrix[0].size();
vector<vector<int>> answer(m, vector<int>(n, -1));
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == 1) answer[i][j] = 31;
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == 2) {
bfs(matrix, answer, i, j);
}
}
}
int minTime = 31;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (answer[i][j] != -1) minTime = min(minTime, answer[i][j]);
}
}
return (minTime == 31 ? 0 : minTime);
}
Solution in Java :
import java.util.*;
class Solution {
int dis[][][];
public int solve(int[][] mat) {
int res = Integer.MAX_VALUE;
int n = mat.length, m = mat[0].length;
List<int[]> list = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (mat[i][j] == 2) {
list.add(new int[] {i, j});
}
}
}
dis = new int[n][m][list.size()];
for (int i = 0; i < dis.length; i++) {
for (int j = 0; j < dis[0].length; j++) {
Arrays.fill(dis[i][j], -1);
}
}
for (int i = 0; i < list.size(); i++) {
bfs(mat, list.get(i)[0], list.get(i)[1], i);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
int sum = 0;
for (int k = 0; k < dis[i][j].length; k++) {
if (dis[i][j][k] == -1) {
sum = Integer.MAX_VALUE;
break;
} else {
sum = Math.max(sum, dis[i][j][k]);
}
}
res = Math.min(res, sum);
}
}
return res;
}
public void bfs(int mat[][], int sr, int sc, int index) {
Queue<int[]> q = new LinkedList<>();
q.add(new int[] {sr, sc, 0});
dis[sr][sc][index] = 0;
while (q.size() != 0) {
int pair[] = q.poll();
int r = pair[0], c = pair[1], level = pair[2];
add(q, mat, r + 1, c, level + 1, index);
add(q, mat, r - 1, c, level + 1, index);
add(q, mat, r, c + 1, level + 1, index);
add(q, mat, r, c - 1, level + 1, index);
}
}
public void add(Queue<int[]> q, int mat[][], int i, int j, int level, int index) {
if (i < 0 || j < 0 || i >= mat.length || j >= mat[0].length)
return;
if (mat[i][j] == 1)
return;
if (dis[i][j][index] != -1)
return;
dis[i][j][index] = level;
q.add(new int[] {i, j, level});
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
m, n = len(matrix), len(matrix[0])
distGrid = [[float("inf") for _ in range(n)] for _ in range(m)]
def bfs(i, j):
queue = deque([(i, j, 0)])
visited = [[False for _ in range(n)] for _ in range(m)]
visited[i][j] = True
while queue:
x, y, dist = queue.popleft()
if distGrid[x][y] == float("inf"):
distGrid[x][y] = 0
distGrid[x][y] = max(distGrid[x][y], dist)
for r, c in [(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)]:
if 0 <= r < m and 0 <= c < n and not visited[r][c] and matrix[r][c] != 1:
visited[r][c] = True
queue.append((r, c, dist + 1))
for i in range(m):
for j in range(n):
if matrix[i][j] == 2:
bfs(i, j)
res = min(min(row) for row in distGrid)
return res if res != float("inf") else 0
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