Farthest Point From Water - Amazon Top Interview Questions
Problem Statement :
You are given a matrix matrix of 0s and 1s where 0 represents water and 1 represents land. Find the land with the largest Manhattan distance from water and return this distance. If there is no land or no water in the board, return -1.
Constraints
n, m ≤ 100 where n and m are the number of rows and columns in matrix
Example 1
Input
matrix = [
[1, 1, 0],
[1, 1, 0],
[0, 0, 1]
]
Output
2
Explanation
grid[0][0] has a Manhattan distance of 2 to water.
Example 2
Input
matrix = [
[1, 1],
[1, 1]
]
Output
-1
Explanation
There is no water in this grid.
Example 3
Input
matrix = [
[1, 1, 1, 1],
[0, 1, 1, 1],
[0, 0, 1, 0]
]
Output
3
Explanation
Land at grid[0][2] has a Manhattan distance of 3 to nearest water.
Solution :
Solution in C++ :
int solve(vector<vector<int>>& matrix) { // Time and Space: O(N * M)
int height = matrix.size();
if (height == 0) return -1;
int width = matrix[0].size();
if (width == 0) return -1;
queue<pair<int, int>> cell_queue; // Push all zeroes into queue
for (int row = 0; row < height; row++) {
for (int col = 0; col < width; col++) {
if (matrix[row][col] == 0) cell_queue.push(make_pair(row, col));
}
}
int dir[5] = {1, 0, -1, 0, 1};
int max_dist = -1;
int curr_dist = 0;
while (!cell_queue.empty()) { // BFS
int size = cell_queue.size();
curr_dist++;
for (int i = 0; i < size; i++) {
int row = cell_queue.front().first;
int col = cell_queue.front().second;
cell_queue.pop();
for (int d = 1; d <= 4; d++) {
int r = row + dir[d - 1];
int c = col + dir[d];
if (r < 0 || c < 0 || r >= height || c >= width || matrix[r][c] == 0) continue;
matrix[r][c] = 0; // Set cell to 0, so we don't count it anymore
max_dist = curr_dist;
cell_queue.push(make_pair(r, c));
}
}
}
return max_dist;
}
Solution in Java :
import java.util.*;
class Solution {
static class Point {
int x;
int y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
}
public int solve(int[][] matrix) {
ArrayDeque<Point> q = new ArrayDeque<>();
int n = matrix.length;
int m = matrix[0].length;
// Stores distance of coordinate
int[][] dis = new int[n][m];
// Direction coordinates
int[] dx = {0, 1, 0, -1};
int[] dy = {1, 0, -1, 0};
int oo = Integer.MAX_VALUE;
int cnt0 = 0;
// Multisource BFS , add all 0s to the queue
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (matrix[i][j] == 0) {
q.add(new Point(i, j));
matrix[i][j] = -1;
dis[i][j] = 0;
++cnt0;
} else
dis[i][j] = oo;
}
}
// Return -1 if there is no 0 or all all 0s
if (cnt0 == 0 || cnt0 == n * m)
return -1;
int ans = -oo;
while (!q.isEmpty()) {
Point p = q.pollFirst();
matrix[p.x][p.y] = -1;
for (int i = 0; i < 4; ++i) {
int newX = p.x + dx[i];
int newY = p.y + dy[i];
if (newX >= 0 && newX < n && newY >= 0 && newY < m && matrix[newX][newY] != -1) {
q.add(new Point(newX, newY));
dis[newX][newY] = Math.min(dis[newX][newY], 1 + dis[p.x][p.y]);
}
}
}
// Find maximum distance of any 1 to closest 0
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
ans = Math.max(ans, dis[i][j]);
}
}
return ans;
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
m, n = len(matrix), len(matrix[0])
def get(y, x):
if 0 <= y < m and 0 <= x < n:
return matrix[y][x]
return math.inf
for y in range(m):
for x in range(n):
if matrix[y][x] != 0:
matrix[y][x] = min(get(y - 1, x) + 1, get(y, x - 1) + 1)
for y in range(m - 1, -1, -1):
for x in range(n - 1, -1, -1):
if matrix[y][x] != 0:
matrix[y][x] = min(matrix[y][x], get(y + 1, x) + 1, get(y, x + 1) + 1)
max_dist = max(val for row in matrix for val in row)
return -1 if max_dist in (math.inf, 0) else max_dist
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