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 :



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                        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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