## 01-Matrix

###### DESCRIPTION (credit Leetcode.com)

You are given an m x n binary matrix grid where each cell contains either a 0 or a 1.

Write a function that returns a matrix of the same dimensions where each cell contains the distance to the nearest 0 in the original matrix. The distance between two adjacent cells is 1. If there is no 0 in the grid, return -1 for each cell.

###### EXAMPLES

**Example 1:**

`mat = [ [1, 0, 1], [0, 1, 0], [1, 1, 1], ]`

Output:

`[ [1, 0, 1], [0, 1, 0], [1, 2, 1], ]`

Run your code to see results here

## Explanation

Since this question involves finding the shortest distance between each cell and the nearest 0, this is a good candidate for a **breadth-first search (BFS)** traversal.

The key idea here is to first enqueue all cells with value 0 into the queue. This allows us to gradually find the distance of each cell to the nearest 0 using a BFS traversal.

Let's look at example input grid to help visualize the process:

`[ [1, 0, 1], [0, 1, 0], [1, 1, 1], ]`

#### Step 1: Initialize the Queue

We'll start by initializing the output grid with all cells set to -1 to indicate that the distance to the nearest 0 is unknown. We'll use BFS to gradually update the distance of each cell in the output grid.

Next, we can initialize the BFS queue with the positions of all cells that have a value of 0. We can also update the distances of these cells in the output grid to 0 directly (doing so also serves the purpose of marking these cells as visited).

`# the coordinates of cells with value 0 queue = [ (0, 1), (1, 0), (1, 2) ]`

`# setting the distances of cells with value 0 to 0 in the output grid output = [ [-1, 0, -1], [0, -1, 0], [-1, -1, -1], ]`

#### Step 2: Perform BFS Traversal (Distance 1)

Next, we'll start the BFS traversal with all the cells with value 0. All the neighbors of the cells currently in the BFS queue that currently have a value of -1 are at distance 1 from the nearest 0. We'll update the distances of these neighbors in the output grid to 1, and enqueue them into the BFS queue.

`queue = [ (0, 0), (0, 2), (1, 1), (2, 0), (2, 2) ]`

`output = [ [1, 0, 1], [0, 1, 0], [1, -1, 1], ]`

#### Step 3: Perform BFS Traversal (Distance 2)

Now, all the neighbors of the cells currently in the BFS queue that have a value of -1 are at distance of 2 from the nearest 0. We'll update the distances of these neighbors in the output grid to 2, and enqueue them into the BFS queue.

`queue = [ (2, 1) ]`

`output = [ [1, 0, 1], [0, 1, 0], [1, 2, 1], ]`

#### Step 4: Perform BFS Traversal (Distance 3)

At this step, all the neighbors of the cells in the BFS queue have been visited already, so we don't need to enqueue any more cells.

`queue = []`

`output = [ [1, 0, 1], [0, 1, 0], [1, 2, 1], ]`

#### Step 5: Return the Output Grid

Since the queue is now empty, we can terminate the BFS traversal and return the output grid, which contains the distances of each cell to the nearest 0.

`[ [1, 0, 1], [0, 1, 0], [1, 2, 1], ]`

## Solution

from collections import dequeclass Solution:def updateMatrix(self, mat):rows, cols = len(mat), len(mat[0])output = [[-1] * cols for _ in range(rows)]queue = deque()# Step 1: Initialize the queue with all the 0 cells# set their distance to 0 in the output gridfor r in range(rows):for c in range(cols):if mat[r][c] == 0:queue.append((r, c))output[r][c] = 0directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]# Step 2: Perform BFS traversaldistance = 1while queue:for _ in range(len(queue)):r, c = queue.popleft()for dr, dc in directions:nr, nc = r + dr, c + dcif 0 <= nr < rows and 0 <= nc < cols:if output[nr][nc] == -1:output[nr][nc] = distancequeue.append((nr, nc))distance += 1# Step 5: Return the output gridreturn output

## Complexity Analysis

**Time Complexity:** O(m * n), where m is the number of rows and n is the number of columns in the input matrix. We visit each cell at most once, and each cell is enqueued at most once.

**Space Complexity:** O(m * n), where m is the number of rows and n is the number of columns in the input matrix. The space required for the output matrix is O(m * n), and the space required for the queue can be at most O(m * n) in the worst case (e.g., when all cells are 0 and are enqueued at the start).

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