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# 01 Matrix LeetCode Solution – Queslers

## Problem – 01 Matrix

Given an `m x n` binary matrix `mat`, return the distance of the nearest `0` for each cell.

The distance between two adjacent cells is `1`.

Example 1:

``````Input: mat = [[0,0,0],[0,1,0],[0,0,0]]
Output: [[0,0,0],[0,1,0],[0,0,0]]``````

Example 2:

``````Input: mat = [[0,0,0],[0,1,0],[1,1,1]]
Output: [[0,0,0],[0,1,0],[1,2,1]]``````

Constraints:

• `m == mat.length`
• `n == mat[i].length`
• `1 <= m, n <= 104`
• `1 <= m * n <= 104`
• `mat[i][j]` is either `0` or `1`.
• There is at least one `0` in `mat`.

### 01 Matrix LeetCode Solution in Java

``````public class Solution {
public int[][] updateMatrix(int[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (matrix[i][j] == 0) {
queue.offer(new int[] {i, j});
}
else {
matrix[i][j] = Integer.MAX_VALUE;
}
}
}

int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

while (!queue.isEmpty()) {
int[] cell = queue.poll();
for (int[] d : dirs) {
int r = cell[0] + d[0];
int c = cell[1] + d[1];
if (r < 0 || r >= m || c < 0 || c >= n ||
matrix[r][c] <= matrix[cell[0]][cell[1]] + 1) continue;
matrix[r][c] = matrix[cell[0]][cell[1]] + 1;
}
}

return matrix;
}
}
``````

### 01 Matrix LeetCode Solution in Python

``````def updateMatrix(self, matrix: List[List[int]]) -> List[List[int]]:
# BFS helper #
def bfs(node):
from collections import deque
q = deque()
i, j = node
q.append(((i,j), 0)) # d (dist to a zero) = 0 initially
visited = set()
dirs = [(1,0), (-1,0), (0,1), (0,-1)]
while q:
for i in range(len(q)):
coor, d = q.popleft()
x, y = coor
# if a zero nei is found
if matrix[x][y] == 0:
return d
# investiagte neighbours
for dir in dirs:
newX, newY = x+dir[0], y+dir[1]
# within bounds:
if newX >= 0 and newX <= len(matrix)-1 and \
newY >= 0 and newY <= len(matrix[0])-1:
# not seen:
if (newX, newY) not in visited:
q.append(((newX, newY), d+1))
return -1

# main logic #
'''
steps:
- itertate over matrix to find cells = 1
- pass cells equaling 1 to a bfs to find the closest 0 to them
- update matrix
'''
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == 1:
d = bfs((i,j)) # d = closest dist to a 0
matrix[i][j] = d # update M with d
return matrix
``````

### 01 Matrix LeetCode Solution in C++

``````class Solution {
public:

bool isvalid(int i,int j,int m,int n)
{
if(i==m||j==n||j<0||i<0)
return false;
return true;
}

vector<vector<int>> dir={{1,0},{0,1},{0,-1},{-1,0}};
vector<vector<int>> updateMatrix(vector<vector<int>>& matrix)
{
queue<pair<int,int>> q;
int m=matrix.size();
int n=matrix[0].size();
vector<vector<int>> dis(m,vector<int>(n,-1));
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
if(matrix[i][j]==0)
{
q.push({i,j});
dis[i][j]=0;
}
}
while(!q.empty())
{
pair<int,int> curr=q.front();
q.pop();
for(auto& x:dir)
{
int a=curr.first+x[0];
int b=curr.second+x[1];
if(isvalid(a,b,m,n)&&dis[a][b]==-1)
{
q.push({a,b});
dis[a][b]=dis[curr.first][curr.second]+1;
}
}
}
return dis;
}
};
``````
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