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;
        
        Queue<int[]> queue = new LinkedList<>();
        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;
                queue.add(new int[] {r, c});
                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
                    visited.add(coor)
                    # 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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