01 Matrix

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1. Example 1:

Input: 0 0 0 0 1 0 0 0 0

Output: 0 0 0 0 1 0 0 0 0

Example 2:

Input: 0 0 0 0 1 0 1 1 1

Output: 0 0 0 0 1 0 1 2 1

Solution: BFS

• We create a array of directions to help updating neighbor nodes
• Push all 0 as start position to queue
• Mark all 1 as INT_MAX
• Start updating neighbors of nodes in queue.
• If neighbor node's value is larger, then we update its value
• Repeat this until the queue is empty

    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
        queue<pair<int, int>> q;

        // 1. Push all 0 as start position to queue
        // 2. Mark all 1 as INT_MAX
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (matrix[i][j] == 0) q.push({i, j});
                else matrix[i][j] = INT_MAX;
            }
        }

        // 2. Start from zeros, update node in all direction
        // 3. Update neighbor that has larger values
        // 4. Then push valid neighbors to queue
        while (!q.empty()) {
            auto t = q.front(); q.pop();
            for (auto dir : dirs) {
                int x = t.first + dir[0], y = t.second + dir[1];
                if (x < 0 || x >= m || y < 0 || y >= n ||
                matrix[x][y] <= matrix[t.first][t.second]) continue;
                matrix[x][y] = matrix[t.first][t.second] + 1;
                q.push({x, y});
            }
        }
        return matrix;
    }