542. 01 Matrix

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/*
 * @lc app=leetcode id=542 lang=cpp
 *
 * [542] 01 Matrix
 *
 * https://leetcode.com/problems/01-matrix/description/
 *
 * algorithms
 * Medium (49.81%)
 * Likes:    9993
 * Dislikes: 434
 * Total Accepted:    696.7K
 * Total Submissions: 1.4M
 * Testcase Example:  '[[0,0,0],[0,1,0],[0,0,0]]'
 *
 * Given an m x n binary matrix mat, return the distance of the nearest 0 for
 * each cell.
 *
 * The distance between two cells sharing a common edge 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 <= 10^4
 * 1 <= m * n <= 10^4
 * mat[i][j] is either 0 or 1.
 * There is at least one 0 in mat.
 *
 *
 *
 * Note: This question is the same as 1765:
 * https://leetcode.com/problems/map-of-highest-peak/
 *
 */

// @lc code=start
class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& mat)
    {
        int m = mat.size(), n = mat[0].size();
        queue<vector<int>> q;
        set<pair<int, int>> used;
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (mat[i][j] == 0)
                {
                    q.push({i, j, 0});
                    used.insert({i, j});
                }
            }
        }

        vector<vector<int>> res = mat;
        vector<vector<int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        while (!q.empty())
        {
            auto node = q.front(); q.pop();
            for (const auto& dir : directions)
            {
                int new_i = node[0] + dir[0];
                int new_j = node[1] + dir[1];
                int dist = node[2] + 1;

                if (new_i < 0 || new_j < 0 || new_i >= m || new_j >= n || used.count({new_i, new_j})) continue;
                q.push({new_i, new_j, dist});
                used.insert({new_i, new_j});
                res[new_i][new_j] = dist;
            }
        }
        return res;
    }
};
// @lc code=end

• T: $O(M \cdot N)$
• S: $O(M \cdot N)$