notes/OJ notes/pages/Leetcode 01-Matrix.md

Leetcode 01-Matrix

2022-07-17 03:15

Algorithms:

#algorithm #BFS

Data structures:

#DS #vector_2d

Difficulty:

#coding_problem #difficulty-medium

Additional tags:

#leetcode #CS_list_need_understanding

Revisions:

N/A

Related topics:

Links:

• Link to problem

---

Problem

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.

Examples

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.

Thoughts

[!summary] This is a #BFS problem, because it needs to find a smallest distance

Why not DFS

I tried with DFS, but

1. it is not suitable for finding smallest distance
2. and is easy to go into a infinite loop.
3. Also, it is hard to determine whether to revisit (update) the distance.

BFS

Start searching from 0s, because the search direction matters.

pseudo code:

• Initialization stage:

  • add every 0 to queue
  • make every 1 a infinite large number, (10001 this case)

• while queue is not empty

  • check for neighbors
    • if OOB (Out of Bound), skip
    • if the value of neighbor's distance is higher than the node, update it, and add it to queue(also update his neighbors)

Solution

class Solution {
  const int MAX = 10002;

public:
  vector<vector<int>> updateMatrix(vector<vector<int>> &mat) {
    // Shouldn't use DFS, since DFS is likely to run into a loop
    queue<pair<int, int>> todo;
    int m = mat.size(), n = mat[0].size();
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        if (mat[i][j] == 0) {
          todo.push({i, j});
        } else {
          mat[i][j] = MAX;
        }
      }
    }

    int offset[] = {-1, 1};
    int x, y;
    int newX, newY;
    int dist;
    while (!todo.empty()) {
      x = todo.front().first;
      y = todo.front().second;
      todo.pop();
      dist = mat[x][y];

      for (int i : offset) {
        newX = x + i;
        if (newX < m && newX >= 0) {
          if (mat[newX][y] > dist + 1) {
            mat[newX][y] = dist + 1;
            todo.push({newX, y});
          }
        }
        newY = y + i;
        if (newY < n && newY >= 0) {
          if (mat[x][newY] > dist + 1) {
            mat[x][newY] = dist + 1;
            todo.push({x, newY});
          }
        }
      }
    }
    return mat;
  }
};