542. 01 Matrix LeetCode Solution

In this guide, you will get 542. 01 Matrix LeetCode Solution with the best time and space complexity. The solution to Matrix problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Matrix solution in C++
Matrix solution in Java
Matrix solution in Python
Additional Resources

Problem Statement of Matrix

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 <= 104
1 <= m * n <= 104
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/

Complexity Analysis

Time Complexity: O(mn)
Space Complexity: O(mn)

542. 01 Matrix LeetCode Solution in C++

class Solution {
 public:
  vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
    constexpr int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    const int m = mat.size();
    const int n = mat[0].size();
    queue<pair<int, int>> q;
    vector<vector<bool>> seen(m, vector<bool>(n));

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (mat[i][j] == 0) {
          q.emplace(i, j);
          seen[i][j] = true;
        }

    while (!q.empty()) {
      const auto [i, j] = q.front();
      q.pop();
      for (const auto& [dx, dy] : dirs) {
        const int x = i + dx;
        const int y = j + dy;
        if (x < 0 || x == m || y < 0 || y == n)
          continue;
        if (seen[x][y])
          continue;
        mat[x][y] = mat[i][j] + 1;
        q.emplace(x, y);
        seen[x][y] = true;
      }
    }

    return mat;
  }
};
/* code provided by PROGIEZ */

542. 01 Matrix LeetCode Solution in Java

class Solution {
  public int[][] updateMatrix(int[][] mat) {
    final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    final int m = mat.length;
    final int n = mat[0].length;
    Queue<Pair<Integer, Integer>> q = new ArrayDeque<>();
    boolean[][] seen = new boolean[m][n];

    for (int i = 0; i < m; ++i)
      for (int j = 0; j < n; ++j)
        if (mat[i][j] == 0) {
          q.offer(new Pair<>(i, j));
          seen[i][j] = true;
        }

    while (!q.isEmpty()) {
      final int i = q.peek().getKey();
      final int j = q.poll().getValue();
      for (int[] dir : dirs) {
        final int x = i + dir[0];
        final int y = j + dir[1];
        if (x < 0 || x == m || y < 0 || y == n)
          continue;
        if (seen[x][y])
          continue;
        mat[x][y] = mat[i][j] + 1;
        q.offer(new Pair<>(x, y));
        seen[x][y] = true;
      }
    }

    return mat;
  }
}
// code provided by PROGIEZ

542. 01 Matrix LeetCode Solution in Python

class Solution:
  def updateMatrix(self, mat: list[list[int]]) -> list[list[int]]:
    dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))
    m = len(mat)
    n = len(mat[0])
    q = collections.deque()
    seen = [[False] * n for _ in range(m)]

    for i in range(m):
      for j in range(n):
        if mat[i][j] == 0:
          q.append((i, j))
          seen[i][j] = True

    while q:
      i, j = q.popleft()
      for dx, dy in dirs:
        x = i + dx
        y = j + dy
        if x < 0 or x == m or y < 0 or y == n:
          continue
        if seen[x][y]:
          continue
        mat[x][y] = mat[i][j] + 1
        q.append((x, y))
        seen[x][y] = True

    return mat
 # code by PROGIEZ