Shortest path in a grid with obstacles elimination

Array Breadth-First Search Matrix

46.1% acceptance

Feb 25, 2026

4828

You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step. Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1.

Solution

Rust

Time O(n * m)

Space O(n * m)

LeetCode

solution.rs

impl Solution {
  pub fn shortest_path(grid: Vec<Vec<i32>>, k: i32) -> i32 {
    let m = grid.len();
    let n = grid[0].len();
    if m == 1 && n == 1 { return 0; }
    // BFS state: (row, col, remaining_k)
    let mut visited = vec![vec![vec![false; (k + 1) as usize]; n]; m];
    let mut queue = std::collections::VecDeque::new();
    queue.push_back((0usize, 0usize, k, 0i32));
    visited[0][0][k as usize] = true;

    let dirs = [(0i32, 1i32), (0, -1), (1, 0), (-1, 0)];
    while let Some((r, c, rem, steps)) = queue.pop_front() {
      for (dr, dc) in dirs {
        let nr = r as i32 + dr;
        let nc = c as i32 + dc;
        if nr < 0 || nr >= m as i32 || nc < 0 || nc >= n as i32 { continue; }
        let nr = nr as usize;
        let nc = nc as usize;
        let is_obstacle = grid[nr][nc] == 1;
        let new_rem = if is_obstacle { rem - 1 } else { rem };
        if new_rem < 0 { continue; }
        if nr == m - 1 && nc == n - 1 { return steps + 1; }
        if !visited[nr][nc][new_rem as usize] {
          visited[nr][nc][new_rem as usize] = true;
          queue.push_back((nr, nc, new_rem, steps + 1));
        }
      }
    }
    -1
  }
}