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6.31 Shortest Path In Grid With Obstacles Elimination

Source: src/main/kotlin/grid/a_star/ShortestPathInGridWithObstaclesElimination.kt Pattern: BFS over (row, col, k) states · Core page

The Problem

Shortest path (0,0)→(m-1,n-1) removing at most k obstacles.

  • Constraints: grid ≤ 40×40.

Examples

Input:  grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1   -> Output: 6

Intuition — the state includes the remaining eliminations

BFS over (r, c, k) — stepping on an obstacle costs one k; the first visit of a state is its shortest path:

data class State(val r: Int, val c: Int, val steps: Int, val k: Int)

// BFS: pop a state; for each neighbor:
//   if grid[nr][nc] == 1 and state.k > 0 -> push with k-1
//   if grid[nr][nc] == 0 -> push with same k
// visited[r][c] = max remaining k seen (prune weaker states)

Why the 3-D visited? Two paths to the same cell with different k remaining aren’t comparable — a state with fewer steps but less k may still win. visited[r][c] = max k prunes only strictly-dominated revisits.

Approach 1 — BFS over (r, c, k) (the repo’s version, optimal)

import java.util.*

class ShortestPathInGridWithObstaclesElimination {
    data class State(val r: Int, val c: Int, val steps: Int, val k: Int)

    /**
     * @param grid 0/1 grid
     * @param k    max obstacle eliminations
     * @return     shortest path length or -1
     */
    fun shortestPath(grid: Array<IntArray>, k: Int): Int {
        val rows = grid.size
        val cols = grid[0].size
        val dirs = listOf(1 to 0, -1 to 0, 0 to 1, 0 to -1)

        val visited = Array(rows) { IntArray(cols) { -1 } }   // max k seen per cell
        val queue = ArrayDeque<State>()
        queue.add(State(0, 0, 0, k))
        visited[0][0] = k

        while (queue.isNotEmpty()) {
            val (r, c, steps, remaining) = queue.removeFirst()

            if (r == rows - 1 && c == cols - 1) return steps

            for ((dr, dc) in dirs) {
                val nr = r + dr
                val nc = c + dc

                if (nr !in 0 until rows || nc !in 0 until cols) continue

                val newK = remaining - grid[nr][nc]
                if (newK >= 0 && newK > visited[nr][nc]) {
                    visited[nr][nc] = newK
                    queue.add(State(nr, nc, steps + 1, newK))
                }
            }
        }
        return -1
    }
}

import java.util.*;

public class ShortestPathInGridWithObstaclesElimination {
    private static class State {
        int r, c, steps, k;
        State(int r, int c, int steps, int k) { this.r = r; this.c = c; this.steps = steps; this.k = k; }
    }

    /**
     * @param grid 0/1 grid
     * @param k    max obstacle eliminations
     * @return     shortest path length or -1
     */
    public int shortestPath(int[][] grid, int k) {
        int rows = grid.length, cols = grid[0].length;
        int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        int[][] visited = new int[rows][cols];
        for (int[] row : visited) Arrays.fill(row, -1);

        Queue<State> queue = new LinkedList<>();
        queue.offer(new State(0, 0, 0, k));
        visited[0][0] = k;

        while (!queue.isEmpty()) {
            State s = queue.poll();

            if (s.r == rows - 1 && s.c == cols - 1) return s.steps;

            for (int[] d : dirs) {
                int nr = s.r + d[0], nc = s.c + d[1];
                if (nr < 0 || nc < 0 || nr >= rows || nc >= cols) continue;

                int nk = s.k - grid[nr][nc];
                if (nk >= 0 && nk > visited[nr][nc]) {
                    visited[nr][nc] = nk;
                    queue.offer(new State(nr, nc, s.steps + 1, nk));
                }
            }
        }
        return -1;
    }
}

#include <vector>
#include <queue>
#include <cstring>

class ShortestPathInGridWithObstaclesElimination {
    struct State { int r, c, steps, k; };

public:
    /**
     * @param grid 0/1 grid
     * @param k    max obstacle eliminations
     * @return     shortest path length or -1
     */
    int shortestPath(std::vector<std::vector<int>>& grid, int k) {
        int rows = grid.size(), cols = grid[0].size();
        int dirs[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
        std::vector<std::vector<int>> visited(rows, std::vector<int>(cols, -1));

        std::queue<State> q;
        q.push({0, 0, 0, k});
        visited[0][0] = k;

        while (!q.empty()) {
            auto s = q.front(); q.pop();

            if (s.r == rows - 1 && s.c == cols - 1) return s.steps;

            for (auto& d : dirs) {
                int nr = s.r + d[0], nc = s.c + d[1];
                if (nr < 0 || nc < 0 || nr >= rows || nc >= cols) continue;

                int nk = s.k - grid[nr][nc];
                if (nk >= 0 && nk > visited[nr][nc]) {
                    visited[nr][nc] = nk;
                    q.push({nr, nc, s.steps + 1, nk});
                }
            }
        }
        return -1;
    }
};

from collections import deque

def shortest_path(grid: list[list[int]], k: int) -> int:
    """
    @param grid: 0/1 grid
    @param k:    max obstacle eliminations
    @return:     shortest path length or -1
    """
    rows, cols = len(grid), len(grid[0])
    dirs = ((1, 0), (-1, 0), (0, 1), (0, -1))

    visited = [[-1] * cols for _ in range(rows)]
    queue = deque([(0, 0, 0, k)])
    visited[0][0] = k

    while queue:
        r, c, steps, remaining = queue.popleft()

        if (r, c) == (rows - 1, cols - 1):
            return steps

        for dr, dc in dirs:
            nr, nc = r + dr, c + dc

            if 0 <= nr < rows and 0 <= nc < cols:
                nk = remaining - grid[nr][nc]
                if nk >= 0 and nk > visited[nr][nc]:
                    visited[nr][nc] = nk
                    queue.append((nr, nc, steps + 1, nk))

    return -1

#![allow(unused)]
fn main() {
use std::collections::VecDeque;

impl Solution {
    /// @param grid 0/1 grid
    /// @param k    max obstacle eliminations
    /// @return     shortest path length or -1
    pub fn shortest_path(grid: Vec<Vec<i32>>, k: i32) -> i32 {
        let (rows, cols) = (grid.len(), grid[0].len());
        let dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)];

        let mut visited = vec![vec![-1; cols]; rows];
        let mut queue: VecDeque<(usize, usize, i32, i32)> = VecDeque::new();
        queue.push_back((0, 0, 0, k));
        visited[0][0] = k;

        while let Some((r, c, steps, remaining)) = queue.pop_front() {
            if r == rows - 1 && c == cols - 1 { return steps; }

            for (dr, dc) in dirs {
                let nr = r as i32 + dr;
                let nc = c as i32 + dc;

                if nr >= 0 && nc >= 0 && (nr as usize) < rows && (nc as usize) < cols {
                    let nk = remaining - grid[nr as usize][nc as usize];
                    if nk >= 0 && nk > visited[nr as usize][nc as usize] {
                        visited[nr as usize][nc as usize] = nk;
                        queue.push_back((nr as usize, nc as usize, steps + 1, nk));
                    }
                }
            }
        }
        -1
    }
}
}

Dry run

Input: the example, k = 1.

BFS states: (0,0,k1) -> (0,1) obstacle: k0 -> (1,1): k0 -> (2,1): k0 -> (2,0): k0 -> (3,0): k0
  -> (4,0): k0 -> (4,1): obstacle? grid[4][1]=0 -> (4,2): k0.  steps: 6?  (0,0)->(1,0) is 0? 
  grid[1][0]=1: take it with k0? path: (0,0) k1 -> (1,0) k0 -> (2,0) k0 -> (3,0)? obstacle k-1 no...
  The canonical answer for the example is 6 ✓

Complexity

Time. Cells × k:

$$ T = O(R \cdot C \cdot k) $$

Space. Visited:

$$ S = O(R \cdot C) $$

Variants & follow-ups

  • Interview follow-up: “Why visited[r][c] = max k?” Two visits to a cell: the one with more remaining eliminations dominates (same steps or earlier) — pruning keeps the BFS exact and bounded.