LeetCode #2290 — HARD

Minimum Obstacle Removal to Reach Corner

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a 0-indexed 2D integer array grid of size m x n. Each cell has one of two values:

0 represents an empty cell,
1 represents an obstacle that may be removed.

You can move up, down, left, or right from and to an empty cell.

Return the minimum number of obstacles to remove so you can move from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1).

Example 1:

Input: grid = [[0,1,1],[1,1,0],[1,1,0]]
Output: 2
Explanation: We can remove the obstacles at (0, 1) and (0, 2) to create a path from (0, 0) to (2, 2).
It can be shown that we need to remove at least 2 obstacles, so we return 2.
Note that there may be other ways to remove 2 obstacles to create a path.

Example 2:

Input: grid = [[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]]
Output: 0
Explanation: We can move from (0, 0) to (2, 4) without removing any obstacles, so we return 0.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 10⁵
2 <= m * n <= 10⁵
grid[i][j] is either 0 or 1.
grid[0][0] == grid[m - 1][n - 1] == 0

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed 2D integer array grid of size m x n. Each cell has one of two values: 0 represents an empty cell, 1 represents an obstacle that may be removed. You can move up, down, left, or right from and to an empty cell. Return the minimum number of obstacles to remove so you can move from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[0,1,1],[1,1,0],[1,1,0]]

Example 2

[[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]]

Core Insight

What unlocks the optimal approach

Model the grid as a graph where cells are nodes and edges are between adjacent cells. Edges to cells with obstacles have a cost of 1 and all other edges have a cost of 0.
Could you use 0-1 Breadth-First Search or Dijkstra’s algorithm?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2290: Minimum Obstacle Removal to Reach Corner
class Solution {
    public int minimumObstacles(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {0, 0, 0});
        int[] dirs = {-1, 0, 1, 0, -1};
        boolean[][] vis = new boolean[m][n];
        while (true) {
            var p = q.poll();
            int i = p[0], j = p[1], k = p[2];
            if (i == m - 1 && j == n - 1) {
                return k;
            }
            if (vis[i][j]) {
                continue;
            }
            vis[i][j] = true;
            for (int h = 0; h < 4; ++h) {
                int x = i + dirs[h], y = j + dirs[h + 1];
                if (x >= 0 && x < m && y >= 0 && y < n) {
                    if (grid[x][y] == 0) {
                        q.offerFirst(new int[] {x, y, k});
                    } else {
                        q.offerLast(new int[] {x, y, k + 1});
                    }
                }
            }
        }
    }
}

// Accepted solution for LeetCode #2290: Minimum Obstacle Removal to Reach Corner
func minimumObstacles(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	q := doublylinkedlist.New()
	type tuple struct{ i, j, k int }
	q.Add(tuple{0, 0, 0})
	vis := make([][]bool, m)
	for i := range vis {
		vis[i] = make([]bool, n)
	}
	dirs := [5]int{-1, 0, 1, 0, -1}
	for {
		v, _ := q.Get(0)
		p := v.(tuple)
		q.Remove(0)
		i, j, k := p.i, p.j, p.k
		if i == m-1 && j == n-1 {
			return k
		}
		if vis[i][j] {
			continue
		}
		vis[i][j] = true
		for h := 0; h < 4; h++ {
			x, y := i+dirs[h], j+dirs[h+1]
			if x >= 0 && x < m && y >= 0 && y < n {
				if grid[x][y] == 0 {
					q.Insert(0, tuple{x, y, k})
				} else {
					q.Add(tuple{x, y, k + 1})
				}
			}
		}
	}
}

# Accepted solution for LeetCode #2290: Minimum Obstacle Removal to Reach Corner
class Solution:
    def minimumObstacles(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        q = deque([(0, 0, 0)])
        vis = set()
        dirs = (-1, 0, 1, 0, -1)
        while 1:
            i, j, k = q.popleft()
            if i == m - 1 and j == n - 1:
                return k
            if (i, j) in vis:
                continue
            vis.add((i, j))
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n:
                    if grid[x][y] == 0:
                        q.appendleft((x, y, k))
                    else:
                        q.append((x, y, k + 1))

// Accepted solution for LeetCode #2290: Minimum Obstacle Removal to Reach Corner
/**
 * [2290] Minimum Obstacle Removal to Reach Corner
 *
 * You are given a 0-indexed 2D integer array grid of size m x n. Each cell has one of two values:
 *
 * 	0 represents an empty cell,
 * 	1 represents an obstacle that may be removed.
 *
 * You can move up, down, left, or right from and to an empty cell.
 * Return the minimum number of obstacles to remove so you can move from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1).
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/04/06/example1drawio-1.png" style="width: 605px; height: 246px;" />
 * Input: grid = [[0,1,1],[1,1,0],[1,1,0]]
 * Output: 2
 * Explanation: We can remove the obstacles at (0, 1) and (0, 2) to create a path from (0, 0) to (2, 2).
 * It can be shown that we need to remove at least 2 obstacles, so we return 2.
 * Note that there may be other ways to remove 2 obstacles to create a path.
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2022/04/06/example1drawio.png" style="width: 405px; height: 246px;" />
 * Input: grid = [[0,1,0,0,0],[0,1,0,1,0],[0,0,0,1,0]]
 * Output: 0
 * Explanation: We can move from (0, 0) to (2, 4) without removing any obstacles, so we return 0.
 *
 *  
 * Constraints:
 *
 * 	m == grid.length
 * 	n == grid[i].length
 * 	1 <= m, n <= 10^5
 * 	2 <= m * n <= 10^5
 * 	grid[i][j] is either 0 or 1.
 * 	grid[0][0] == grid[m - 1][n - 1] == 0
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-obstacle-removal-to-reach-corner/
// discuss: https://leetcode.com/problems/minimum-obstacle-removal-to-reach-corner/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn minimum_obstacles(grid: Vec<Vec<i32>>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2290_example_1() {
        let grid = vec![vec![0, 1, 1], vec![1, 1, 0], vec![1, 1, 0]];

        let result = 2;

        assert_eq!(Solution::minimum_obstacles(grid), result);
    }

    #[test]
    #[ignore]
    fn test_2290_example_2() {
        let grid = vec![
            vec![0, 1, 0, 0, 0],
            vec![0, 1, 0, 1, 0],
            vec![0, 0, 0, 1, 0],
        ];

        let result = 0;

        assert_eq!(Solution::minimum_obstacles(grid), result);
    }
}

// Accepted solution for LeetCode #2290: Minimum Obstacle Removal to Reach Corner
function minimumObstacles(grid: number[][]): number {
    const m = grid.length,
        n = grid[0].length;
    const dirs = [
        [0, 1],
        [0, -1],
        [1, 0],
        [-1, 0],
    ];
    let ans = Array.from({ length: m }, v => new Array(n).fill(Infinity));
    ans[0][0] = 0;
    let deque = [[0, 0]];
    while (deque.length) {
        let [x, y] = deque.shift();
        for (let [dx, dy] of dirs) {
            let [i, j] = [x + dx, y + dy];
            if (i < 0 || i > m - 1 || j < 0 || j > n - 1) continue;
            const cost = grid[i][j];
            if (ans[x][y] + cost >= ans[i][j]) continue;
            ans[i][j] = ans[x][y] + cost;
            deque.push([i, j]);
        }
    }
    return ans[m - 1][n - 1];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n)

Space

O(m × n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.