#2290 - Minimum Obstacle Removal to Reach Corner

Hard • View on LeetCode • View on GitHub

Array Matrix Breadth-First Search Heap Priority Queue Graph Shortest Path

Problem Description

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).

Solution

/**
 * @param {number[][]} grid
 * @return {number}
 */
var minimumObstacles = function(grid) {
  const rows = grid.length;
  const cols = grid[0].length;
  const directions = [[0, 1], [0, -1], [1, 0], [-1, 0]];
  const dp = new Array(rows).fill().map(() => new Array(cols).fill(Number.MAX_SAFE_INTEGER));
  const queue = [[0, 0, 0]];
  let result = rows * cols + 1;

  while (queue.length > 0) {
    const levelSize = queue.length;
    for (let l = 0; l < levelSize; l++) {
      const [i, j, cost] = queue.shift();

      if (i === rows - 1 && j === cols - 1) {
        result = Math.min(result, cost);
        continue;
      }

      if (cost > dp[i][j]) {
        continue;
      }

      for (const [dx, dy] of directions) {
        const newI = i + dx;
        const newJ = j + dy;

        if (isValid(newI, newJ, cost)) {
          queue.push([newI, newJ, dp[newI][newJ]]);
        }
      }
    }
  }

  return result;

  function isValid(i, j, cost) {
    const valid = i >= 0 && j >= 0 && i < rows && j < cols && dp[i][j] > cost + grid[i][j];
    if (valid) dp[i][j] = cost + grid[i][j];
    return valid;
  }
};