#317 - Shortest Distance from All Buildings

Problem Description

You are given an m x n grid grid of values 0, 1, or 2, where:

each 0 marks an empty land that you can pass by freely,
each 1 marks a building that you cannot pass through, and
each 2 marks an obstacle that you cannot pass through.

You want to build a house on an empty land that reaches all buildings in the shortest total travel distance. You can only move up, down, left, and right.

Return the shortest travel distance for such a house. If it is not possible to build such a house according to the above rules, return -1.

The total travel distance is the sum of the distances between the houses of the friends and the meeting point.

Solution

/**
 * @param {number[][]} grid
 * @return {number}
 */
var shortestDistance = function(grid) {
  const rows = grid.length;
  const cols = grid[0].length;
  const distances = new Array(rows).fill().map(() => new Array(cols).fill(0));
  const reachCounts = new Array(rows).fill().map(() => new Array(cols).fill(0));
  let buildingCount = 0;

  function bfs(startRow, startCol) {
    const queue = [[startRow, startCol]];
    const visited = new Array(rows).fill().map(() => new Array(cols).fill(false));
    visited[startRow][startCol] = true;
    let distance = 0;

    const directions = [[-1, 0], [1, 0], [0, -1], [0, 1]];

    while (queue.length) {
      const size = queue.length;
      distance++;
      for (let i = 0; i < size; i++) {
        const [row, col] = queue.shift();
        for (const [dr, dc] of directions) {
          const newRow = row + dr;
          const newCol = col + dc;
          if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols
              && !visited[newRow][newCol] && grid[newRow][newCol] !== 1
              && grid[newRow][newCol] !== 2) {
            queue.push([newRow, newCol]);
            visited[newRow][newCol] = true;
            distances[newRow][newCol] += distance;
            reachCounts[newRow][newCol]++;
          }
        }
      }
    }
  }

  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      if (grid[i][j] === 1) {
        buildingCount++;
        bfs(i, j);
      }
    }
  }

  let minDistance = Infinity;
  for (let i = 0; i < rows; i++) {
    for (let j = 0; j < cols; j++) {
      if (grid[i][j] === 0 && reachCounts[i][j] === buildingCount) {
        minDistance = Math.min(minDistance, distances[i][j]);
      }
    }
  }

  return minDistance === Infinity ? -1 : minDistance;
};