317. Shortest Distance from All Buildings
package LeetCode_317 import java.util.* /** * 317. Shortest Distance from All Buildings * (Prime) * You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. * You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where: 1. Each 0 marks an empty land which you can pass by freely. 2. Each 1 marks a building which you cannot pass through. 3. Each 2 marks an obstacle which you cannot pass through. Example: Input: [[1,0,2,0,1],[0,0,0,0,0],[0,0,1,0,0]] 1 - 0 - 2 - 0 - 1 | | | | | 0 - 0 - 0 - 0 - 0 | | | | | 0 - 0 - 1 - 0 - 0 Output: 7 Explanation: Given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2), the point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7. Note: There will be at least one building. If it is not possible to build such house according to the above rules, return -1. * */ class Solution { /* *solution:BFS, do bfs for each building, * Time complexity:O(m^2*n^2), Space complexity:O(mn) * */ fun shortestDistance(grid: Array<IntArray>?): Int { if (grid == null || grid.isEmpty()) { return -1 } var buildingCount = 0 val m = grid.size val n = grid[0].size //save distance of each x,y to building val distance = Array(m) { IntArray(n) } //save how many building x,y can reach val reach = Array(m) { IntArray(n) } //4 directions val direction = intArrayOf(0, -1, 0, 1, 0) for (i in 0 until m) { for (j in 0 until n) { //start bfs when meet building if (grid[i][j] == 1) { buildingCount++ val queue = LinkedList<Pair<Int, Int>>() val visited = Array(m) { BooleanArray(n) } visited[i][j] = true var level = 0 queue.offer(Pair(i, j)) while (queue.isNotEmpty()) { val size = queue.size for (i in 0 until size) { val cur = queue.pop() val curX = cur.first val curY = cur.second //update the distance for x,y distance[curX][curY] += level //update the count of reach to 1 for x,y reach[curX][curY]++ for (d in 0 until 4) { val nextX = curX + direction[d] val nextY = curY + direction[d + 1] if (nextX >= 0 && nextX < m && nextY >= 0 && nextY < n && !visited[nextX][nextY]) { visited[nextX][nextY] = true queue.offer(Pair(nextX, nextY)) } } } level++ } } } } var shortest = Int.MAX_VALUE /* * checking each x,y is 0, if current reach count equal to total building count, compare for shortest one * */ for (i in 0 until m) { for (j in 0 until n) { if (grid[i][j] == 0 && reach[i][j] == buildingCount) { shortest = Math.min(shortest, distance[i][j]) } } } return if (shortest == Int.MAX_VALUE) -1 else shortest } }