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| 1 | +package g0801_0900.s0864_shortest_path_to_get_all_keys |
| 2 | + |
| 3 | +import java.util.* |
| 4 | + |
| 5 | +class Solution { |
| 6 | + private var m = 0 |
| 7 | + private var n = 0 |
| 8 | + fun shortestPathAllKeys(stringGrid: Array<String>): Int { |
| 9 | + // strategy: BFS + masking |
| 10 | + m = stringGrid.size |
| 11 | + n = stringGrid[0].length |
| 12 | + val grid = Array(m) { CharArray(n) } |
| 13 | + var index = 0 |
| 14 | + // convert to char Array |
| 15 | + for (s in stringGrid) { |
| 16 | + grid[index++] = s.toCharArray() |
| 17 | + } |
| 18 | + // number of keys |
| 19 | + var count = 0 |
| 20 | + val q: Queue<IntArray> = LinkedList() |
| 21 | + for (i in 0 until m) { |
| 22 | + for (j in 0 until n) { |
| 23 | + // find starting position |
| 24 | + if (grid[i][j] == '@') { |
| 25 | + q.add(intArrayOf(i, j, 0)) |
| 26 | + } |
| 27 | + // count number of keys |
| 28 | + if (grid[i][j] in 'a'..'f') { |
| 29 | + count++ |
| 30 | + } |
| 31 | + } |
| 32 | + } |
| 33 | + val dx = intArrayOf(-1, 0, 1, 0) |
| 34 | + val dy = intArrayOf(0, -1, 0, 1) |
| 35 | + // this is the amt of keys we need |
| 36 | + val target = (1 shl count) - 1 |
| 37 | + // keep track of position and current state |
| 38 | + val visited = Array(m) { |
| 39 | + Array(n) { |
| 40 | + BooleanArray(target + 1) |
| 41 | + } |
| 42 | + } |
| 43 | + // set initial position and state to true |
| 44 | + visited[q.peek()[0]][q.peek()[1]][0] = true |
| 45 | + var steps = 0 |
| 46 | + while (!q.isEmpty()) { |
| 47 | + // use size to make sure everything is on one level |
| 48 | + var size = q.size |
| 49 | + while (--size >= 0) { |
| 50 | + val curr = q.poll() |
| 51 | + val x = curr[0] |
| 52 | + val y = curr[1] |
| 53 | + val state = curr[2] |
| 54 | + // found all keys |
| 55 | + if (state == target) { |
| 56 | + return steps |
| 57 | + } |
| 58 | + for (i in 0..3) { |
| 59 | + val nx = x + dx[i] |
| 60 | + val ny = y + dy[i] |
| 61 | + // use new state so we don't mess up current state |
| 62 | + var nState = state |
| 63 | + // out of bounds or reached wall |
| 64 | + if (!inBounds(nx, ny) || grid[nx][ny] == '#') { |
| 65 | + continue |
| 66 | + } |
| 67 | + // found key |
| 68 | + // use OR to add key to our current state because if we already had the key the |
| 69 | + // digit would still be 1/true |
| 70 | + if (grid[nx][ny] in 'a'..'f') { |
| 71 | + // bit mask our found key |
| 72 | + nState = state or (1 shl grid[nx][ny] - 'a') |
| 73 | + } |
| 74 | + // found lock |
| 75 | + // use & to see if we have the key |
| 76 | + // 0 means that the digit we are looking at is 0 |
| 77 | + // need a 1 at the digit spot which means there is a key there |
| 78 | + if (('A' > grid[nx][ny] || grid[nx][ny] > 'F' || nState and (1 shl grid[nx][ny] - 'A') != 0) |
| 79 | + && !visited[nx][ny][nState] |
| 80 | + ) { |
| 81 | + q.add(intArrayOf(nx, ny, nState)) |
| 82 | + visited[nx][ny][nState] = true |
| 83 | + } |
| 84 | + } |
| 85 | + } |
| 86 | + steps++ |
| 87 | + } |
| 88 | + return -1 |
| 89 | + } |
| 90 | + |
| 91 | + private fun inBounds(x: Int, y: Int): Boolean { |
| 92 | + return x in 0 until m && y >= 0 && y < n |
| 93 | + } |
| 94 | +} |
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