The "Minimum Path Cost in a Hidden Grid" problem presents a grid where each cell has a non-negative integer cost, but the grid's structure is hidden and can only be discovered by moving step by step. You start at a given cell, and can move in four directions: up, down, left, or right. Your goal is to find the minimum total cost to reach a target cell from the starting position, where the total cost is the sum of the costs of all visited cells (including both the start and end cells).
You interact with the grid using a provided interface:
GridMaster.canMove(direction): Returns true if you can move in the specified direction.GridMaster.move(direction): Moves you in the specified direction and returns the cost of the new cell.GridMaster.isTarget(): Returns true if the current cell is the target.At first glance, this problem appears similar to classic shortest path problems like Dijkstra's algorithm. However, the twist is that the grid is "hidden": you do not know its layout, size, or the cost of each cell in advance. You can only discover the grid by exploring it step by step using the provided interface.
A brute-force approach would be to perform a complete exploration (like BFS or DFS), recording every cell and its cost, and then running a shortest path algorithm on the discovered map. However, since you can only move one cell at a time and must backtrack physically through the interface, efficiency is crucial.
The challenge is twofold:
The solution consists of two main phases: exploration and pathfinding.
Design Choices:
Suppose the hidden grid looks like this (unknown to you at first):
[1, 2, 100]
[1, 3, 1 ]
[100, 1, 1 ]
The start is at (1,0) with cost 1, and the target is at (2,2) with cost 1.
canMove to check validity.move to enter the cell, record its cost and coordinates, and recursively explore further.isTarget returns true, record the coordinates as the target.Brute-force Approach:
This is efficient and feasible for the problem's constraints.
The "Minimum Path Cost in a Hidden Grid" problem blends exploration and pathfinding: you must first discover the grid using DFS and backtracking, then apply Dijkstra's algorithm to find the minimum-cost path. The key insight is to treat the problem in two phases—mapping and shortest path—using efficient data structures for both. This strategy ensures you systematically uncover the grid and find the optimal path, even without knowing the grid's structure in advance.
# Assume GridMaster interface is provided as per Leetcode
class Solution:
def findShortestPath(self, master: 'GridMaster') -> int:
from collections import deque
import heapq
# Directions and their opposites for backtracking
dirs = ['U', 'D', 'L', 'R']
moves = {'U': (-1, 0), 'D': (1, 0), 'L': (0, -1), 'R': (0, 1)}
rev = {'U': 'D', 'D': 'U', 'L': 'R', 'R': 'L'}
grid = {}
target = [None, None]
def dfs(x, y):
if master.isTarget():
target[0], target[1] = x, y
grid[(x, y)] = master.getCost() if hasattr(master, "getCost") else 0 # fallback if getCost is not present
for d in dirs:
dx, dy = moves[d]
nx, ny = x + dx, y + dy
if (nx, ny) not in grid and master.canMove(d):
cost = master.move(d)
grid[(nx, ny)] = cost
dfs(nx, ny)
master.move(rev[d]) # backtrack
# The starting cell is at (0,0)
grid[(0, 0)] = master.getCost() if hasattr(master, "getCost") else 0
dfs(0, 0)
if target[0] is None:
return -1
# Dijkstra on discovered grid
heap = [(grid[(0, 0)], 0, 0)]
visited = set()
while heap:
cost, x, y = heapq.heappop(heap)
if (x, y) == (target[0], target[1]):
return cost
if (x, y) in visited:
continue
visited.add((x, y))
for d in dirs:
dx, dy = moves[d]
nx, ny = x + dx, y + dy
if (nx, ny) in grid and (nx, ny) not in visited:
heapq.heappush(heap, (cost + grid[(nx, ny)], nx, ny))
return -1
// Assume GridMaster interface is provided as per Leetcode
class Solution {
public:
int findShortestPath(GridMaster &master) {
// Directions and their opposites for backtracking
vector<char> dirs = {'U', 'D', 'L', 'R'};
vector<int> dx = {-1, 1, 0, 0};
vector<int> dy = {0, 0, -1, 1};
unordered_map<char, char> rev = {{'U','D'}, {'D','U'}, {'L','R'}, {'R','L'}};
map<pair<int,int>, int> grid;
pair<int,int> target = {INT_MAX, INT_MAX};
function<void(int, int)> dfs = [&](int x, int y) {
if (master.isTarget()) {
target = {x, y};
}
for (int i = 0; i < 4; ++i) {
int nx = x + dx[i], ny = y + dy[i];
if (grid.count({nx, ny}) == 0 && master.canMove(dirs[i])) {
int cost = master.move(dirs[i]);
grid[{nx, ny}] = cost;
dfs(nx, ny);
master.move(rev[dirs[i]]);
}
}
};
grid[{0,0}] = 0;
dfs(0, 0);
if (target.first == INT_MAX) return -1;
// Dijkstra
set<pair<int, pair<int,int>>> pq; // {cost, {x, y}}
pq.insert({grid[{0,0}], {0,0}});
set<pair<int,int>> visited;
while (!pq.empty()) {
auto [cost, pos] = *pq.begin(); pq.erase(pq.begin());
int x = pos.first, y = pos.second;
if (make_pair(x, y) == target) return cost;
if (visited.count({x, y})) continue;
visited.insert({x, y});
for (int i = 0; i < 4; ++i) {
int nx = x + dx[i], ny = y + dy[i];
if (grid.count({nx, ny}) && !visited.count({nx, ny})) {
pq.insert({cost + grid[{nx, ny}], {nx, ny}});
}
}
}
return -1;
}
};
// Assume GridMaster interface is provided as per Leetcode
class Solution {
private static final char[] DIRS = {'U', 'D', 'L', 'R'};
private static final int[] DX = {-1, 1, 0, 0};
private static final int[] DY = {0, 0, -1, 1};
private static final Map<Character, Character> REV = Map.of('U','D', 'D','U', 'L','R', 'R','L');
private Map<String, Integer> grid = new HashMap<>();
private int targetX = Integer.MAX_VALUE, targetY = Integer.MAX_VALUE;
public int findShortestPath(GridMaster master) {
dfs(0, 0, master);
if (targetX == Integer.MAX_VALUE) return -1;
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> Integer.compare(a[0], b[0]));
pq.offer(new int[]{grid.get("0,0"), 0, 0});
Set<String> visited = new HashSet<>();
while (!pq.isEmpty()) {
int[] curr = pq.poll();
int cost = curr[0], x = curr[1], y = curr[2];
String key = x + "," + y;
if (x == targetX && y == targetY) return cost;
if (visited.contains(key)) continue;
visited.add(key);
for (int i = 0; i < 4; ++i) {
int nx = x + DX[i], ny = y + DY[i];
String nkey = nx + "," + ny;
if (grid.containsKey(nkey) && !visited.contains(nkey)) {
pq.offer(new int[]{cost + grid.get(nkey), nx, ny});
}
}
}
return -1;
}
private void dfs(int x, int y, GridMaster master) {
String key = x + "," + y;
if (master.isTarget()) {
targetX = x;
targetY = y;
}
grid.put(key, grid.getOrDefault(key, 0));
for (int i = 0; i < 4; ++i) {
int nx = x + DX[i], ny = y + DY[i];
String nkey = nx + "," + ny;
if (!grid.containsKey(nkey) && master.canMove(DIRS[i])) {
int cost = master.move(DIRS[i]);
grid.put(nkey, cost);
dfs(nx, ny, master);
master.move(REV.get(DIRS[i]));
}
}
}
}
// Assume GridMaster interface is provided as per Leetcode
var findShortestPath = function(master) {
const dirs = ['U', 'D', 'L', 'R'];
const moves = {'U': [-1, 0], 'D': [1, 0], 'L': [0, -1], 'R': [0, 1]};
const rev = {'U': 'D', 'D': 'U', 'L': 'R', 'R': 'L'};
const grid = {};
let target = null;
function dfs(x, y) {
if (master.isTarget()) {
target = [x, y];
}
grid[x + ',' + y] = grid[x + ',' + y] || 0;
for (let d of dirs) {
const [dx, dy] = moves[d];
const nx = x + dx, ny = y + dy;
if ((nx + ',' + ny) in grid) continue;
if (master.canMove(d)) {
let cost = master.move(d);
grid[nx + ',' + ny] = cost;
dfs(nx, ny);
master.move(rev[d]);
}
}
}
grid['0,0'] = 0;
dfs(0, 0);
if (!target) return -1;
// Dijkstra's algorithm
let heap = [[grid['0,0'], 0, 0]];
const visited = new Set();
while (heap.length) {
heap.sort((a, b) => a[0] - b[0]);
let [cost, x, y] = heap.shift();
if (x === target[0] && y === target[1]) return cost;
let key = x + ',' + y;
if (visited.has(key)) continue;
visited.add(key);
for (let d of dirs) {
const [dx, dy] = moves[d];
const nx = x + dx, ny = y + dy;
let nkey = nx + ',' + ny;
if (nkey in grid && !visited.has(nkey)) {
heap.push([cost + grid[nkey], nx, ny]);
}
}
}
return -1;
};