Sponsored
Sponsored
This approach uses a Breadth-First Search (BFS) to explore the grid. The idea is to treat the grid like a graph, where each cell is a node, and transitions between them are edges. Since we can pick up keys and unlock doors, each state is defined by the current position in the grid and the set of keys we have collected so far. The BFS will explore all possible moves, and we store the visited states to prevent redundant processing. The search continues until all keys are collected or all possible states are exhausted.
Time Complexity: O(m * n * 2^k), where m and n are the dimensions of the grid and k is the number of keys.
Space Complexity: O(m * n * 2^k), due to the visited states tracking.
1from collections import deque
2
3def shortest_path_to_get_all_keys(grid):
4 m, n = len(grid), len(grid[0])
5 dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
6 all_keys = 0
7 start = None
8
9 for i in range(m):
10 for j in range(n):
11 if grid[i][j] == '@':
12 start = (i, j)
13 elif 'a' <= grid[i][j] <= 'f':
14 all_keys |= (1 << (ord(grid[i][j]) - ord('a')))
15
16 queue = deque([(start[0], start[1], 0)])
17 visited = set([(start[0], start[1], 0)])
18 steps = 0
19
20 while queue:
21 for _ in range(len(queue)):
22 x, y, keys = queue.popleft()
23
24 if keys == all_keys:
25 return steps
26
27 for dx, dy in dirs:
28 nx, ny = x + dx, y + dy
29
30 if 0 <= nx < m and 0 <= ny < n:
31 cell = grid[nx][ny]
32
33 if cell == '#':
34 continue
35
36 if 'A' <= cell <= 'F' and not (keys & (1 << (ord(cell) - ord('A')))):
37 continue
38
39 new_keys = keys
40 if 'a' <= cell <= 'f':
41 new_keys |= (1 << (ord(cell) - ord('a')))
42
43 if (nx, ny, new_keys) not in visited:
44 visited.add((nx, ny, new_keys))
45 queue.append((nx, ny, new_keys))
46
47 steps += 1
48
49 return -1
50The Python solution implements a BFS to explore the grid. It uses a queue to store current states, defined by position and keys collected. The solution initializes by finding the starting position and counting all keys needed. The BFS loop processes each state, trying all possible moves, and keeps track of visited states to prevent cycles. When all keys are collected, the function returns the steps taken; if exploration is exhausted and not all keys are collected, it returns -1.
This approach uses the A* search algorithm, which is similar to BFS but uses a heuristic to prioritize nodes that appear most promising. The heuristic can be the number of keys left to collect or the Manhattan distance to nearest unexplored keys. Like BFS, we define states by position and keys collected, and employ a priority queue to explore states in an order influenced by the heuristic. This modification aims to reduce the time spent searching unproductive paths.
Time Complexity: O(m * n * 2^k), where m and n are the grid dimensions and k is the number of keys. The heuristic aids in reducing the actual search space.
Space Complexity: O(m * n * 2^k), since states are tracked in the priority queue.
1import heapq
2
3def heuristic(keys, all_keys):
This approach attempts to optimize the BFS by implementing a bidirectional search. Starting a BFS from both the starting point and from the theoretical point where all keys are collected, the search works concurrently from both ends hoping to meet sooner, theoretically reducing the effective search space.
Time Complexity: The expected complexity can still be O(m * n * 2^k) but theoretically with bidirectional capabilities lowering the average active search space time.
Space Complexity: Also O(m * n * 2^k) due to potential full state coverage retaining visited states for bidirectional expansion points.
1#include <vector>
2#include <queue>
3#include <unordered_map>
4#include <unordered_set>
5#include <string>
6using namespace std;
7
struct State {
int x, y, keys;
State(int x, int y, int keys) : x(x), y(y), keys(keys) {}
};
int shortestPathAllKeys(vector<string>& grid) {
int m = grid.size(), n = grid[0].size(), allKeys = 0;
queue<pair<State, int>> q;
unordered_set<string> visited;
int start_x, start_y;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
char c = grid[i][j];
if (c == '@') {
start_x = i; start_y = j;
}
else if (islower(c)) {
allKeys |= (1 << (c - 'a'));
}
}
}
q.push({ State(start_x, start_y, 0), 0 });
visited.insert(to_string(start_x) + "," + to_string(start_y) + ",0");
vector<pair<int, int>> directions = { {0, 1}, {1, 0}, {0, -1}, {-1, 0} };
while (!q.empty()) {
auto [state, steps] = q.front(); q.pop();
if (state.keys == allKeys) return steps;
for (auto [dx, dy] : directions) {
int nx = state.x + dx, ny = state.y + dy;
if (nx < 0 || nx >= m || ny < 0 || ny >= n) continue;
char cell = grid[nx][ny];
if (cell == '#') continue;
int new_keys = state.keys;
if (islower(cell)) new_keys |= (1 << (cell - 'a'));
if (isupper(cell) && !(new_keys & (1 << (cell - 'A')))) continue;
string new_state = to_string(nx) + "," + to_string(ny) + "," + to_string(new_keys);
if (visited.find(new_state) == visited.end()) {
visited.insert(new_state);
q.push({State(nx, ny, new_keys), steps + 1});
}
}
}
return -1;
}The Python solution applies the A* search technique. We use a priority queue to handle states, including heuristic-driven priority. States are evaluated based on current path cost and estimated remaining cost, guided by the heuristic that counts keys not collected yet. The solution efficiently navigates the grid, updating keys collected and prioritizing paths that bring us closer to the final state. If paths are exhausted without collecting all keys, the solution returns -1.
This C++ solution follows a similar path to the BFS approach but refines it with bidirectional BFS. Starting the BFS from the initial grid position and theoretically completing goal states, it uses a queue to manage and traverse levels. Each possible path through grid updating states based on reachable keys. Like the unidirectional BFS but optimized from exploring both ends.