Shortest Path to Get All Keys
Identifying Problem Isomorphism
“Shortest Path to Get All Keys” can be mapped to “Keys and Rooms”.
Here’s the reasoning:
Both problems involve exploring a given structure (a grid in “Shortest Path to Get All Keys” and a list of rooms in “Keys and Rooms”) with constraints related to keys. In both cases, keys are necessary to progress further in the exploration.
The “Shortest Path to Get All Keys” problem asks for the shortest path to collect all keys in a grid where some cells are locked and can only be accessed when we have the corresponding key. Similarly, in the “Keys and Rooms” problem, some rooms can be accessed only when we have the key to that room.
The difference is in the complexity and dimension. “Shortest Path to Get All Keys” is a more complex problem as it involves finding the shortest path in a 2D grid with multiple keys and locks, which can be approached by using techniques like BFS with a state to track the keys collected so far. On the other hand, “Keys and Rooms” is a simpler problem where you just have to ensure if you can visit all the rooms starting from room 0, typically solved using Depth-First Search (DFS) or BFS.
Problem Classification
The given problem is a combination of a Grid-based Navigation problem and a Key-Lock puzzle.
What Components:
- You start at a given position on a 2D grid.
- The grid contains walls that you cannot pass, empty cells where you can move freely, keys represented by lowercase letters that you can pick up, and locks represented by uppercase letters that you can only pass if you have the corresponding key.
- The goal is to pick up all the keys in the minimum number of moves.
The problem can be classified in the following way:
Navigation: You have to move around the grid while avoiding walls and locks you can’t open.
Pathfinding: The goal is not only to reach a certain point, but to reach all keys. This introduces an element of pathfinding, as you need to find the shortest path that covers all keys.
Graph Theory: The grid can be thought of as a graph, where each cell is a node. Moving from one cell to another is equivalent to traversing an edge. The keys and locks add a layer of complexity to the graph, as they dynamically change the graph’s connectivity.
State Space Search: The problem involves searching through the space of all possible states (positions on the grid and sets of keys you have picked up) to find the path to the goal state (having all keys). This suggests that techniques like breadth-first search, depth-first search, or A* search might be applicable.
Dynamic Programming: The subproblem of finding the shortest path to a key given the current state (position and keys you have already picked up) can potentially overlap with other subproblems. This suggests that dynamic programming might be a useful technique for solving this problem efficiently.
Puzzle: This is a type of puzzle problem, as it involves finding a solution to a complex scenario with many interacting parts.
Problem Simplification and Explanation
In the problem, you can imagine yourself in a maze-like house with several rooms. The house is unusual because it has specific doors that can only be opened with a specific key, and you can’t climb over these doors or break them down. The keys and the doors they open are unique to each other, like a lock and its key in the real world.
Each room in this house is represented by a cell in the grid. Some rooms are locked (represented by uppercase letters), some rooms contain keys (represented by lowercase letters), and some rooms are empty (represented by a ‘.’). Some parts of the house are blocked off by walls (represented by ‘#’), which you can’t pass through. There’s also a special room where you start your exploration of the house (represented by ‘@’).
Your task is to navigate this house, collecting all the keys in the least number of steps. Each step corresponds to moving from one room to an adjacent room in any of the four cardinal directions: north, east, south, and west. You can’t walk through walls or locked doors, but you can pick up any key you find by walking into the room that contains it. Picking up a key allows you to pass through the door it opens.
If you pick up a key, it remains in your possession for the rest of your exploration; you don’t leave it in the lock after opening a door. If you come across a locked door for which you have the key, you can open it and pass through. But if you encounter a door for which you don’t yet have the key, you must backtrack and explore other parts of the house until you find the right key.
To reiterate, the goal is to pick up all the keys in the house, not necessarily to open all the doors. If you can pick up all the keys without opening some of the doors, that’s fine. If it’s impossible to pick up all the keys (for example, if a key is located behind a door for which there is no corresponding key), you should indicate that by returning -1.
The main challenge is to figure out the most efficient route through the house that allows you to pick up all the keys. This involves making strategic decisions about which parts of the house to explore first, which keys to pick up in which order, and when to backtrack.
Thought Process
Given the problem statement, here are some insights that can guide us toward a solution:
Grid traversal: The problem describes a grid and movement in four cardinal directions. This is a clue that we’ll likely need to traverse the grid, and common ways to do this include depth-first search (DFS) and breadth-first search (BFS). Here, since we want to find the minimum number of moves, BFS is more suitable.
Keys and locks: The problem mentions keys and locks. This is a hint that we’ll need a way to keep track of which keys we’ve picked up, which keys we still need, and which locks we can open.
Grid size and number of keys: The constraints limit the size of the grid and the number of keys. This suggests that our solution can afford to be fairly computational, as the number of possible states (each cell, with each possible set of keys) is manageable.
Impossible situations: The problem asks us to return -1 if it’s impossible to pick up all keys. This suggests that our solution needs a way to detect when we’ve exhausted all possibilities without success.
Given these insights, a possible approach is as follows:
- Use BFS to traverse the grid. For each cell, consider all adjacent cells that can be reached in one move.
- To keep track of keys, use a bitmask. Each bit in the mask represents whether we have a certain key.
- In BFS, the ‘visited’ concept will be a combination of the cell and the current keys possessed, as it’s possible to visit the same cell with different keys.
- If at any point we’ve picked up all keys, return the number of moves taken to get there.
- If the BFS finishes without finding all keys, return -1.
Here’s a Python solution using this approach:
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In this code, nx and ny represent the next cell to move to, dx and dy represent the change in x and y coordinates to move to the next cell, and (1 << (ord(grid[nx][ny].lower()) - ord('a'))) is the bitmask operation to update the keys collected.
Solution Approach and Analysis
Model the Grid as a Graph: You can think of each cell in the grid as a node in a graph, with an edge to each of its neighbors (up, down, left, and right). This lets you apply graph traversal algorithms to explore the grid. Walls are like impassable terrain, while keys and locks are special nodes that you interact with in specific ways.
Initialize Variables: You’ll need variables to store your current position, the state of your keys, and the number of steps taken. Initially, your position is the starting point ‘@’, you have no keys, and you’ve taken zero steps.
Apply Breadth-First Search (BFS): Begin a BFS from your current position. BFS works by visiting all nodes at distance 1, then all nodes at distance 2, and so on, which makes it ideal for finding the shortest path in an unweighted graph.
Bitmasking for Key Tracking: Each time you reach a cell with a key, you’ll want to “pick up” that key by changing your key state. You can use bitmasking for this: each bit in a 6-bit number represents whether you have a certain key. For instance, if you pick up key ‘a’, you can represent this as 000001 in binary, or 1 in decimal.
Modify BFS for Key and Lock Interaction: Normally in BFS, you’d mark a node as visited once you reach it to avoid visiting it again. But in this case, you might need to visit the same cell multiple times with different sets of keys. So, instead of marking a cell as visited, you mark a cell as visited with a certain set of keys. You can visit it again if you have a different set of keys.
End Condition: Your BFS continues until either all cells are visited with all possible sets of keys, or you’ve visited a cell with all keys. If you finish the BFS without finding all keys, return -1.
Track the Number of Moves: Each time you move to a new cell in the BFS, increment a counter that tracks the number of moves. When you find all keys, this counter will hold the minimum number of moves required.
Let’s go through an example:
Let’s take the first example from the problem statement: grid = ["@.a..","###.#",“b.A.B”]
Our steps would be as follows:
Initialize our variables. Our starting position is (0,0), and we have no keys (represented as 00 in binary).
Begin BFS. The only option is to move right. This puts us at (0,1) with no keys, after 1 move.
Continue BFS. We can only move right again. This puts us at (0,2) with key ‘a’ (represented as 01 in binary), after 2 moves.
Continue BFS. We can now move either right or down. Let’s say we move right. This puts us at (0,3) with key ‘a’, after 3 moves.
Continue BFS. We can only move right. This puts us at (0,4) with key ‘a’, after 4 moves.
Continue BFS. We can only move down. This puts us at (1,4) with key ‘a’, after 5 moves.
Continue BFS. We can only move left. This puts us at (1,3) with key ‘a’, after 6 moves.
Continue BFS. We can only move down. This puts us at (2,3) with key ‘b’ (represented as
11 in binary, since we now have both keys), after 7 moves.
- Continue BFS. We can only move left. This puts us at (2,2) with keys ‘a’ and ‘b’, after 8 moves.
At this point, we’ve picked up all keys, so we can stop. Our counter says it took 8 moves, so we return 8.
Identification of Applicable Theoretical Concepts
There are several mathematical and algorithmic concepts that are relevant to this problem:
Graph Theory: The problem can be modeled as a graph where each cell is a node and edges connect adjacent cells. You can use graph traversal algorithms to navigate the grid.
Breadth-First Search (BFS): BFS is a graph traversal algorithm that can be used to find the shortest path in an unweighted graph, which is suitable for this problem where each move costs the same regardless of direction.
Depth-First Search (DFS): Alternatively, DFS is another graph traversal algorithm that could be used to explore the grid. However, it might not be as efficient as BFS in finding the shortest path.
Bitmasking: Given the constraint that there are at most 6 keys, we can use a technique called bitmasking to keep track of which keys have been picked up. Bitmasking involves using the binary representation of an integer to represent a set of options. In this case, a 6-bit integer can represent the state of all keys.
State Space Search: This problem involves searching the space of all possible states to find a path to the goal. Each state includes the current position and the keys that have been picked up. BFS or DFS can be used to search the state space.
Dynamic Programming (DP): The problem has overlapping subproblems, which suggests that DP might be useful. For example, the shortest path to a particular cell with a certain set of keys might be a subproblem common to multiple different paths.
Queue Data Structure: This can be used in BFS to keep track of the cells to be visited.
By leveraging these concepts, we can devise an efficient approach to solve this problem. BFS combined with bitmasking and state tracking seems to be a viable strategy. Starting from the initial position, we can use BFS to explore the grid, at each step checking whether we’ve reached a cell with a key, and if we can pick up that key (based on the alphabetic order constraint). We use bitmasking to update the state of the keys and continue the BFS until we’ve picked up all the keys or have exhausted all reachable cells.
Constraints
Given the problem statement and constraints, here are some conditions that we can take advantage of to find an efficient solution:
Grid size: The grid is relatively small, with maximum dimensions of 30x30. This means we can use computationally heavier algorithms such as depth-first search or breadth-first search without worrying too much about time or space complexity.
Limited Keys: The problem statement mentions that there are at most 6 different keys in the grid. This is a very small number, which means we can use bitmasking to keep track of the keys we’ve collected. Bitmasking is a technique where we use the bits of an integer to represent a set of options. In this case, we can use a 6-bit integer to represent the state of all keys.
Start Point: The grid has exactly one starting point ‘@’, which serves as a root for our search algorithm.
The keys and locks are in alphabetic order: This is a crucial detail. It implies that the keys will be found in the order they appear in the alphabet. This reduces the complexity of the problem since we do not have to consider scenarios where we might find key ‘b’ before key ‘a’. We can use this information to ensure that our search is proceeding in the right direction.
The number of keys equals the number of locks: This simplifies the problem because we know that for every lock we encounter, there is exactly one key that can open it. We don’t have to consider scenarios where there are extra keys or locks.
One move consists of walking one space in one of the four cardinal directions: This limits the possible actions we can take at any given point, which makes the search space more manageable.
By exploiting these characteristics, we can develop a more efficient solution to the problem. For example, using breadth-first search with a queue that stores not only the current position but also the state of the keys allows us to explore all possible states efficiently. Bitmasking allows us to handle the key state easily, and the fact that keys and locks are in alphabetic order guides our search in the right direction.
Language Agnostic Coding Drills
Consider the following piece of complex software code.
Your mission is to deconstruct this code into the smallest possible learning units, each corresponding to a separate coding concept. Consider these concepts as unique coding drills that can be individually implemented and later assembled into the final solution.
Dissect the code and identify each distinct concept it contains. Remember, this process should be language-agnostic and generally applicable to most modern programming languages.
Once you’ve identified these coding concepts or drills, list them out in order of increasing difficulty. Provide a brief description of each concept and why it is classified at its particular difficulty level.
Next, describe the problem-solving approach that would lead from the problem statement to the final solution. Think about how each of these coding drills contributes to the overall solution. Elucidate the step-by-step process involved in using these drills to solve the problem. Please refrain from writing any actual code; we’re focusing on understanding the process and strategy.
Targeted Drills in Python
Now that you’ve identified and ordered the coding concepts from a complex software code in the previous exercise, let’s focus on creating Python-based coding drills for each of those concepts.
Begin by writing a separate piece of Python code that encapsulates each identified concept. These individual drills should illustrate how to implement each concept in Python. Please ensure that these are suitable even for those with a basic understanding of Python.
In addition to the general concepts, identify and write coding drills for any problem-specific concepts that might be needed to create a solution. Describe why these drills are essential for our problem.
Once all drills have been coded, describe how these pieces can be integrated together in the right order to solve the initial problem. Each drill should contribute to building up to the final solution.
Remember, the goal is to not only to write these drills but also to ensure that they can be cohesively assembled into one comprehensive solution.
Q&A
It’s good that you have a working A* solver and it’s even better that you’re striving to improve it. In the context of A* algorithm, the heuristic function plays an integral role in performance. Your aim is to estimate the cost from the current node to the goal node.
For A* algorithm to give an optimal solution, the heuristic function needs to be admissible, meaning it should never overestimate the cost to reach the goal. This is where some of your attempted heuristics might have gone wrong.
The manhattan distance, average distance, maximum distance, and closest key distance heuristics can potentially overestimate the cost in some scenarios, especially considering the constraints of the problem (e.g., the fact that you have to pick up keys in a certain order).
It’s also worth noting that, in addition to being admissible, an ideal heuristic function for A* would also be consistent (or monotonic). This means that for any node N and any successor P of N, the estimated cost of reaching the goal from N is no greater than the cost of getting to P plus the estimated cost of reaching the goal from P.
Finding the right heuristic for a problem can be challenging, and sometimes, like in your case, it’s not immediately clear what the best heuristic would be. A heuristic of zero (which basically makes A* behave like Dijkstra’s algorithm) is always admissible and consistent, but it won’t help speed up the search.
One possible approach for your problem is to use a heuristic that takes into account the cost of reaching the remaining keys, based on the current state of keys that you have. This could involve precomputing and storing the shortest path distance between all pairs of keys, and then, during the A* search, using these distances to estimate the cost of reaching the remaining keys.
However, this still may not provide a perfect heuristic, because the order in which you pick up the keys can greatly affect the overall cost, and this is not something easily accounted for in a simple heuristic function.
A* is guaranteed to find an optimal solution even with a heuristic of zero, as long as you’re using a good strategy for breaking ties between nodes with equal path costs. Remember that the heuristic function is just a tool for performance optimization, and sometimes, the nature of the problem makes it hard to find an effective heuristic.
If you’re still interested in improving your solution, you might want to look into other search algorithms or techniques that could be more suited to this problem, such as Iterative Deepening A* (IDA*) or techniques related to Constraint Satisfaction Problems (CSPs).
The solution says to use BFS !! Why can’t we use DFS? (i.e backtracking) for this problem ? Are DFS and backtracking same?
The reason BFS (Breadth-First Search) is often chosen for these kinds of problems is because BFS guarantees to find the shortest path in a graph where all edges have the same weight, which is the case in this problem (each move from one cell to another can be considered as an edge with weight 1). This is because BFS explores all nodes at the current depth (or distance) before proceeding to nodes at the next depth level.
On the other hand, DFS (Depth-First Search) goes as deep as possible along each branch before backtracking. This could lead it down a very long path, even if a shorter one exists. Therefore, DFS does not necessarily find the shortest path.
In terms of whether DFS and backtracking are the same, they are closely related but not exactly the same. DFS is a type of traversal algorithm that goes as far as possible down a path before retreating, while backtracking is a general algorithm for finding all (or some) solutions to computational problems, particularly constraint satisfaction problems, by incrementally building candidates to the solutions, and abandoning a candidate as soon as it is determined that the candidate cannot possibly be extended to a valid solution.
The classic DFS can be considered a kind of backtracking algorithm, but not all backtracking algorithms are DFS. For instance, some backtracking algorithms might not explore all branches of the search tree, because they can reject some branches based on specific criteria.
Therefore, while both DFS and backtracking involve exploring and retreating, they are not the same. DFS is a specific kind of graph traversal (which can be used in a backtracking algorithm), and backtracking is a more general algorithmic strategy.
should be tagged with memoization as well, since it can be solved with BFS and memoization without using a bitMask.
Yes, you’re correct. In fact, memoization is an integral part of this problem solution, even if it isn’t always explicitly labeled as such. When we use a data structure (like a 3D array in this case) to keep track of states we’ve already visited, we’re effectively using memoization to remember past results and avoid redundant work.
The bitmask serves a specific function as part of the memoization strategy in this problem: it provides a compact way to represent the collection of keys we’ve picked up. However, there are certainly other ways to represent the same information. For instance, instead of a bitmask, we could use a set or list to track the keys we’ve picked up.
Using a bitmask has some benefits in this context because it allows us to represent the state of all keys in a single integer (using each bit in the integer to represent the presence or absence of a corresponding key). This can be more space-efficient than using a separate variable for each key, and it also allows us to use bitwise operations to quickly and efficiently manipulate the key collection (e.g., adding a key, checking if a key is present, etc.).
But regardless of how we choose to represent the key collection, the underlying principle of memoization remains the same: we’re remembering past states to avoid redoing the same work.
So, yes, tagging the problem with ‘memoization’ is quite appropriate, as it’s a key aspect of the problem solution, even though the specific method of memoization (i.e., using a bitmask or some other data structure to represent the keys) may vary.
What is the last testcase? I’m getting TLE and I can’t see the input.
TLE doesn’t make sense btw, my time complexity is O(m⋅n⋅2k)O(m \cdot n \cdot 2^k)O(m⋅n⋅2 k ) What am I missing :‘c
Here’s my code ._.
class Solution { public: vector<vector> directions = {{0, +1}, {+1, 0}, {0, -1}, {-1, 0}};
int count_keys(vector<string>& grid, int m, int n) {
set<char> s;
for(int i = 0; i < m; i++) for(int j = 0; j < n; j++) if(islower(grid[i][j])) s.insert(grid[i][j]);
return s.size();
}
pair<int, int> find_start_pos(vector<string>& grid, int m, int n) {
for(int i = 0; i < m; i++) for(int j = 0; j < n; j++) if(grid[i][j] == '@') return {i, j};
return {-1, -1};
}
int key_map(set<char>& keys) {
int res = 0;
for(char k : keys) {
res += 1 * (k == 'a');
res += 2 * (k == 'b');
res += 4 * (k == 'c');
res += 8 * (k == 'd');
res += 16 * (k == 'e');
res += 32 * (k == 'f');
}
return res;
}
int solve(vector<string>& grid, int num_of_keys, int i, int j, int m, int n) {
int res = 1e5;
vector<vector<vector<bool>>> visited(m, vector<vector<bool>>(n, vector<bool>(64)));
queue<pair<vector<int>, set<char>>> q;
q.push({{i, j, 0, 0}, {}});
while(!q.empty()) {
int ii = q.front().first[0];
int jj = q.front().first[1];
int id = q.front().first[2];
int sc = q.front().first[3];
set<char> keys = q.front().second;
q.pop();
if(ii < 0 || jj < 0 || ii >= m || jj >= n) continue;
if(grid[ii][jj] == '#') continue;
if(islower(grid[ii][jj])) keys.insert(grid[ii][jj]);
if(keys.size() == num_of_keys) {
res = min(res, sc);
continue;
}
if(isupper(grid[ii][jj])) {
if(keys.find(tolower(grid[ii][jj])) != keys.end()) grid[ii][jj] = '.';
else continue;
}
if(visited[ii][jj][id]) continue;
else visited[ii][jj][id] = true;
for(vector<int> dir : directions) q.push({{ii + dir[0], jj + dir[1], key_map(keys), sc + 1}, keys});
}
return res;
}
int shortestPathAllKeys(vector<string>& grid) {
int m = grid.size();
int n = grid[0].size();
pair<int, int> start = find_start_pos(grid, m, n);
int num_of_keys = count_keys(grid, m, n);
grid[start.first][start.second] = '.';
int res = solve(grid, num_of_keys, start.first, start.second, m, n);
return res >= 1e5 ? -1 : res;
}
};
I replaced the type for allKeys to be Set instead of int. I am unable to figure out whats wrong with my code. Can someone please help?
class Solution { public int shortestPathAllKeys(String[] grid) { int m = grid.length, n = grid[0].length(); Queue queue = new LinkedList<>(); int[][] moves = new int[][] {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
// seen['key'] is only for BFS with key state equals 'keys'
Map<Set<Character>, Set<Pair<Integer, Integer>>> seen = new HashMap<>();
Set<Character> keySet = new HashSet<>();
Set<Character> lockSet = new HashSet<>();
int startR = -1, startC = -1;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
char cell = grid[i].charAt(j);
if (cell >= 'a' && cell <= 'f') {
keySet.add(cell);
}
if (cell >= 'A' && cell <= 'F') {
lockSet.add(cell);
}
if (cell == '@') {
startR = i;
startC = j;
}
}
}
// [row, column, distance, key state]
queue.offer(new Node(startR, startC, 0, new HashSet<>()));
seen.put(new HashSet<>(), new HashSet<>());
seen.get(new HashSet<>()).add(new Pair<>(startR, startC));
while (!queue.isEmpty()) {
Node cur = queue.poll();
int curR = cur.x, curC = cur.y, dist = cur.distance;
Set<Character> keys = cur.keys;
for (int[] move : moves) {
int newR = curR + move[0], newC = curC + move[1];
// If this cell (newR, newC) is reachable.
if (newR >= 0 && newR < m && newC >= 0 && newC < n && grid[newR].charAt(newC) != '#') {
char cell = grid[newR].charAt(newC);
// If it is a key:
if (keySet.contains(cell)) {
// If we have collected it before, no need to revisit this cell.
if (keys.contains(cell)) {
continue;
}
// Otherwise, we can walk to this cell and pick it up.
Set<Character> newKeys = new HashSet<>(keys);
newKeys.add(cell);
// If we collect all keys, return dist + 1.
// Otherwise, just add this state to seen and queue.
if (newKeys.equals(keySet)) {
return dist + 1;
}
seen.putIfAbsent(newKeys, new HashSet<>());
seen.get(newKeys).add(new Pair<>(newR, newC));
queue.offer(new Node(newR, newC, dist + 1, newKeys));
}
// If it is a lock and we don't have its key, continue.
if (lockSet.contains(cell) && (!keys.contains(cell - 'A'))) {
continue;
}
// We can walk to this cell if we haven't been here before with the same key state.
else if (!seen.get(keys).contains(new Pair<>(newR, newC))) {
seen.get(keys).add(new Pair<>(newR, newC));
queue.offer(new Node(newR, newC, dist + 1, keys));
}
}
}
}
return -1;
}
}
class Node { int x; int y; int distance; Set keys;
public Node(int x, int y, int distance, Set<Character> keys) {
this.x = x;
this.y = y;
this.distance = distance;
this.keys = keys;
}
}
Your code overall has a correct approach towards the problem. But there seems to be a minor issue in your code which is causing the error. The issue lies in your lock-checking logic in this line:
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Here, you’re checking if the cell is a lock by checking if it’s in the lockSet. If it is a lock, you’re checking if we have its key by checking if keys contains cell - 'A'.
The problem is that cell - 'A' will give you an integer value, not a character. If the cell contains a lock ‘A’, then cell - 'A' will be 0, not ‘a’. This is why the key is never found in keys, even if you have collected it.
To fix this, you should convert the lock to the key properly:
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In this corrected line, Character.toLowerCase(cell) will convert the lock character (‘A’-‘F’) to the corresponding key character (‘a’-‘f’).
That should resolve the issue. This issue also brings to attention the need for thorough testing and debugging in these types of problems, which often have very nuanced requirements that can lead to such bugs. It’s often useful to write test cases that cover various scenarios to ensure your code works as expected in all situations.
class Solution { public: int solve(vector&grid,int x,int y,int keyMask) { int n=grid.size(); int m=grid[0].size();
queue<pair<pair<int,int>,pair<int,int»>q;
// push the start into queue
// 1. number of moves taken // 2. mask keys taken // 3. x pos // 4. y pos
q.push({{0,0},{x,y}});
vector<vector<vector<int>>> vis(n+1, vector<vector<int>>(m+1, vector<int>(keyMask+1,0)));
// mark the start as visited
vis[x][y][0]=1;
// 4 directions
vector<vector<int>>dir={{-1,0},{1,0},{0,-1},{0,1}};
// keys gathered
unordered_map<char,bool>keyMap;
keyMap['A']=0;
keyMap['B']=1;
keyMap['C']=2;
keyMap['D']=3;
keyMap['E']=4;
keyMap['F']=5;
keyMap['a']=0;
keyMap['b']=1;
keyMap['c']=2;
keyMap['d']=3;
keyMap['e']=4;
keyMap['f']=5;
while(!q.empty())
{
pair<pair<int,int>,pair<int,int>>tp=q.front();
q.pop();
int moves=tp.first.first;
int mask=tp.first.second;
int currX=tp.second.first;
int currY=tp.second.second;
vis[currX][currY][mask]=1;
cout<<mask<<" ";
// we have gathered all the keys
if(mask==keyMask)return moves;
for(int i=0;i<4;i++)
{
int nextX=currX+dir[i][0];
int nextY=currY+dir[i][1];
if(nextX>=0 && nextX<n && nextY>=0 && nextY<m && vis[nextX][nextY][mask]==0)
{
if(grid[nextX][nextY]>='A' && grid[nextX][nextY]<='F')
{
// it is a lock
int bit=keyMap[grid[nextX][nextY]];
int hasKey=mask&(1<<bit);
if(hasKey==1)
{
q.push({{moves+1,mask},{nextX,nextY}});
}
}
else if(grid[nextX][nextY]>='a' && grid[nextX][nextY]<='f')
{
int bit=keyMap[grid[nextX][nextY]];
int newMask=mask|(1<<bit);
q.push({{moves+1,newMask},{nextX,nextY}});
}
else if(grid[nextX][nextY]=='.')
{
q.push({{moves+1,mask},{nextX,nextY}});
}
}
}
}
return -1; } int shortestPathAllKeys(vector& grid) {
int n=grid.size();
int m=grid[0].size();
int numberOfKeys=0;
int reqMask=0;
int startx,starty;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(grid[i][j]=='@')
{
startx=i;
starty=j;
}
if(grid[i][j]=='a')
reqMask=reqMask|(1<<0);
else if(grid[i][j]=='b')
reqMask=reqMask|(1<<1);
else if(grid[i][j]=='c')
reqMask=reqMask|(1<<2);
else if(grid[i][j]=='d')
reqMask=reqMask|(1<<3);
else if(grid[i][j]=='e')
reqMask=reqMask|(1<<4);
else if(grid[i][j]=='f')
reqMask=reqMask|(1<<5);
}
}
// cout«reqMask; int res=solve(grid,startx,starty,reqMask);
return res; } };
can anyone tell me where am I going wrong?
Your implementation is mostly correct, but it seems that you have a bug related to the condition of encountering a wall # in the grid.
In your BFS, you have conditions to handle lower-case letters (which represent keys), and upper-case letters (which represent locks), and also the situation where the grid cell is a ‘.’ (which means a free space). However, you don’t have a condition to handle ‘#’ (which represents a wall). Your code tries to pass through walls, which is not allowed in the problem.
To fix this bug, you need to add an additional condition in your BFS loop to continue to the next iteration if the character in the new position is a ‘#’.
Here is the updated BFS part:
| |
In the updated code, I also move vis[nextX][nextY][mask] = 1; to each condition. This is because you only want to mark a state as visited when you actually add it to the queue.
Another thing to note is that in the condition if(hasKey==1), you should check if(hasKey != 0). This is because hasKey might not be 1 when it’s true. It can be any non-zero integer.
I haven’t yet written my solution. Can anyone please check my intuition :
Let there be k keys. Then we can collect keys k! way. Some of them may not even possible to collect. If none of these are possible then -1. With the help of BFS we can find the key in the particular order if possible. To find the ith key in a particular order, if the path compulsorily requires a lock to open whose key is not yet found then that particular order is not possible. TC: (k! mn)
Your intuition is on the right track, but the problem is a bit more complex than that due to the spatial dependencies of the keys and locks.
Your approach considers permutations of key orders, which is a good starting point. However, the issue is that even if we find a valid order of keys, the paths to collect them might overlap or intersect in a way that makes that order invalid or inefficient.
For instance, imagine a situation where the optimal path to collect keys is A->B->C, but the direct path from A to B is blocked by the C-lock. In that case, the optimal path would be A->C->B, and simply permuting the order of keys without considering the locks and spatial arrangement would not solve the problem correctly.
A more comprehensive approach to this problem is to use breadth-first search (BFS) in combination with a bit mask to track the state of collected keys.
Here’s the general idea:
Traverse the grid once to find the start position, the positions of all the keys, and the total number of keys.
Start a BFS from the start position. The state in the BFS queue is a combination of the current cell position and a “key state”, which is an integer where the i-th bit is 1 if the i-th key has been collected, and 0 otherwise.
For each state (current cell and key state), iterate its four neighboring cells. If a neighbor is reachable (empty or with a key), and it hasn’t been visited before with the same key state, then it’s a valid new state and add it into the queue.
If a neighboring cell contains a key that hasn’t been collected, then change the key state accordingly.
If all keys have been collected (i.e., the key state is (1 « total number of keys) - 1), then we’ve found a valid path. Return its length, which is the level of BFS.
If no valid path is found after BFS ends, return -1.
The time complexity of this approach is O((2^k) * mn), where m and n are the dimensions of the grid, and k is the number of keys. This is because in the worst case, we might have to visit every cell with each possible key state.
need help with my code, I used DFS/backtracking and this is giving -1 for input ["@…a",".###A",“b.BCc”]…. can anyone explain what’s wrong?
class Solution { public: int row, col, ans = INT_MAX, c = 0; void dfs(vector& grid, int i, int j, unordered_set& keys, vector& vis, int dist) { if (i < 0 || j < 0 || i >= row || j >= col || vis[i][j] || grid[i][j] == ‘#’) { return; } if (grid[i][j] >= ‘A’ && grid[i][j] <= ‘Z’) { char keyChar = tolower(grid[i][j]); if (keys.find(keyChar) == keys.end()) return; } vis[i][j] = true; if (grid[i][j] >= ‘a’ && grid[i][j] <= ‘z’) keys.insert(grid[i][j]); if (keys.size() == c) { ans = min(ans, dist); } dfs(grid, i, j + 1, keys, vis, dist + 1); dfs(grid, i, j - 1, keys, vis, dist + 1); dfs(grid, i + 1, j, keys, vis, dist + 1); dfs(grid, i - 1, j, keys, vis, dist + 1); vis[i][j] = false; if (grid[i][j] >= ‘a’ && grid[i][j] <= ‘z’) keys.erase(grid[i][j]); } int shortestPathAllKeys(vector& grid) { int i = 0, j = 0, a = 0, b = 0; for (auto& x : grid) { j = 0; for (auto& k : x) { if (k >= ‘a’ && k <= ‘z’) { c++; } if (k == ‘@’) { a = i; b = j; } j++; } i++; } row = grid.size(); col = grid[0].size(); unordered_set key; vector vis(row, vector(col, false)); dfs(grid, a, b, key, vis, 0); return ans == INT_MAX ? -1 : ans; } };
Your code uses Depth First Search (DFS) for solving the problem. While DFS is a good approach for certain problems, in this case, it can lead to inefficiencies or even incorrect results due to its nature of exhaustively searching one path before backtracking.
One issue in your code lies in the backtracking section, where you mark the node as unvisited and also remove the key (if it’s a key node). The issue here is that you might have another path that also goes through this node and requires this key. Because you remove the key upon backtracking, the other path wouldn’t be able to proceed when it’s actually a valid path.
Let’s consider the case ["@…a",".###A",“b.BCc”]. Your DFS will go like:
- Start at ‘@’, get ‘a’, can’t proceed further because ‘A’ is blocking the way, start backtracking.
- Backtrack and remove ‘a’ from keys, then try to go to ‘b’ path.
- You can’t proceed on ‘b’ path because ‘a’ is not in keys anymore.
In order to correct the behavior, you could consider the BFS approach, because BFS would check all paths of a certain length before proceeding to the next length, ensuring it explores all possible orders of key collection.
Additionally, consider using a bitmask for the state of keys. This will allow you to mark each cell as visited for each possible key state, which will handle the case where the same cell needs to be visited multiple times with different keys collected.
Finally, you should be careful with handling uppercase and lowercase letters. Uppercase letters ‘A’-‘F’ represent locks, and lowercase ‘a’-‘f’ represent keys. If you encounter a lock, you should check if you have the corresponding key before proceeding, otherwise, you should ignore this path.
grid = ["@.a.c",".##C#",“b.A.B”] my ans = 10 leetcode ans = 8
can someone please make me understand how the answer is 8?
my logic in java :
public int shortestPathAllKeys(String[] grid) {
int m = grid.length;
int n = grid[0].length();
boolean[][] visited = new boolean[m][n];;
Set<Character> foundKeys = new HashSet<>();
Queue<int[]> q = new LinkedList<>();
int[][] dirs = new int[][]{{0,1},{1,0},{0,-1},{-1,0}};
int startI = 0;
int startJ = 0;
int count = 0;
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(grid[i].charAt(j) == '@'){
q.add(new int[]{i, j, 0});
visited[i][j] = true;
}
if(grid[i].charAt(j) >= 'a' && grid[i].charAt(j) <= 'z'){
count++;
}
}
}
while(!q.isEmpty() && count > 0){
int[] top = q.remove();
int i = top[0];
int j = top[1];
int path = top[2];
if(grid[i].charAt(j) >= 'a' && grid[i].charAt(j) <= 'z' && !foundKeys.contains(grid[i].charAt(j))){
count--;
visited = new boolean[m][n];
visited[i][j] = true;
foundKeys.add(grid[i].charAt(j));
q.clear();
}
for(int[] dir : dirs){
int newI = i + dir[0];
int newJ = j + dir[1];
if(newI < 0 || newI >= m || newJ < 0 || newJ >= n || visited[newI][newJ]
|| grid[newI].charAt(newJ) == '#'){
continue;
}
if(grid[newI].charAt(newJ) >= 'A' && grid[newI].charAt(newJ) <= 'Z'
&& !foundKeys.contains(Character.toLowerCase(grid[newI].charAt(newJ)))){
continue;
}
q.add(new int[]{newI, newJ, path + 1});
visited[newI][newJ] = true;
}
}
if(count > 0){
return -1;
}
return q.remove()[2] - 1;
}
The mistake in your approach lies in the part where you clear your queue when you find a key. What this does is discard all the other possible routes that you could have taken. So, when you reach a key, you’re essentially saying, “this is the only path I care about now,” which could make you overlook shorter paths.
Let’s take the example grid = ["@.a.c",".##C#",“b.A.B”].
Your BFS will proceed as follows:
- From ‘@’ to ‘a’ - 2 moves.
- From ‘a’ to ‘b’ - 4 moves. (total: 2 + 4 = 6)
- From ‘b’ to ‘c’ - 4 moves. (total: 6 + 4 = 10)
But the shortest path is:
- From ‘@’ to ‘a’ - 2 moves.
- From ‘a’ to ‘c’ - 2 moves. (total: 2 + 2 = 4)
- From ‘c’ to ‘b’ - 4 moves. (total: 4 + 4 = 8)
This shorter path gets overlooked in your approach because the queue gets cleared when ‘a’ is discovered, and all the paths from ‘a’ that were in the queue are lost.
Instead of clearing the queue and the visited array every time you find a key, a better approach would be to use a separate visited array for each state of keys. A “state of keys” refers to the set of keys that you currently have - for example, having keys ‘a’ and ‘b’ would be one state, and having keys ‘a’, ‘b’, and ‘c’ would be another state. You can represent these states as an integer, where the i-th bit is 1 if you have the i-th key and 0 otherwise.
By using a separate visited array for each state, you can make sure that you don’t visit the same cell in the same state more than once, but you can still visit the same cell in a different state. This allows you to explore all possible paths and find the shortest one.
DFS does not necessarily yield shortest path in an undirected graph
Yes, you are correct. Depth-First Search (DFS) does not guarantee the shortest path in an undirected graph with unweighted edges or a directed/undirected graph with weighted edges. DFS is more suited for tasks like checking connectedness or connectivity in a graph, path finding in a maze, or performing a task that involves exploring all paths or options until a condition is met.
The nature of DFS is such that it explores one path deeply until it cannot go any further before backtracking and exploring other paths. This means that it does not always find the shortest path first, as it may have chosen to explore a longer path to start with.
For finding the shortest path, especially in weighted graphs, algorithms like Dijkstra’s or Bellman-Ford’s are more appropriate. In the case of unweighted graphs, Breadth-First Search (BFS) is typically used to find the shortest path, as it naturally checks all vertices at distance d from the start before it checks any vertices at distance d+1. This ensures that if a path exists of length d, BFS finds it before it finds any path of length greater than d.
In the context of your problem, it seems like you’re trying to find the shortest path that involves collecting certain items (keys), which means you’re looking for the shortest path given certain conditions. This is a more complex problem that might require modifications to typical shortest path algorithms or might require more advanced pathfinding algorithms. In this case, the A* algorithm could be a good fit as it allows for heuristic-based searching which can be adapted to account for your keys.
BFS is a graph traversal algorithm that visits all the vertices of a graph in breadth-first order. It is a suitable choice for finding the shortest path between two vertices of a graph, which is applicable to this problem. However, a regular BFS cannot solve this problem. Let’s examine the issue using the example shown above.
Indeed, the problem at hand is more complicated than what can be solved by a regular Breadth-First Search (BFS) because it’s essentially a “Shortest Path with Keys and Doors” problem, a variant of the typical shortest path problem. BFS alone is not sufficient to solve this problem as the order in which we collect the keys significantly influences the total path length.
Consider the example grid: ["@.a.c",".##C#","b.A.B"]
In this case, the BFS might reach the lock ‘C’ before collecting the corresponding key ‘c’ if it chooses to go right first instead of down, resulting in a blocked path. Even if we have visited a certain cell, we might need to visit it again later when we have more keys. Therefore, the “visited” state of a cell is not only determined by its coordinates, but also the keys we have when we visit it.
To tackle this, we need to add another dimension to the problem to keep track of the keys we have. Instead of marking [i][j] as visited, we should mark [i][j][state] as visited, where state is a bitmask representing the set of keys we currently have. This way, we allow revisiting a cell as long as we have a new key when we visit it.
Hence, we should use a modified version of BFS that incorporates state (keys obtained so far) into the search. This variant is often referred to as multi-state BFS. In the BFS queue, instead of only storing the coordinates of each cell, we should also store the state of keys we have at that point. For instance, each element in the queue could be a tuple or a structure containing the coordinates of the cell and the keys we have when we visit it.
We can easily identify the shortest path, but when we apply a regular BFS, we encounter a problem. When we reach grid[0][4] and collect key a, we discover that there is no way back since revisiting a cell is not allowed in BFS. Hence, our BFS must differentiate between different key-holding states. Within the same state, we cannot revisit a cell. However, two searchings between different key-holding states are entirely independent. For instance, we start with having no keys, and the current states are denoted as (have key a, have key b) = (0, 0). We pick up key a, and our state changes to (1, 0). Consequently, the following BFS is based on the new state, implying that the cells visited in the previous state (0, 0) can be visited again.
Exactly, that’s the key insight!
You have correctly identified that BFS needs to incorporate “state” as well, where each state represents the current set of keys we hold. With each state, the grid can appear different, with some paths (behind the doors) being accessible with certain keys but not others. This makes each BFS exploration of the grid with a different state somewhat independent of the others, and so a cell that has been visited in one state may need to be revisited in another state.
Therefore, instead of simply marking a cell as visited, we need to mark a cell as visited in a specific state. For instance, we can represent the state as a bitmask, where each bit represents whether we hold a specific key or not.
By doing this, we can avoid the problem you mentioned about being unable to return after collecting a key. For example, after we’ve collected the ‘a’ key in your scenario, the state changes from (0, 0) to (1, 0). We can then revisit the cells that were visited in the (0, 0) state, allowing us to return and collect other keys, or unlock doors.
Therefore, to solve this problem, you’d need to implement a BFS with state, often referred to as a multi-state BFS. It’s a more advanced concept and more challenging to implement, but it’s capable of solving complex problems like this one.
In summary, we will perform BFS on multiple identical grids, each representing a distinct key-holding state. When we pick up a new key, we will switch to the grid corresponding to the new state and perform BFS on that grid.
Absolutely! You’ve got it. You can think of it as each unique set of keys corresponding to a different “layer” of the grid, and when you obtain a new key, you’re moving to a new layer where previously blocked paths might now be open.
So, in your BFS, you’ll need to maintain not just the current cell, but also the current state (i.e., the set of keys you have). When you find a key, you’ll generate a new state and continue BFS from there.
To avoid cycles within the same layer, you’ll need to keep track of visited states at each cell, not just visited cells. This way, you won’t repeat a path that has already been taken within the same layer.
Finally, keep in mind that each layer is independent of the others. In other words, visiting a cell in one layer doesn’t prevent you from visiting it in another layer. This is why you need to track visited cells for each layer separately.
It’s a more complex approach than standard BFS, but it’s a powerful technique for solving this kind of problem.
How do we update each BFS?
Similar to a regular BFS where we use (row, col, distance) to represent the distance of each cell. In this case, we also need to include the key-holding states as an additional parameter, so the states in the queue are in the format of (row, col, key-holding state, distance). In order to save all states with their key-holding states, we can initialize a hash map of sets seen, which tracks the visited cells in each state. Specifically, seen[key-holding states] represents all the visited cells with this key-holding state.
Yes, you’re correct! Including the key-holding state in your BFS queue and visited set allows you to keep track of which cells you’ve visited in each key-holding state.
This means each queue element can be a structure or a tuple containing four items: the current row, the current column, the current key-holding state, and the distance travelled so far.
When you visit a cell, you add it to the visited set for the current key-holding state. The key-holding state can be represented as an integer or a binary number if there are up to 10 keys (since there are 10 bits in an integer). Each bit in the binary representation of the state can correspond to whether or not you have a particular key.
For example, if there are 3 keys (‘a’, ‘b’, ‘c’), and you have keys ‘a’ and ‘c’, you can represent this state as the binary number 101, or the integer 5.
Then, as you’re doing your BFS, when you find a key, you add it to your current key-holding state and continue the BFS from the current cell with the new key-holding state.
This way, you’re effectively doing a separate BFS for each key-holding state, which allows you to find the shortest path to collect all keys.
How can we represent each key-holding state?
Yes, you are correct. Bit manipulation is an efficient way to keep track of states in this problem. Each bit in the integer keys represents a key. If the bit is 1, it means we have the key; if it’s 0, we don’t have the key.
We can easily add a key to our current state by using the bitwise OR operator | and bitwise shift operator <<. For example, if we pick up the key ‘b’, we can add it to our state like this: keys = keys | (1 << ('b' - 'a')). This shifts a 1 to the left by ('b' - 'a') positions, resulting in a number where only the bit corresponding to ‘b’ is 1. We then use the bitwise OR operator to add this to keys.
The advantage of this approach is that it’s very efficient in terms of memory and computation, because bitwise operations are usually very fast. It also allows us to easily compare states and check whether we’ve visited a certain state before.
In conclusion, in this problem, we need to keep track of both the location in the grid and the current state of keys collected. We can do this by maintaining a queue of (row, col, keys) for BFS and a set of visited states to ensure we don’t visit the same state more than once. This ensures we always find the shortest path to collect all keys.
How can we determine if we can visit a locked cell?
Similar to updating key-holding state when we pick up a new key, we need to check if we have collected the corresponding key when encountering a lock. To achieve this, we check if the ithi^{th}i th bit of keys is set to 1. For example, suppose that we encounter the lock C with the state as key = 7, we can verify that we have collected the key c by checking the value of key & (1 « (ord(‘C’) - ord(‘A’))). The result is 1, which denotes that we have collected the key c and can visit this cell.
Yes, you’re correct. When you encounter a locked door, you can use bitwise AND & to check if you have the corresponding key.
The expression (1 << (ord('C') - ord('A'))) generates a number that only has the bit corresponding to ‘C’ set to 1. By performing a bitwise AND operation with keys, we get a number that has all bits set to 0, except possibly the one corresponding to ‘C’.
If the result is not 0, it means the bit in keys corresponding to ‘C’ was set to 1, so we have the key for ‘C’. If the result is 0, it means the bit in keys corresponding to ‘C’ was set to 0, so we don’t have the key for ‘C’, and we can’t pass through the door.
So, the condition if (keys & (1 << (ord(lock) - ord('A'))) != 0) checks whether we can pass through a locked door. If it’s true, we can pass through the door; if it’s false, we can’t.
This makes use of the property of bitwise AND: if both bits are 1, the result is 1; otherwise, the result is 0. So this operation can effectively check whether a particular bit in keys is set to 1, indicating we have the corresponding key.