1368. Minimum Cost to Make at Least One Valid Path in a Grid


Problem Description

The task is to find the minimum cost to establish at least one valid path in an m x n grid. This path must start at the top left cell (0, 0) and end at the bottom right cell (m - 1, n - 1). Each cell in the grid contains a sign which directs you to the next cell to visit. The possible signs are:

  • 1: Move right to the cell (i, j + 1)
  • 2: Move left to the cell (i, j - 1)
  • 3: Move down to the cell (i + 1, j)
  • 4: Move up to the cell (i - 1, j)

Some signs might point outside of the grid boundaries, which are to be considered invalid directions for the purpose of a valid path. The cost to change a sign in any cell is 1, and each sign can be changed only once.

The objective is to determine the minimum cost to alter the signs in such a way that at least one valid path from the top left to the bottom right cell exists.

Intuition

To solve this problem, we use a breadth-first search (BFS) approach, but with a slight tweak. This problem can be thought of as traversing a graph where each cell represents a node, and the signs are the directed edges to the neighboring nodes. The challenge, however, lies in the fact that these directed edges can be altered at a cost.

The modified BFS algorithm uses a double-ended queue (deque) to keep track of the cells to be visited. This is crucial as the queue can have elements added to both its front and back, which helps in maintaining the order of traversal based on the cost associated with moving to a particular cell.

Here's the intuition behind the BFS traversal:

  1. We begin at the starting cell (0, 0) with an initial cost of 0.
  2. As we visit a cell, we look at all possible directions we could move from that cell.
  3. If the direction aligns with the arrow currently in the cell, we can move to that cell at no additional cost. In this case, we add this cell at the front of the deque to prioritize it.
  4. If the direction doesn't align with the arrow, we would need to change the sign with a cost of 1. For these cells, we add them to the back of the deque as they represent potential paths but at a higher cost.
  5. Each cell is visited only once to ensure minimal traversal cost, thus, we maintain a visited set.
  6. This process continues until we reach the destination cell (m - 1, n - 1) or there are no more cells left to visit in the deque.
  7. The cost associated with the first visit to the destination cell is the minimum cost needed to make at least one valid path, which is what we return.

By always choosing to traverse in the indicated direction without any cost, we ensure that we are taking advantage of the free moves as much as possible before incurring any additional costs. The visited set prevents us from revisiting cells and possibly entering a loop.

Learn more about Breadth-First Search, Graph, Shortest Path and Heap (Priority Queue) patterns.

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Solution Approach

The provided reference code implements a BFS strategy using a deque. Here's a step-by-step breakdown of the solution's implementation:

  1. Initialize the dimensions m (rows) and n (columns) of the grid.
  2. Define an array dirs to represent the possible directions of travel based on the grid's signs: [[0, 0], [0, 1], [0, -1], [1, 0], [-1, 0]]. Each sub-array corresponds to the deltas for the row and column indices when moving in each of the four directions.
  3. Create a double-ended queue q, initialized with a tuple containing the row index 0, column index 0, and the initial cost 0.
  4. Create a set vis to keep track of visited cells and prevent revisiting them.
  5. Start a loop to continue processing until the deque q is empty. Each iteration will handle one cell's visit:
    • Dequeue an element (i, j, d) from the front of q, where i and j are the current cell's indices, and d is the cost to reach this cell.
    • If the cell (i, j) has already been visited, skip processing it to avoid loops and unnecessary cost increments.
    • Mark the current cell (i, j) as visited by adding it to vis.
    • If the cell (i, j) is the destination (m-1, n-1), return the cost d because a valid path has been found.
    • For each possible direction k (1 to 4), calculate the indices of the adjacent cell (x, y) by adding the directional deltas to the current indices i and j.
  6. Check if the new cell (x, y) is within the bounds of the grid:
    • If the sign at the current cell grid[i][j] indicates the direction k, meaning no sign change is needed, add (x, y, d) to the front of q, not increasing the cost, as it's a free move in the desired direction.
    • If the sign does not match, add (x, y, d + 1) to the back of q, increasing the cost by 1 to account for the change of sign.
  7. If no valid path is found by the end of the traversal, return -1.

The use of a deque allows for efficient addition of cells to either the front or back, depending on whether a cost is incurred. By prioritizing the moves that don't require sign changes (zero cost), the algorithm ensures the minimum cost path is found first. The set vis prevents revisiting and recalculating paths for positions that have already been evaluated, which optimizes the search. This approach effectively treats the grid as a graph and considers the costs of edges dynamically, resulting in a clever application of the BFS algorithm adapted for weighted pathfinding.

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Example Walkthrough

Let's assume we have a 3x3 grid, where the signs in the cells are arranged as follows:

11 3 4
24 2 1
33 1 3

Now let's walk through the solution approach with this grid:

  1. We start at the top-left cell (0, 0) with an initial cost of 0.
  2. From here, the sign 1 tells us to move right to cell (0, 1). This is a valid move with no cost so we add (0, 1) to the front of our deque.
  3. Now we consider the cell (0, 1) with the sign 3, which tells us to move down. The target cell (1, 1) has not been visited, so we add (1, 1) to the front of the deque again without any additional cost.
  4. At cell (1, 1), the sign 2 indicates moving left to (1, 0). This cell has not been visited, so we add (1, 0) to the front of the deque.
  5. Next, we visit cell (1, 0) which contains the sign 4, telling us to move up to (0, 0). However, this has been visited, so we don't add anything to our deque.
  6. Our next cell would be (0, 1) again, but since it's visited, we ignore it.
  7. Continuing this process, we encounter cell (1, 1) again, which we've visited, so we move on.
  8. Now, we get to cell (1, 0) and from here, the direction 4 (up) is not valid since it would take us outside the grid or into visited cells. At this point, we need to consider changing the direction, so we will add the neighboring cells (1, 1) to the right and (2, 0) down to the back of the deque with an added cost of 1.
  9. Cell (1, 1) won't be processed as it's already visited, so we look at cell (2, 0) with the sign 3 which leads us down to (2, 1). There's no cost for moving down since the sign matches. We add (2, 1) to the front of the deque.
  10. From cell (2, 1), we move right to (2, 2) as indicated by the 1 sign. As this is a free move, it is added to the front of the deque.
  11. Finally, we arrive at the bottom right cell (2, 2) and our destination is reached without needing further sign changes. The cost at this point is 1.

The minimum cost required to establish at least one valid path in this grid is 1, which entails changing one sign. By strategically using a deque and a visited set, we conducted a breadth-first search that prioritized no-cost moves over those requiring a sign change.

Not Sure What to Study? Take the 2-min Quiz:

What's the output of running the following function using the following tree as input?

1def serialize(root):
2    res = []
3    def dfs(root):
4        if not root:
5            res.append('x')
6            return
7        res.append(root.val)
8        dfs(root.left)
9        dfs(root.right)
10    dfs(root)
11    return ' '.join(res)
12
1import java.util.StringJoiner;
2
3public static String serialize(Node root) {
4    StringJoiner res = new StringJoiner(" ");
5    serializeDFS(root, res);
6    return res.toString();
7}
8
9private static void serializeDFS(Node root, StringJoiner result) {
10    if (root == null) {
11        result.add("x");
12        return;
13    }
14    result.add(Integer.toString(root.val));
15    serializeDFS(root.left, result);
16    serializeDFS(root.right, result);
17}
18
1function serialize(root) {
2    let res = [];
3    serialize_dfs(root, res);
4    return res.join(" ");
5}
6
7function serialize_dfs(root, res) {
8    if (!root) {
9        res.push("x");
10        return;
11    }
12    res.push(root.val);
13    serialize_dfs(root.left, res);
14    serialize_dfs(root.right, res);
15}
16

Python Solution

1from collections import deque
2
3class Solution:
4    def minCost(self, grid: List[List[int]]) -> int:
5        # Initialize rows and cols with the dimensions of the grid
6        rows, cols = len(grid), len(grid[0])
7      
8        # Define the direction vectors for right, left, down, up respectively
9        directions = [[0, 0], [0, 1], [0, -1], [1, 0], [-1, 0]]
10      
11        # Initialize queue with the starting point and the current cost (0)
12        queue = deque([(0, 0, 0)])
13      
14        # Set to maintain visited cells
15        visited = set()
16      
17        # Iterate until the queue is empty
18        while queue:
19            # Pop the cell from queue with its current cost
20            i, j, cost = queue.popleft()
21          
22            # Check if the cell is already visited to avoid redundancy
23            if (i, j) in visited:
24                continue
25          
26            # Mark the current cell as visited
27            visited.add((i, j))
28          
29            # Check if we have reached the bottom right corner, return cost if true
30            if i == rows - 1 and j == cols - 1:
31                return cost
32          
33            # Explore all possible directions from the current cell
34            for k in range(1, 5):
35                x, y = i + directions[k][0], j + directions[k][1]
36              
37                # Check for valid cell coordinates
38                if 0 <= x < rows and 0 <= y < cols:
39                    # If the current direction matches the arrow in the grid cell, no cost is added
40                    if grid[i][j] == k:
41                        queue.appendleft((x, y, cost))
42                    # If the direction is different, add a cost of 1 to switch the arrow
43                    else:
44                        queue.append((x, y, cost + 1))
45      
46        # If no path found to the bottom right corner return -1
47        return -1
48

Java Solution

1class Solution {
2    public int minCost(int[][] grid) {
3        // m holds the number of rows in the grid.
4        int numRows = grid.length;
5        // n holds the number of columns in the grid.
6        int numCols = grid[0].length;
7        // vis holds information whether a cell has been visited.
8        boolean[][] visited = new boolean[numRows][numCols];
9        // Queue for performing BFS with modifications for 0-cost moves.
10        Deque<int[]> queue = new ArrayDeque<>();
11        // Starting by adding the top-left cell with 0 cost.
12        queue.offer(new int[] {0, 0, 0});
13        // dirs are used to navigate throughout the grid. (right, left, down, up)
14        int[][] directions = {{0, 0}, {0, 1}, {0, -1}, {1, 0}, {-1, 0}};
15      
16        // BFS starts here
17        while (!queue.isEmpty()) {
18            // Dequeue a cell info from the queue.
19            int[] position = queue.poll();
20            // i and j hold the current cell row and column, d holds the current cost.
21            int i = position[0], j = position[1], cost = position[2];
22
23            // If we've reached the bottom-right cell, return the cost.
24            if (i == numRows - 1 && j == numCols - 1) {
25                return cost;
26            }
27            // If this cell is already visited, skip it.
28            if (visited[i][j]) {
29                continue;
30            }
31            // Mark the cell as visited.
32            visited[i][j] = true;
33
34            // Explore all possible directions from the current cell.
35            for (int k = 1; k <= 4; ++k) {
36                int newX = i + directions[k][0], newY = j + directions[k][1];
37                // Check the validity of the new cell coordinates.
38                if (newX >= 0 && newX < numRows && newY >= 0 && newY < numCols) {
39                    // If the current direction is the same as the arrow in this cell (no cost to move here).
40                    if (grid[i][j] == k) {
41                        // Add the new cell at the front of the queue to explore it sooner (as it's no cost).
42                        queue.offerFirst(new int[] {newX, newY, cost});
43                    } else {
44                        // Otherwise, add the new cell at the end of the queue and increase the cost by 1.
45                        queue.offer(new int[] {newX, newY, cost + 1});
46                    }
47                }
48            }
49        }
50        // If the queue is empty and we didn't reach the bottom-right cell, return -1 as it's not possible.
51        return -1;
52    }
53}
54

C++ Solution

1class Solution {
2public:
3    int minCost(vector<vector<int>>& grid) {
4        // Get dimensions of the grid
5        int rows = grid.size(), cols = grid[0].size();
6
7        // Initialize a 2D vector to keep track of visited cells
8        vector<vector<bool>> visited(rows, vector<bool>(cols, false));
9
10        // Define directions according to the grid value (1: right, 2: left, 3: down, 4: up)
11        vector<vector<int>> directions = {{0, 0}, {0, 1}, {0, -1}, {1, 0}, {-1, 0}};
12
13        // Queue to perform the BFS, holding pairs of (cell_index, current_cost)
14        deque<pair<int, int>> queue;
15        queue.push_back({0, 0}); // Start from the top-left corner with 0 cost
16
17        // While there are elements in the queue
18        while (!queue.empty()) {
19            // Get the front element
20            auto current = queue.front();
21            queue.pop_front();
22
23            // Calculate row and column from the cell index
24            int row = current.first / cols, col = current.first % cols;
25            int cost = current.second;
26
27            // If we've reached the bottom-right corner, return the current cost
28            if (row == rows - 1 && col == cols - 1) return cost;
29
30            // If the current cell is already visited, skip it
31            if (visited[row][col]) continue;
32
33            // Mark the current cell as visited
34            visited[row][col] = true;
35
36            // Check all four adjacent cells
37            for (int k = 1; k <= 4; ++k) {
38                int newRow = row + directions[k][0], newCol = col + directions[k][1];
39
40                // If the new cell is within the grid bounds
41                if (newRow >= 0 && newRow < rows && newCol >= 0 && newCol < cols) {
42                    // Calculate the new cell index
43                    int newIndex = newRow * cols + newCol;
44
45                    // If the grid indicates the current direction, add to the front of the queue without increasing cost
46                    if (grid[row][col] == k)
47                        queue.push_front({newIndex, cost});
48                    // Otherwise, the direction is not towards the grid arrow - increase the cost and add to the back of the queue
49                    else
50                        queue.push_back({newIndex, cost + 1});
51                }
52            }
53        }
54
55        // Return -1 if the bottom-right corner could not be reached
56        return -1;
57    }
58};
59

Typescript Solution

1function minCost(grid: number[][]): number {
2    // Dimensions of the grid.
3    const rows = grid.length,
4        cols = grid[0].length;
5      
6    // Initialize the answer grid with infinity, indicating no paths have been taken.
7    let costGrid = Array.from({ length: rows }, () => new Array(cols).fill(Infinity));
8    costGrid[0][0] = 0; // Starting point cost is zero.
9
10    // Queue for BFS traversal, starting from the top-left cell.
11    let queue: [number, number][] = [[0, 0]];
12
13    // Directions for right, left, down, up.
14    const directions = [
15        [0, 1],  // right
16        [0, -1], // left
17        [1, 0],  // down
18        [-1, 0], // up
19    ];
20
21    // Perform BFS on the grid.
22    while (queue.length) {
23        // Current position.
24        let [x, y] = queue.shift()!;
25
26        // Explore all possible directions.
27        for (let step = 1; step < 5; step++) {
28            // Get the next direction.
29            let [dx, dy] = directions[step - 1];
30            // Compute the new position.
31            let [i, j] = [x + dx, y + dy];
32
33            // Check boundary conditions.
34            if (i < 0 || i >= rows || j < 0 || j >= cols) continue;
35
36            // Calculate the cost to enter the new position.
37            let cost = ~~(grid[x][y] != step) + costGrid[x][y];
38
39            // Skip if the new cost is not less than the current cost at position (i, j).
40            if (cost >= costGrid[i][j]) continue;
41
42            // Update the cost at the new position.
43            costGrid[i][j] = cost;
44
45            // Prioritize movement in the direction the arrow points by placing it at the front of the queue.
46            if (grid[x][y] == step) {
47                queue.unshift([i, j]);
48            } else { // Otherwise, enqueue normally.
49                queue.push([i, j]);
50            }
51        }
52    }
53
54    // Return the minimum cost to reach the bottom right corner.
55    return costGrid[rows - 1][cols - 1];
56}
57
Fast Track Your Learning with Our Quick Skills Quiz:

The three-steps of Depth First Search are:

  1. Identify states;
  2. Draw the state-space tree;
  3. DFS on the state-space tree.

Time and Space Complexity

The provided code solves the problem using a Breadth First Search (BFS) algorithm with a slight optimization. The algorithm has two modes of queue operations: normally numbers are appended to the end of the queue, but when moving in the direction the grid arrow points to, it's added to the front. This has the potential to reduce the number of steps needed to reach the end.

Time Complexity:

The time complexity is O(m * n), where m is the number of rows and n is the number of columns in the grid. This is because in the worst-case scenario, each cell is visited only once due to the use of the visited (vis) set. Even though there is a nested loop to iterate over directions, they only add constant work at each node, so it does not affect the overall linear complexity with respect to the number of cells.

Space Complexity:

The space complexity is also O(m * n) because of the visited set (vis) storing up to m * n unique cell positions to ensure cells are not revisited. Additionally, the queue (q) in the worst case may also contain elements from all the cells in the grid when they are being processed sequentially. Hence, the overall space complexity remains O(m * n).

Learn more about how to find time and space complexity quickly.


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