Imagine you’re navigating a grid where each cell has a sign directing you to the next cell. Your mission? Make at least one valid path from the top-left corner to the bottom-right corner, but it might cost you! Each time you change a sign, you incur a cost of 1. Our goal is to minimize this cost to create a valid path. Let’s dive into the problem, explore a solution, and dissect it step-by-step! 🤓

📜 Problem Statement

You are given an m x n grid. Each cell in the grid has a direction sign indicating where to move next:

1. 1 → Move right.
2. 2 → Move left.
3. 3 → Move down.
4. 4 → Move up.

You start at the top-left cell (0, 0) and aim to reach the bottom-right cell (m - 1, n - 1). A valid path is one that follows the grid’s directions.

Rules:

• You can change the direction of a cell’s sign, but it costs 1 per change.
• Your goal is to minimize the cost to make at least one valid path from the start to the destination.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100
• 1 <= grid[i][j] <= 4

🔍 Examples

Example 1:

Input:

grid = [
  [1, 1, 1, 1],
  [2, 2, 2, 2],
  [1, 1, 1, 1],
  [2, 2, 2, 2]
]

Output:

3

Explanation:

• Follow the path (0, 0) → (0, 1) → (0, 2) → (0, 3).
• Change the sign at (0, 3) to move down (cost = 1).
• Continue to (1, 3) → (1, 2) → (1, 1) → (1, 0).
• Change the sign at (1, 0) to move down (cost = 1).
• Finally, change the sign at (2, 3) to move down (cost = 1).
• Total cost: 3.

Example 2:

Input:

grid = [
  [1, 1, 3],
  [3, 2, 2],
  [1, 1, 4]
]

Output:

0

Explanation:

• The path (0, 0) → (0, 1) → (0, 2) → (1, 2) → (2, 2) is valid without any changes.

Example 3:

Input:

grid = [
  [1, 2],
  [4, 3]
]

Output:

1

Explanation:

• Change the direction of (0, 1) to move down.
• Follow (0, 0) → (0, 1) → (1, 1).

💡 Solution Overview

This problem revolves around navigating a grid 🗺️, where each cell contains a direction ➡️⬅️⬇️⬆️ pointing to where you should ideally go next. Think of it as a map with arrows 🧭, but the paths it suggests may not always lead to your destination 🏁.

To ensure you can travel from the starting cell 🟩 at the top-left corner (0, 0) to the destination cell 🟥 at the bottom-right corner (m-1, n-1), you might need to change the directions in some cells 🛠️.

The twist? Changing a direction comes with a cost 💰 of 1, while following the arrow as is costs nothing 🆓. The task is to figure out the minimum cost 💵 to create at least one valid path from start to finish.

Solution Intuition 💡

This problem can be visualized as finding the shortest path 🚗 in a weighted graph 🎢, where:

1. Each cell in the grid is a node 🟨.
2. Moving between nodes involves a cost, determined by whether the direction needs to be changed or not 🔄.
3. The objective is to minimize the total cost 💸 to navigate from the top-left to the bottom-right corner.

Breaking Down the Approach 🛠️

To solve this problem efficiently, we need to think systematically about exploring the grid 🗺️ while minimizing costs:

1. Understanding Directions 🚦

Each cell has a predefined direction encoded as:

• 1 → Move Right ➡️
• 2 → Move Left ⬅️
• 3 → Move Down ⬇️
• 4 → Move Up ⬆️

From any given cell, you can either follow its arrow 🏹 (incurring no cost 🆓) or modify the arrow 🔄 to move to a different neighboring cell (incurring a cost of 1 💰).

2. Core Idea: Breadth-First Search (BFS) 🔍

The shortest path in a grid-like structure can often be solved using BFS. In this scenario:

• We explore all neighboring cells of the current cell 📍.
• If a neighbor can be reached by following the current direction, we prioritize it (cost = 0 🟢).
• If a neighbor requires changing the direction, we explore it later (cost = 1 🔴).

This ensures that we visit all possible paths 🌟, starting with those that cost less.

3. Using a Priority Queue ⏳

To manage the order in which cells are visited based on their cost:

• We use a priority queue 🗂️ (think of it as a min-heap 🌐).
• Cells are added to the queue along with their current cumulative cost. The queue always prioritizes cells with the smallest cost 🔑.

For example:

• If a cell (i, j) has a cost of 2 💸 and another cell (x, y) has a cost of 3, (i, j) will be explored first 🚀.

4. Steps to Reach the Solution 🚶

• Start at the top-left corner (0, 0) with a cost of 0 🟢.
• At each step:
  • Check all four possible neighbors (up ⬆️, down ⬇️, left ⬅️, right ➡️).
  • If the direction from the current cell matches the neighbor’s position, add it to the queue with the same cost (no change needed ✅).
  • If the direction doesn’t match, add the neighbor to the queue with an increased cost of 1 🔄.
• Continue this process until you reach the destination 🏁 (m-1, n-1).
• Keep track of visited cells 🛑 to avoid redundant work.

5. When to Stop 🛑

The process stops as soon as you reach the bottom-right corner 🟥. At this point, the cost accumulated represents the minimum cost 💰 required to make at least one valid path.

Visualization of the Solution 🔎

Let’s walk through an example grid 🌟:

grid = [[1, 1, 3], 
        [3, 2, 2], 
        [1, 1, 4]]

1. Start at (0, 0) with a cost of 0 🟢. The direction is 1 (right ➡️), so move to (0, 1) without any additional cost.
2. From (0, 1), follow the direction 1 (right ➡️) to (0, 2), still at cost 0 🆓.
3. At (0, 2), the direction is 3 (down ⬇️). This aligns with the arrow pointing down to (1, 2), so no modification is needed ✅.
4. Finally, move from (1, 2) down ⬇️ to the destination (2, 2) at no extra cost 🎉.

This path ensures the minimum cost 💰 to reach the destination.

Why BFS Works Well Here 🤔

BFS is perfect for this scenario because it explores all possible paths in layers of increasing cost:

1. Paths with cost = 0 🟢 are explored first.
2. If the destination isn’t reached, paths with cost = 1 🔴 are considered next.
3. This ensures we find the minimum cost 💵 efficiently without unnecessary exploration.

Using a priority queue ⏳ enhances BFS by always prioritizing the least costly path, avoiding unnecessary exploration of higher-cost paths until absolutely needed 🚀.

Key Concepts Recap 📝

1. Graph Representation 🎢: Each cell is a node 🟨, and moving to a neighbor has a weight of 0 🟢 or 1 🔴.
2. Shortest Path 🛣️: BFS with a priority queue ensures that paths are explored in the order of increasing cost.
3. Optimization 🚀: By following the natural direction whenever possible, we minimize the total modifications required 💰.

🧑‍💻 Python Implementation

Here’s the complete code:

from collections import deque

class Solution:
    def minCost(self, grid: list[list[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        directions = [[0, 0], [0, 1], [0, -1], [1, 0], [-1, 0]]  # Right, Left, Down, Up
        queue = deque([(0, 0, 0)])  # (row, col, cost)
        visited = set()

        while queue:
            i, j, cost = queue.popleft()

            if (i, j) in visited:
                continue
            visited.add((i, j))

            if i == rows - 1 and j == cols - 1:
                return cost

            for k in range(1, 5):
                x, y = i + directions[k][0], j + directions[k][1]
                if 0 <= x < rows and 0 <= y < cols:
                    if grid[i][j] == k:
                        queue.appendleft((x, y, cost))
                    else:
                        queue.append((x, y, cost + 1))

        return -1

🔍 Complexity Analysis

Time Complexity:

• Each cell is processed once, leading to O(m * n) operations.
• Directional checks for each cell are constant.
• Overall: O(m * n).

Space Complexity:

• Space is used for the queue and the visited set: O(m * n).

🛠 Example Walkthrough

Let’s walk through Example 1 in detail with larger numbers:

Input:

grid = [
  [1, 1, 1, 1],
  [2, 2, 2, 2],
  [1, 1, 1, 1],
  [2, 2, 2, 2]
]

1. Start at (0, 0). The initial cost is 0.
2. Move right (0, 0) → (0, 1) → (0, 2) → (0, 3).
   • Change the sign at (0, 3) to move down. Cost: 1.
3. Move down (0, 3) → (1, 3).
   • Change the sign at (1, 3) to move left. Cost: 2.
4. Continue (1, 3) → (1, 2) → (1, 1) → (1, 0).
   • Change the sign at (1, 0) to move down. Cost: 3.
5. Move (1, 0) → (2, 0) → (2, 1) → (2, 2) → (2, 3).
6. Change the sign at (2, 3) to move down. Cost: 4.
7. Finally, reach (3, 3).

🧩 Edge Cases

1. Single-cell grid:

   grid = [[1]]
   Output: 0
   

2. Straight-line path:

   grid = [[1, 1, 1, 1]]
   Output: 0
   

3. Maximum grid size:
   Ensure efficiency with 100 x 100 grids.

🎉 Conclusion

This problem challenges us to think about optimization and graph traversal. Using BFS with a priority queue ensures minimal cost while exploring all possible paths. Whether you’re navigating a grid or solving a real-world logistics problem, this approach showcases the power of systematic exploration.