Asked in: DPWORLD
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All Testcases Passed ✔
Solution
from functools import lru_cache
from sys import setrecursionlimit
setrecursionlimit(10**6)
def numPaths(warehouse):
// ... rest of solution available after purchaseExplanation
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To solve the problem of finding the number of distinct paths in an n x m warehouse matrix from the top-left corner to the bottom-right corner, moving only right or down and avoiding blocked cells, you need a clear plan based on dynamic programming principles. The goal is to efficiently count how many ways exist to travel through open cells (marked as 1) while respecting movement restrictions and obstacles.
Step 1: Understand the Problem and Constraints
You are given a grid where each cell is either open (1) or blocked (0). The forklift operator can only move right or down from any cell. The task is to find the total number of distinct paths starting from the top-left cell (0,0) to the bottom-right cell (n-1,m-1). Since the number of paths can become very large, you will return the answer modulo 10^9 + 7.
Key points to note:
- Movements are restricted to only right or down (no diagonal or left/up moves).
- You cannot pass through or land on blocked cells (cells with value 0).
- The start and end cells must be open for any path to exist.
- Grid size can be up to 1000x1000, so an efficient O(n*m) solution is necessary.
Step 2: Think About the Nature of the Problem
This problem is a classic grid path counting problem with obstacles. If there were no obstacles, the total number of paths from the top-left to bottom-right would be a combinatorial problem, easily calculated by binomial coefficients (choosing when to move right or down). However, obstacles complicate this because certain cells cannot be visited.
Dynamic programming is well-suited here because the number of paths to reach a particular cell depends on the number of ways to reach the cell directly above it and the cell directly to the left, given that those cells are accessible.
Step 3: Define Your Subproblem and State
Create a 2D array dp of the same size as the warehouse matrix. Each dp[i][j] will store the number of distinct paths to reach the cell at position (i,j).
The relation is:
- If warehouse[i][j] == 0 (blocked), then dp[i][j] = 0 because the cell cannot be reached or passed through.
- Otherwise, dp[i][j] = dp[i-1][j] + dp[i][j-1], representing paths from above and from the left.
Step 4: Handle Boundary Conditions
- The starting cell dp[0][0] is 1 if warehouse[0][0] == 1, since the operator starts there. If the start is blocked, dp[0][0] = 0 and the answer is immediately 0 because no path can start.
- For the first row (i=0), dp[0][j] = dp[0][j-1] if warehouse[0][j] is open; otherwise 0. Movement is only from left since up is out of bounds.
- For the first column (j=0), dp[i][0] = dp[i-1][0] if warehouse[i][0] is open; otherwise 0. Movement is only from above.
Step 5: Fill the dp Table
Iterate over all cells row by row, column by column. For each cell:
- Check if it’s blocked. If yes, dp[i][j] = 0.
- If not blocked, sum up paths from the top and left neighbors (if those neighbors exist and are not blocked).
- Always apply modulo operation to avoid integer overflow and keep the numbers within limits.
Step 6: Result Interpretation
Once the dp table is filled, dp[n-1][m-1] will contain the total number of ways to reach the bottom-right cell from the top-left cell, considering all the paths that pass only through open cells and moving only right or down.
Step 7: Optimize for Constraints
- Since n and m can be as large as 1000, the O(n*m) time complexity of filling the dp table is efficient and suitable.
- Use modulo arithmetic during every addition to avoid overflow and adhere to problem constraints.
Step 8: Edge Cases to Consider
- The starting cell or ending cell is blocked (no paths exist).
- Rows or columns where all cells are blocked, making the path impossible beyond a certain point.
- A grid with a single cell that is open (the answer is 1).
- Large grids with a complex pattern of blocked cells.
Summary:
By recognizing this as a grid path counting problem with obstacles, you naturally arrive at a dynamic programming solution. The main challenge is correctly setting up the dp relations, carefully handling boundaries, and respecting blocked cells by setting dp to zero in those positions. After processing all cells, the value at the bottom-right corner dp[n-1][m-1] will give you the count of all valid paths modulo 10^9+7.
This approach ensures that you count every possible distinct path accurately while adhering to the movement constraints and avoiding obstacles efficiently within the given time and space constraints.
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