Count Matrix Paths Through Ordered Checkpoints

Start at (rows-1,0) and finish at (rows-1,columns-1). Every move advances one column and changes the row by -1, 0, or +1 while remaining in bounds.
Count paths that visit every listed checkpoint in the given order. Checkpoints have strictly increasing columns and may include either endpoint. Return the count modulo 1,000,000,007.

Function
countCheckpointPaths(rows: int, columns: int, checkpoints: int[][]) → int

Examples
Example 1
rows = 3
columns = 4
checkpoints = [[1,1],[2,3]]
return = 2
Two choices reach the first checkpoint and two more steps finish at the required endpoint.
Example 2
rows = 1
columns = 5
checkpoints = []
return = 1
With one row, there is exactly one path.
Example 3
rows = 2
columns = 2
checkpoints = [[0,1]]
return = 0
The checkpoint conflicts with the required bottom-right destination, so no path qualifies.

Constraints
1 <= rows, columns <= 1000.
Checkpoint coordinates are valid and columns are strictly increasing.

Treat each column as one layer. Keep an array dp of size rows, where dp[r] is the number of ways to be at row r in the current column. Start with dp[rows-1] = 1 at column 0. To advance, new[r] = dp[r-1] + dp[r] + dp[r+1], bounds checked, mod 1,000,000,007. That's a layered BFS that merges counts instead of visiting nodes, and it runs in O(rows * columns). For checkpoints, after computing a column that holds a checkpoint, zero every cell except the checkpoint row. That forces the path through it. Order is automatic because columns strictly increase and you sweep left to right. The pitfall is forgetting the end: read dp[rows-1] at the last column, and if a checkpoint in that column sits elsewhere, you get 0, which is Example 3. Also handle a checkpoint at column 0 that isn't the start row. Apply the mod on every addition. If you freeze live, StealthCoder is the hedge, but the zero-out trick is the whole problem.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.