0095MediumMathDynamic ProgrammingCombinatorics

Grid Courier Routes

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A bicycle courier works a planned district shaped as m streets running east-west crossed by n streets running north-south, giving m rows and n columns of intersections.

The courier waits at the intersection in the northwest corner, row 1 and column 1, and must deliver to the intersection in the southeast corner, row m and column n. Traffic rules allow exactly two kinds of step: ride one block east, landing on the next column of the same row, or ride one block south, landing on the next row of the same column.

Two routes count as different when the sequence of steps differs. Return how many different routes reach the delivery point.

Examples

Example 1

Input: m = 4, n = 5
Output: 35

Any route takes 3 southbound blocks and 4 eastbound blocks in some order, and there are 35 such orders.

Example 2

Input: m = 1, n = 6
Output: 1

A single row leaves no room to ride south, so the only route is five straight eastbound blocks.

Example 3

Input: m = 2, n = 2
Output: 2

The courier either rides east then south, or south then east.

Constraints

1 <= m <= 100
1 <= n <= 100
The district is laid out so that the number of different routes is at most 10^12.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_routes(m: int, n: int) -> int:

Java

public int countRoutes(int m, int n)

The intended approach

Time O(m * n)Space O(n)

The number of routes into an intersection is the number of routes into the intersection directly north of it plus the number into the one directly west of it, since those are the only two places the last block could have started. Sweep the grid row by row while keeping a single array of route counts for the current row, seeded with ones for the top row where only eastbound riding is possible.

Pythonentry point: count_routes

def count_routes(m, n):
    row = [1] * n
    for _ in range(m - 1):
        for col in range(1, n):
            row[col] = row[col] + row[col - 1]
    return row[n - 1]

The obvious version

Plain recursion that branches on each possible first block and adds up the completions, with no memoisation. Exponential, but transparently correct.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_routes(m, n):
    if m == 1 or n == 1:
        return 1
    return bf_count_routes(m - 1, n) + bf_count_routes(m, n - 1)

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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