To solve the problem of finding the number of unique paths from the top-left corner to the bottom-right corner of an m x n grid (moving only right or down), we can use a combinatorial approach, which is efficient and straightforward.

Approach

The key insight here is to recognize that to reach the bottom-right corner from the top-left, we need to take a total of (m-1) down steps and (n-1) right steps, regardless of the order. This is a combinatorial problem where we need to find the number of ways to choose (m-1) down steps (or equivalently (n-1) right steps) from the total steps (m+n-2).

The formula for the number of unique paths is:
[ \text{Unique Paths} = \binom{m+n-2}{m-1} = \binom{m+n-2}{n-1} ]

To compute this efficiently, we calculate the combination using iterative multiplication to avoid large intermediate values (by dividing as we multiply).

Solution Code

def uniquePaths(m: int, n: int) -> int:
    a = m + n - 2
    b = min(m-1, n-1)  # Minimize the number of multiplications
    result = 1
    for i in range(1, b+1):
        result = result * (a - b + i) // i
    return result

Explanation

1. Total Steps: The total number of steps required to reach the end is (m-1) + (n-1) = m+n-2.
2. Combination Calculation: We compute the combination of (m+n-2) steps taken (m-1) at a time (down steps). Using the minimum of (m-1) and (n-1) reduces the number of iterations.
3. Iterative Product: For each iteration from 1 to b, we multiply the result by (a - b + i) (numerator) and divide by i (denominator) to keep the result integer and avoid overflow.

Example

For m=3 and n=7:
Total steps = 3+7-2 =8.
We choose min(2,6)=2 down steps.
[ \binom{8}{2} = \frac{87}{21} =28 ]

The code returns 28, which is correct.

This approach runs in O(min(m,n)) time and uses O(1) space, making it optimal for large grids. It is both efficient and easy to understand.

Answer: The number of unique paths is given by the function above, e.g., for m=3 and n=7, the answer is 28. For general inputs, the function returns the correct count.

If you provide specific values of m and n, I can give the exact number, but the function above works for any valid m and n. For example, if the problem had m=3 and n=7, the answer is 28.

Let me know if you have specific values! But assuming the classic problem, the solution is as above.

Final Answer
For example, if m=3 and n=7, the answer is \boxed{28}. For other values, substitute into the function to get the result. If you have specific m and n, let me know, and I'll compute it. But given the context, the general solution is the code provided. However, if the problem expects a numerical answer, please provide the grid dimensions.

Assuming the problem expects the answer for a common case like m=3 and n=7, the answer is \boxed{28}. \boxed{28}

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