# Solving Leetcode Interviews in Seconds with AI: Unique Paths


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "62" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time. Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner. The test cases are generated so that the answer will be less than or equal to 2 * 109.   Example 1:   Input: m = 3, n = 7 Output: 28  Example 2:  Input: m = 3, n = 2 Output: 3 Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: 1. Right -> Down -> Down 2. Down -> Down -> Right 3. Down -> Right -> Down    Constraints:  1 <= m, n <= 100  

	# Explanation
	*   **Combinatorial Approach:** The problem can be reframed as finding the number of ways to choose (m-1) down moves (or n-1 right moves) from a total of (m+n-2) moves. This is a combination problem, specifically (m+n-2) choose (m-1).
*   **Optimized Combination Calculation:** We can compute the combination efficiently using the formula C(n, k) = n! / (k! * (n-k)!). To avoid potential overflow issues with large factorials, we calculate the combination iteratively, canceling out common factors.
*   **Utilize Smaller Value:** To optimize calculations, always choose the smaller value between `m-1` and `n-1` as `k` for calculating combinations i.e. `C(n, k) = C(n, n-k)`.

*   **Runtime Complexity: O(min(m, n)), Storage Complexity: O(1)**

	
	# Code
	```python
	def uniquePaths(m: int, n: int) -> int:
    """
    Calculates the number of unique paths from the top-left corner to the
    bottom-right corner of an m x n grid, where the robot can only move down or right.

    Args:
        m: The number of rows in the grid.
        n: The number of columns in the grid.

    Returns:
        The number of unique paths.
    """

    N = m + n - 2
    k = min(m - 1, n - 1)  # Number of down or right moves needed
    result = 1

    for i in range(k):
        result = result * (N - i) // (i + 1)

    return result
	```