# Solving Leetcode Interviews in Seconds with AI: Minimum Path Sum


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "64" 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
	> Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time.   Example 1:   Input: grid = [[1,3,1],[1,5,1],[4,2,1]] Output: 7 Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.  Example 2:  Input: grid = [[1,2,3],[4,5,6]] Output: 12    Constraints:  m == grid.length n == grid[i].length 1 <= m, n <= 200 0 <= grid[i][j] <= 200  

	# Explanation
	Here's the breakdown of the approach and the Python code:

*   **Dynamic Programming:** The core idea is to use dynamic programming. We build a table `dp` of the same size as the grid, where `dp[i][j]` stores the minimum path sum to reach cell `(i, j)`.
*   **Base Case and Recurrence Relation:** The base case is `dp[0][0] = grid[0][0]`. The recurrence relation is: `dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])`. Essentially, the minimum path sum to reach `(i, j)` is the value of `grid[i][j]` plus the minimum of the path sums to reach its top and left neighbors.
*   **Iterative Calculation:**  We fill the `dp` table iteratively, starting from the top-left corner and moving towards the bottom-right corner.

*   **Time and Space Complexity:** O(m\*n) time complexity, O(m\*n) space complexity

	
	# Code
	```python
	def minPathSum(grid):
    """
    Finds the minimum path sum from top-left to bottom-right in a grid.

    Args:
        grid: A list of lists of integers representing the grid.

    Returns:
        The minimum path sum.
    """

    m = len(grid)
    n = len(grid[0])

    # Initialize a DP table to store minimum path sums
    dp = [[0] * n for _ in range(m)]

    # Base case: starting point
    dp[0][0] = grid[0][0]

    # Fill the first row
    for j in range(1, n):
        dp[0][j] = dp[0][j-1] + grid[0][j]

    # Fill the first column
    for i in range(1, m):
        dp[i][0] = dp[i-1][0] + grid[i][0]

    # Fill the rest of the DP table
    for i in range(1, m):
        for j in range(1, n):
            dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])

    # The bottom-right cell contains the minimum path sum
    return dp[m-1][n-1]
	```