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, 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

    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]