Introduction

In this blog post, we will explore how to solve the LeetCode problem "120" 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 triangle array, return the minimum path sum from top to bottom. For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row. Example 1: Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]] Output: 11 Explanation: The triangle looks like: 2 3 4 6 5 7 4 1 8 3 The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above). Example 2: Input: triangle = [[-10]] Output: -10 Constraints: 1 <= triangle.length <= 200 triangle[0].length == 1 triangle[i].length == triangle[i - 1].length + 1 -104 <= triangle[i][j] <= 104 Follow up: Could you do this using only O(n) extra space, where n is the total number of rows in the triangle?

Explanation

Here's the breakdown of the solution and the Python code:

• Approach:

  • Use dynamic programming. Start from the bottom row and work upwards.
  • Store the minimum path sum from each cell to the bottom in an array.
  • For each cell in the current row, the minimum path sum is the cell's value plus the minimum of the two adjacent cells in the row below.

• Complexity: O(n) time, O(n) space, where n is the number of rows in the triangle.

Code

    def minimumTotal(triangle):
    """
    Calculates the minimum path sum from top to bottom in a triangle.

    Args:
        triangle: A list of lists representing the triangle.

    Returns:
        The minimum path sum.
    """

    n = len(triangle)
    dp = triangle[-1]  # Initialize dp with the last row

    for i in range(n - 2, -1, -1):
        for j in range(i + 1):
            dp[j] = triangle[i][j] + min(dp[j], dp[j + 1])

    return dp[0]