Introduction

In this blog post, we will explore how to solve the LeetCode problem "174" 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

The demons had captured the princess and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of m x n rooms laid out in a 2D grid. Our valiant knight was initially positioned in the top-left room and must fight his way through dungeon to rescue the princess. The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately. Some of the rooms are guarded by demons (represented by negative integers), so the knight loses health upon entering these rooms; other rooms are either empty (represented as 0) or contain magic orbs that increase the knight's health (represented by positive integers). To reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step. Return the knight's minimum initial health so that he can rescue the princess. Note that any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned. Example 1: Input: dungeon = [[-2,-3,3],[-5,-10,1],[10,30,-5]] Output: 7 Explanation: The initial health of the knight must be at least 7 if he follows the optimal path: RIGHT-> RIGHT -> DOWN -> DOWN. Example 2: Input: dungeon = [[0]] Output: 1 Constraints: m == dungeon.length n == dungeon[i].length 1 <= m, n <= 200 -1000 <= dungeon[i][j] <= 1000

Explanation

Here's the breakdown of the solution:

• Dynamic Programming (Bottom-Up): We use dynamic programming to solve this problem. We start from the princess's cell (bottom-right) and work our way backward to the knight's starting cell (top-left).
• Minimum Health Calculation: For each cell, we calculate the minimum health required to enter that cell, ensuring the knight survives until the end. This is based on the minimum health needed in the cells below and to the right.

• Base Cases: The base cases are the cells along the bottom row and the rightmost column.

• Runtime & Storage Complexity: O(mn) Time, O(mn) Space

Code

    def calculateMinimumHP(dungeon):
    """
    Calculates the minimum initial health required for the knight to rescue the princess.

    Args:
        dungeon: A 2D list representing the dungeon.

    Returns:
        The minimum initial health required.
    """
    m, n = len(dungeon), len(dungeon[0])
    dp = [[0] * n for _ in range(m)]

    # Initialize the bottom-right cell (princess's cell)
    dp[m - 1][n - 1] = max(1, 1 - dungeon[m - 1][n - 1])

    # Initialize the last row (from right to left)
    for j in range(n - 2, -1, -1):
        dp[m - 1][j] = max(1, dp[m - 1][j + 1] - dungeon[m - 1][j])

    # Initialize the last column (from bottom to top)
    for i in range(m - 2, -1, -1):
        dp[i][n - 1] = max(1, dp[i + 1][n - 1] - dungeon[i][n - 1])

    # Fill the rest of the dp table (from bottom-right to top-left)
    for i in range(m - 2, -1, -1):
        for j in range(n - 2, -1, -1):
            dp[i][j] = max(1, min(dp[i + 1][j], dp[i][j + 1]) - dungeon[i][j])

    return dp[0][0]