Dungeon Game

This problem can be solved by using a dynamic programming approach. The goal is to find the minimum initial health the knight needs to rescue the princess, considering that he can only move rightward or downward and must keep his health above 0.

Step 1: Initialize a 2D DP Array

Create a 2D array dp with dimensions (m + 1) x (n + 1) and initialize it with a large value. The last cell dp[m][n] should be set to 1, as the minimum health required at the destination must be 1.

Step 2: Iterate Through the Grid

Start iterating from the bottom-right corner of the grid (above the princess’s cell) and move towards the top-left corner.

Step 3: Calculate Minimum Health at Each Cell

For each cell (i, j), the minimum health required to move forward is the minimum of the cells to its right (i, j + 1) and below (i + 1, j), minus the value of the current cell dungeon[i][j]. If this value drops below 1, set it to 1.

Step 4: Return the Result

Return the value of the top-left cell dp[0][0], which will give the minimum initial health required.

Here’s the code:

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class Solution:
    def calculateMinimumHP(self, dungeon: List[List[int]]) -> int:
        m, n = len(dungeon), len(dungeon[0])
        dp = [[float('inf')] * (n + 1) for _ in range(m + 1)]
        dp[m][n - 1] = dp[m - 1][n] = 1

        for i in range(m - 1, -1, -1):
            for j in range(n - 1, -1, -1):
                min_health = min(dp[i + 1][j], dp[i][j + 1]) - dungeon[i][j]
                dp[i][j] = max(min_health, 1)

        return dp[0][0]

Key Takeaways

We use dynamic programming to keep track of the minimum health required at each cell.
We start from the destination and move towards the starting point, calculating the minimum health required at each step.
The result is stored in the top-left cell, representing the minimum initial health required for the knight to rescue the princess.

Identifying Problem Isomorphism

“Dungeon Game” is related to “Minimum Path Sum”.

In both problems, we have a grid-like structure and the objective is to move from one point in the grid to another, minimizing or maximizing some criteria. For the “Dungeon Game”, the goal is to minimize the initial health so that the knight is able to get to the princess by surviving all the dungeons. On the other hand, in “Minimum Path Sum”, the task is to find the path from top left to bottom right which minimizes the sum of all numbers along its path.

The primary solution technique for both problems is dynamic programming, as they involve making optimal choices at each step based on the future implications of the current choice.

“Minimum Path Sum” is simpler as it just involves simple path summation, while “Dungeon Game” has more complex rules and conditions.

This is not an exact isomorphism, but mapping based on the abstract structure of the problems and the methods employed in their solutions.

10 Prerequisite LeetCode Problems

“174. Dungeon Game” involves dynamic programming and grid traversal. Here are 10 problems to prepare for this problem:

These cover a variety of dynamic programming and grid-based concepts, which is beneficial in solving the problem “174. Dungeon Game”.

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class Solution:
    def calculateMinimumHP(self, dungeon: List[List[int]]) -> int:
        rows, columns = len(dungeon), len(dungeon[0])
        hp = [[0]*columns for i in range(rows)]

        hp[-1][-1] = max(1, 1-dungeon[-1][-1]) 

        for i in range(rows-2, -1,-1):
            hp[i][-1] = max(1, 
                            hp[i+1][-1] - dungeon[i][-1])  
        for j in range(columns-2, -1, -1): 
            hp[-1][j] = max(1, 
                            hp[-1][j+1] - dungeon[-1][j])

        for i in range(rows-2, -1, -1):
            for j in range(columns-2, -1, -1):                
                hp[i][j] = max(1, min(hp[i+1][j] - dungeon[i][j], 
                                      hp[i][j+1] - dungeon[i][j]) )

        return hp[0][0]

Problem Classification

Dungeon Game - The task is to determine the minimum initial health a player needs to reach a certain point in a dungeon represented as a grid. It’s about planning a path with a minimum starting value, so it’s a Path Planning Problem.

Identifying Problem Isomorphism

“Dungeon Game”, is a dynamic programming problem. It’s about finding the minimum initial health that the knight needs to reach the princess located at the bottom-right corner of a dungeon grid.

This problem could be isomorphic to the “Minimum Path Sum” problem on LeetCode, which is problem 64. In both of these problems, the goal is to find a path from the top-left to the bottom-right of a grid, minimizing some cost along the way.

However, there is a crucial difference that prevents these problems from being directly isomorphic: in the “Dungeon Game” problem, the cost at each cell can be either positive or negative, and the total cost can not fall below 1 at any point. In contrast, in the “Minimum Path Sum” problem, all costs are non-negative.

“Dungeon Game” shares similarities with other grid-based dynamic programming problems (for instance, “Unique Paths II”, LeetCode problem 63, where you have to find a path in a grid with obstacles), but again, the mechanics differ because of the requirement for the knight to maintain a certain minimum health level throughout the journey, which is unique to the “Dungeon Game” problem.

As such, while there might be some similar problem-solving patterns or approaches shared with other problems, the “Dungeon Game” problem has unique constraints that set it apart. Therefore, finding a strictly isomorphic problem might be challenging.

Yes, those are some of the closest problems I could find, given that the “Dungeon Game” problem has some unique constraints.

“Minimum Path Sum” (LeetCode 64) and “Unique Paths II” (LeetCode 63) problems both involve finding a path through a grid, which is a similar setup to the “Dungeon Game”. However, neither of these problems deals with the additional constraints in “Dungeon Game” where the cost of each cell can be either positive or negative, and a minimum total cost needs to be maintained throughout the journey.

While there are likely other problems that involve path-finding through a grid or matrix, the specific conditions of the “Dungeon Game” make it unique. The need to handle both positive and negative costs, and to maintain a minimum total cost, add complexity to this problem that is not seen in most other similar problems.

Therefore, while these problems share similarities in terms of the problem setup (finding a path through a grid) and the use of dynamic programming to solve them, they are not strictly isomorphic to the “Dungeon Game”. The unique constraints in the “Dungeon Game” problem make it a distinctive problem on its own.

Language Agnostic Coding Drills

These drills aim to practice fundamental programming concepts, which are universally applicable across modern programming languages.

Creating a 2D array: Learn to create a 2-dimensional array (a matrix) with a specific number of rows and columns.
Accessing and Modifying 2D array elements: Practice accessing and modifying specific elements in a 2D array. This might involve accessing the last element in a row/column, or changing the value of an element in the middle of the array.
Iteration: Understand how to iterate through 2D arrays. This can be done in different orders: left to right, top to bottom (usual way), right to left, bottom to top (reverse way), diagonally, etc.
Conditional Logic: Learn to use conditional logic (if-else statements) to compare two values and make decisions in your code.
Functions: Practice writing functions that take arguments and return a result. This includes writing a function that takes two numbers and returns their maximum and minimum.
Recursive function: Understand how to write a recursive function, a function that calls itself to solve smaller instances of the same problem.
Working with dictionaries (key-value pairs): Learn how to add elements, access elements, modify elements and remove elements from a dictionary.
Sorting: Practice sorting a list of numbers in ascending or descending order. Additionally, learn how to use a custom comparator to sort complex objects.

The aim of these drills is to understand the core concepts, which remain the same irrespective of the programming language. Once you’re comfortable with these, you can then combine these concepts to solve complex problems. For instance, you might use a 2D array with conditional logic inside a recursive function to navigate a grid.

Targeted Drills in Python

Working with Multidimensional Arrays: Drill: Create a two-dimensional list, access elements, modify them, and print the two-dimensional list.
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matrix = [[0 for _ in range(5)] for _ in range(5)] print(matrix) matrix[2][2] = 1 print(matrix)

Initializing the 2D array with a default value Drill: Write a function to initialize a two-dimensional list of given dimensions with a specified value.

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def initialize_2d(rows, cols, value):
    return [[value]*cols for _ in range(rows)]

print(initialize_2d(3, 3, 0))  # prints: [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

Traversing 2D arrays using nested loops: Drill: Write a function that prints each element in a two-dimensional list.

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def print_2d(matrix):
    for row in matrix:
        for element in row:
            print(element, end=' ')
        print()

matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
print_2d(matrix)

Implementing Max-Min Logic Drill: Write a function that takes a list of integers and returns the maximum and minimum values in the list.

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def max_and_min(numbers):
    return max(numbers), min(numbers)

print(max_and_min([3, 7, 2, 9, 1]))  # prints: (9, 1)

Nested loops in reverse order Drill: Write a function to print a given list in reverse order.

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def print_reverse(lst):
    for i in range(len(lst)-1, -1, -1):
        print(lst[i])

print_reverse([1, 2, 3, 4, 5])

Combining these drills for the final solution

The problem is solved using dynamic programming. We start from the bottom-right of the grid and move towards the top-left, while keeping track of the minimum health needed at every point. We initialize the grid with max(1, 1-dungeon[-1][-1]), and then iteratively fill up the remaining cells. For each cell, we take the minimum of the cell below and to the right, subtract the value of the current cell in the dungeon, and take the maximum of this value and 1 (since the minimum health should be 1). Finally, we return the value of the top-left cell of the hp grid.

These drills are in Python, but the underlying concepts (like 2D array manipulation, loops, and conditionals) are language-agnostic and should be applicable to most modern programming languages.

Let’s break down the Python solution into basic components and create coding drills for each one:

Creating a 2D array with a given value:
Drill: Create a 2D array with size 5x5 and initialize all values as 0.
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rows, cols = 5, 5 array_2d = [[0]*cols for _ in range(rows)] print(array_2d)
Calculating the maximum and minimum:
Drill: Write a function that takes two integers and returns the maximum and minimum of the two.
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def calc_max_min(a, b): return max(a, b), min(a, b) print(calc_max_min(3, 5)) # Output: (5, 3)

Working with 2D arrays in Python:

Drill: Accessing and modifying elements in a 2D array.

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array_2d = [[i+j for i in range(5)] for j in range(5)]
print(array_2d)  # Prints the 2D array

# Modify the last element of the last list
array_2d[-1][-1] = 100
print(array_2d)  # Prints the 2D array with the last element of the last list changed

Iterating over 2D array in reverse:

Drill: Print a 2D array in reverse order, i.e., from the last list to the first list and within each list, from the last element to the first element.

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array_2d = [[i+j for i in range(5)] for j in range(5)]

for row in reversed(array_2d):
    for elem in reversed(row):
        print(elem, end=' ')
    print()

Using conditional logic in Python:

Drill: Write a function that takes an integer and returns the integer if it’s greater than 0; otherwise, returns -1.

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def check_num(num):
    if num > 0:
        return num
    else:
        return -1

print(check_num(5))  # Output: 5
print(check_num(-3))  # Output: -1