What is the time complexity of Dungeon Game?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Dungeon Game?

The space complexity is O(m*n).

What data structures are used to solve Dungeon Game?

This problem uses Array, Dynamic Programming, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Dungeon Game?

Insight 1: The problem can be efficiently solved using dynamic programming, working backwards from the destination to the starting point. Insight 2: Instead of calculating the maximum health the knight can have, we calculate the minimum health he needs at each cell to survive until the end.

Which companies ask LeetCode 174. Dungeon Game in interviews?

This problem is frequently asked at Flipkart.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Dungeon Game - LeetCode 174 Solution

Dungeon Game - Complete Solution Guide

Dungeon Game is LeetCode problem 174, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 i

Detailed Explanation

The problem presents a dungeon represented by a 2D grid, where each cell can contain either demons (negative integers), magic orbs (positive integers), or be empty (0). A knight starts at the top-left corner and needs to reach the princess in the bottom-right corner. The knight can only move right or down. The knight has initial health, and his health decreases when entering a room with demons and increases when entering a room with magic orbs. The goal is to find the minimum initial health the knight needs to start with to rescue the princess without his health dropping to 0 or below at any point.

Solution Approach

The solution employs dynamic programming to determine the minimum health required at each cell. It starts from the destination (bottom-right) and iterates backward to the starting cell (top-left). For each cell, it calculates the minimum health needed to enter that cell, considering the minimum health needed in the next possible cells (right and down). The minimum of the right and down cells determine the minimum health to arrive at either of those positions. The current dungeon value is subtracted from this health, giving the health needed to *enter* the current cell. If this required health is non-positive, it means the knight can survive the current cell with health 1 (since health cannot be zero or negative). The final value at dp[0][0] represents the minimum initial health required.

Step-by-Step Algorithm

Step 1: Determine the dimensions of the dungeon (m x n).
Step 2: Create a DP table `dp` of size (m+1) x (n+1) and initialize all its elements to infinity. The extra row and column are for boundary conditions.
Step 3: Initialize `dp[m][n-1]` and `dp[m-1][n]` to 1. These represent the minimum health needed to enter the last row/column cells next to the princess.
Step 4: Iterate through the dungeon from the bottom-right (excluding the princess's cell) to the top-left. Use nested loops that decrement `i` from `m-1` to 0 and `j` from `n-1` to 0.
Step 5: For each cell (i, j), calculate `min_health_needed_after` as the minimum of `dp[i+1][j]` (health needed moving down) and `dp[i][j+1]` (health needed moving right).
Step 6: Calculate `health_needed_to_enter` as `min_health_needed_after - dungeon[i][j]`. This is the minimum health needed to survive the current cell.
Step 7: If `health_needed_to_enter` is less than or equal to 0, set `dp[i][j]` to 1 (meaning at least 1 health is needed). Otherwise, set `dp[i][j]` to `health_needed_to_enter`.
Step 8: After the loops complete, `dp[0][0]` contains the minimum initial health the knight needs to start with.

Key Insights

Insight 1: The problem can be efficiently solved using dynamic programming, working backwards from the destination to the starting point.
Insight 2: Instead of calculating the maximum health the knight can have, we calculate the minimum health he needs at each cell to survive until the end.
Insight 3: Initialize the DP table with 'infinity' to handle boundary conditions and to ensure the minimum health needed is propagated correctly.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Dynamic Programming, Matrix.

Companies

Asked at: Flipkart.

Dungeon Game | LeetCode 174

Hard

Array Dynamic Programming Matrix

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

class Solution {
    public int calculateMinimumHP(int[][] dungeon) {
        int m = dungeon.length;
        int n = dungeon[0].length;

        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i <= m; i++) {
            for (int j = 0; j <= n; j++) {
                dp[i][j] = Integer.MAX_VALUE;
            }
        }
        dp[m][n - 1] = 1;
        dp[m - 1][n] = 1;

        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                int min_health_needed_after = Math.min(dp[i + 1][j], dp[i][j + 1]);
                int health_needed_to_enter = min_health_needed_after - dungeon[i][j];

                if (health_needed_to_enter <= 0) {
                    dp[i][j] = 1;
                } else {
                    dp[i][j] = health_needed_to_enter;
                }
            }
        }

        return dp[0][0];
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 174: Dungeon Game - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Matrix

Companies

Flipkart

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:174

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Determine the dimensions of the dungeon (m x n).
2
Step 2: Create a DP table `dp` of size (m+1) x (n+1) and initialize all its elements to infinity. The extra row and column are for boundary conditions.
3
Step 3: Initialize `dp[m][n-1]` and `dp[m-1][n]` to 1. These represent the minimum health needed to enter the last row/column cells next to the princess.
4
Step 4: Iterate through the dungeon from the bottom-right (excluding the princess's cell) to the top-left. Use nested loops that decrement `i` from `m-1` to 0 and `j` from `n-1` to 0.
5
Step 5: For each cell (i, j), calculate `min_health_needed_after` as the minimum of `dp[i+1][j]` (health needed moving down) and `dp[i][j+1]` (health needed moving right).
6
Step 6: Calculate `health_needed_to_enter` as `min_health_needed_after - dungeon[i][j]`. This is the minimum health needed to survive the current cell.
7
Step 7: If `health_needed_to_enter` is less than or equal to 0, set `dp[i][j]` to 1 (meaning at least 1 health is needed). Otherwise, set `dp[i][j]` to `health_needed_to_enter`.
8
Step 8: After the loops complete, `dp[0][0]` contains the minimum initial health the knight needs to start with.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The nested loops iterate over each cell of the dungeon grid once, leading to O(m*n) time complexity. The DP table stores the minimum health needed at each cell, resulting in O(m*n) space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach with memoization could be used. This involves defining a recursive function that calculates the minimum health required to reach the destination from a given cell. Memoization is used to store the results of already computed cells to avoid redundant calculations.

Time: O(m*n)Space: O(m*n) due to the memoization table and recursion stack.

Trade-offs: While conceptually similar to DP, recursion with memoization can sometimes be less efficient in practice due to the overhead of function calls and potential stack overflow issues for very large dungeons. DP generally has better constant-time performance.

✨

Key Insights

1Insight 1: The problem can be efficiently solved using dynamic programming, working backwards from the destination to the starting point.
2Insight 2: Instead of calculating the maximum health the knight can have, we calculate the minimum health he needs at each cell to survive until the end.
3Insight 3: Initialize the DP table with 'infinity' to handle boundary conditions and to ensure the minimum health needed is propagated correctly.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Empty dungeon. The code implicitly handles this as the loops won't execute, and the initial values returned would be related to the initialization of DP table.
Edge case 2: Dungeon with all positive values. The knight's health will always increase, and the code correctly determines the minimum initial health needed to be 1 in certain cases or calculates the required health to not drop below 0.
Edge case 3: Dungeon with all negative values. The knight's health will always decrease, and the code computes the minimum initial health to counteract the health loss from the negative values.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Initializing the DP table incorrectly. Setting the base cases wrong or not handling boundary conditions properly can lead to incorrect results.

✗Common mistake 2: Not considering the minimum health needed at each step. Forgetting that the knight's health must always be greater than 0 is a critical error.

✗Common mistake 3: Trying to solve the problem from top-left to bottom-right directly. It's more natural and easier to solve the problem by working backwards.

✗Common mistake 4: Not understanding dynamic programming. Trying to hardcode the solution without realizing it's a DP question and following the correct DP methodology of defining the recurrence relations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation in a network. Imagine a network where nodes consume or produce resources. You want to find the minimum initial resources needed to ensure all nodes can operate without running out of resources.

🏦Real-world scenario 2: Path planning with constraints. Consider a robot navigating a warehouse. Each location might have a cost (energy consumption) or benefit (energy recharge). Finding the optimal path with minimum initial energy mirrors the dungeon game problem.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Dungeon Game"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?