What is the time complexity of Taking Maximum Energy From the Mystic Dungeon?

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

What data structures are used to solve Taking Maximum Energy From the Mystic Dungeon?

This problem uses Array, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Taking Maximum Energy From the Mystic Dungeon?

Insight 1: The key insight is to realize that the problem can be efficiently solved by working backwards from the end of the array and accumulating the energy values. Insight 2: We can modify the original `energy` array in place to store the maximum energy obtainable from each starting position.

Which companies ask LeetCode 3147. Taking Maximum Energy From the Mystic Dungeon in interviews?

This problem is frequently asked at IBM.

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Taking Maximum Energy From the Mystic Dungeon - LeetCode 3147 Solution

Taking Maximum Energy From the Mystic Dungeon - Complete Solution Guide

Taking Maximum Energy From the Mystic Dungeon is LeetCode problem 3147, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

In a mystic dungeon, n magicians are standing in a line. Each magician has an attribute that gives you energy. Some magicians can give you negative energy, which means taking energy from you. You have been cursed in such a way that after absorbing energy from magician i , you will be instantly transported to magician (i + k) . This process will be repeated until you reach the magician where (i + k) does not exist. In other words, you will choose a starting point and then teleport with k jumps un

Detailed Explanation

The problem describes a scenario where a person is cursed and must absorb energy from magicians arranged in a line. The person starts at a chosen magician and then teleports forward by a fixed number of positions 'k' until they reach the end of the line. The goal is to find the starting position that maximizes the total energy gained from all the magicians visited during this process. The energy of a magician can be positive or negative, and you must absorb the energy from each magician you land on.

Solution Approach

The solution uses dynamic programming to efficiently calculate the maximum energy achievable from each possible starting position. It iterates backward from the `n - k - 1` index of the `energy` array to 0. For each index `i`, it adds the energy at index `i + k` to the energy at index `i`. This effectively computes the total energy gained by starting at position `i` and following the teleportation rule. After this process, the `energy` array will hold the total energy attainable from each starting index. The algorithm then simply finds the maximum value in the modified `energy` array, which represents the maximum energy obtainable from any starting position.

Step-by-Step Algorithm

Step 1: Iterate backward through the `energy` array from index `n - k - 1` down to 0. This is because if you start at an index i >= n-k, you'll take that element's value, and be done.
Step 2: For each index `i`, add the energy at index `i + k` to the energy at index `i`. This accumulates the energy values for each possible starting position according to the teleportation rule.
Step 3: After the loop completes, find the maximum value in the `energy` array. This maximum value represents the maximum energy obtainable from any starting position.

Key Insights

Insight 1: The key insight is to realize that the problem can be efficiently solved by working backwards from the end of the array and accumulating the energy values.
Insight 2: We can modify the original `energy` array in place to store the maximum energy obtainable from each starting position.
Insight 3: The maximum energy obtainable will be the largest value in the modified `energy` array.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Prefix Sum.

Companies

Asked at: IBM.

Taking Maximum Energy From the Mystic Dungeon | LeetCode 3147

Medium

Array Prefix Sum

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int maximumEnergy(int[] energy, int k) {
        int n = energy.length;
        for (int i = n - k - 1; i >= 0; i--) {
            energy[i] += energy[i + k];
        }
        int maxEnergy = energy[0];
        for (int i = 1; i < n; i++) {
            if (energy[i] > maxEnergy) {
                maxEnergy = energy[i];
            }
        }
        return maxEnergy;
    }
}

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Problem Info

Description

Problem 3147: Taking Maximum Energy From the Mystic Dungeon - Detailed explanation and solution approaches.

Topics

Array Prefix Sum

Companies

IBM

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3147

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

3 steps

1
Step 1: Iterate backward through the `energy` array from index `n - k - 1` down to 0. This is because if you start at an index i >= n-k, you'll take that element's value, and be done.
2
Step 2: For each index `i`, add the energy at index `i + k` to the energy at index `i`. This accumulates the energy values for each possible starting position according to the teleportation rule.
3
Step 3: After the loop completes, find the maximum value in the `energy` array. This maximum value represents the maximum energy obtainable from any starting position.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because both the accumulation loop and the maximum-finding loop iterate through the array at most once. The space complexity is constant because the algorithm operates in-place, modifying the input array directly without requiring auxiliary data structures proportional to the input size.

📖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

Using a separate array for prefix sums

Instead of modifying the input array, create a separate array to store the accumulated energy values for each starting position. This maintains the original input.

Time: O(n)Space: O(n)

Trade-offs: Using a separate array consumes more space but preserves the original array. This may be preferable if the original array needs to be preserved for later use.

✨

Key Insights

1Insight 1: The key insight is to realize that the problem can be efficiently solved by working backwards from the end of the array and accumulating the energy values.
2Insight 2: We can modify the original `energy` array in place to store the maximum energy obtainable from each starting position.
3Insight 3: The maximum energy obtainable will be the largest value in the modified `energy` array.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: All energy values are negative. The algorithm correctly handles this case by returning the least negative value (which will be the largest) among the possible sums.
Edge case 2: `k` is close to the length of the array. The algorithm still works correctly because the loop condition `i >= 0` ensures that the index `i + k` is always within the bounds of the array when accessed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect loop termination condition. If the loop does not iterate from `n - k - 1` to 0 properly, the energy accumulation will be incorrect.

✗Common mistake 2: Not handling the case where all energy values are negative. The algorithm must correctly identify the least negative sum as the maximum energy.

✅

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: Optimizing resource allocation in a network where resources are distributed along a path with fixed jump distances.

🏦Real-world scenario 2: Analyzing financial investments with periodic returns at fixed intervals to find the most profitable starting point.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: August 27, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Taking Maximum Energy From the Mystic Dungeon"?

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

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