How do you solve Taking Maximum Energy From the Mystic Dungeon on LeetCode?

Taking Maximum Energy From the Mystic Dungeon (LeetCode #3147) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Dynamic programming helps avoid redundant calculations.

What is the brute force solution for Taking Maximum Energy From the Mystic Dungeon?

The brute force approach for Taking Maximum Energy From the Mystic Dungeon is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible starting magician and calculating the total energy gained from that starting point by jumping k positions until the end. While straightforward, this method can be inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Taking Maximum Energy From the Mystic Dungeon?

Taking Maximum Energy From the Mystic Dungeon uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Array.

What are the interview tips for Taking Maximum Energy From the Mystic Dungeon?

Explain your thought process clearly while coding. Discuss the trade-offs of your approaches. Practice similar problems to recognize patterns.

What are common mistakes when solving Taking Maximum Energy From the Mystic Dungeon?

Not considering the base case for the last magician. Forgetting to check bounds when accessing dp[i + k].

#3147

Taking Maximum Energy From the Mystic Dungeon

Medium

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using dynamic programming, we can store the maximum energy that can be gained starting from each magician. This allows us to avoid recalculating energies for previously computed indices, significantly improving efficiency.

⚙️

Algorithm

4 steps

1Step 1: Create a dp array where dp[i] represents the maximum energy starting from magician i.
2Step 2: Initialize dp[n-1] with energy[n-1] since the last magician can only provide their own energy.
3Step 3: Iterate backwards from the second last magician to the first, calculating dp[i] = energy[i] + dp[i + k] if i + k < n.
4Step 4: The result will be the maximum value in the dp array.

solution.py10 lines

1# Full working Python code
2
3def max_energy_optimal(energy, k):
4    n = len(energy)
5    dp = [0] * n
6    dp[n - 1] = energy[n - 1]
7    for i in range(n - 2, -1, -1):
8        dp[i] = energy[i] + (dp[i + k] if (i + k) < n else 0)
9    return max(dp)
10

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once to fill the dp array, and the space complexity is O(n) due to the additional dp array used for storing results.

1Dynamic programming helps avoid redundant calculations.
2Understanding the relationship between indices is crucial for optimization.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.