LeetCode 3147: Taking Maximum Energy From the Mystic Dungeon Solution
Master LeetCode problem 3147 (Taking Maximum Energy From the Mystic Dungeon), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
3147. Taking Maximum Energy From the Mystic Dungeon
Problem Explanation
Explanation:
To maximize the energy gained while moving through the magicians in the dungeon, we can use a dynamic programming approach. We can start from the last magician and work our way backwards, calculating the maximum energy gain we can achieve when starting from each magician. At each step, we calculate the maximum energy gain for the current magician by considering the energy gain from absorbing the magician's energy and the maximum energy gain from the magician that can be reached after making the k jumps.
- Create a dp array to store the maximum energy gain when starting from each magician.
- Initialize the dp array with the values of the energy array.
- Iterate backwards through the magicians starting from the second to last magician.
- For each magician, calculate the maximum energy gain by considering the energy gain from absorbing the magician's energy and adding it to the maximum energy gain from the magician that can be reached after making the k jumps.
- Return the maximum energy gain that can be achieved starting from the first magician.
Time Complexity: O(n) Space Complexity: O(n)
:
Solution Code
class Solution {
public int maxEnergy(int[] energy, int k) {
int n = energy.length;
int[] dp = new int[n];
dp[n - 1] = energy[n - 1];
for (int i = n - 2; i >= 0; i--) {
int maxGain = Integer.MIN_VALUE;
for (int j = i + 1; j <= Math.min(i + k, n - 1); j++) {
maxGain = Math.max(maxGain, dp[j]);
}
dp[i] = energy[i] + maxGain;
}
return dp[0];
}
}Try It Yourself
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