Jump Game VI
Problem Description
You are given a 0-indexed integer array nums and an integer k.
You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive.
You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array.
Return the maximum score you can get.
Example 1:
Input: nums = [1,-1,-2,4,-7,3], k = 2
Output: 7
Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
Example 2:
Input: nums = [10,-5,-2,4,0,3], k = 3
Output: 17
Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.
Example 3:
Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
Output: 0
Constraints:
- 1 <= nums.length, k <= 105
- -104 <= nums[i] <= 104
Solve it on LeetCode
Approaches
1. Brute Force
Intuition:
In the brute force approach, the idea is to explore every possible path by jumping up to k steps forward and choose the path that yields the maximum score. This involves recursively jumping from the current position all the way to the n-1 index and calculating the score for each path.
Code:
- Time Complexity: O(k^n), as each step can branch into up to
kfurther recursive calls. - Space Complexity: O(n), due to the recursion stack that can go as deep as
n.
2. Dynamic Programming with Deque
Intuition:
To achieve better performance, we utilize a deque to maintain a sliding window of size k over the array and store indices. The front of the deque will always represent the maximum possible score we can get at that point. This allows calculating the maximum score at each step in constant time.
Code:
- Time Complexity: O(n), each element is processed at most twice.
- Space Complexity: O(n), due to the storage of the dp array and deque.
3. Dynamic Programming with Priority Queue
Intuition:
This approach uses a priority queue to maintain the maximum scores at each index similar to the deque strategy. The priority queue will always allow us to peek the current maximum score in logarithmic time.
Code:
- Time Complexity: O(n log k), due to the insert and remove operations on the priority queue.
- Space Complexity: O(n), as we store scores and their respective indices.