# 1696-jump-game-vi Try it on leetcode ## 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:

## Solution(Python) ```Python class Solution: def maxResult(self, nums: List[int], k: int) -> int: return self.optimal(nums, k) # Time Complexity: O(n*k) # Space Complexity: O(n*k) def naive(self, nums: List[int], k: int) -> int: n = len(nums) @cache def dfs(i): if i >= n-1: return nums[i] res = float('-inf') for j in range(i+1,min(n-1,i+k)+1): include = dfs(j) res = max(res,nums[i]+include) return res if res != float('-inf') else 0 return dfs(0) # Time Complexity: O(nlogk) # Space Complexity: O(n) def better(self, nums: List[int], k: int) -> int: n = len(nums) dp = [0] * (n+1) q = [] heapq.heapify(q) for i in range(n-1, -1, -1): sum_so_far = float('-inf') while q and q[0][1] > min(n-1,i+k): heapq.heappop(q) if q: sum_so_far = max(sum_so_far, -q[0][0]) dp[i] = nums[i] + (sum_so_far if sum_so_far != float('-inf') else 0) heapq.heappush(q,(-dp[i],i)) return dp[0] # Time Complexity: O(n) # Space Complexity: O(n) def optimal(self, nums: List[int], k: int) -> int: deq, n = deque([0]), len(nums) for i in range(1, n): while deq and deq[0] < i - k: deq.popleft() nums[i] += nums[deq[0]] while deq and nums[i] >= nums[deq[-1]]: deq.pop() deq.append(i) return nums[-1] ```