1696-jump-game-vi

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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

Solution(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]