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[7/9] 1696. Jump Game VI

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작성일 2022.07.08 19:19

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Medium

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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

#Apple, #Mathworks, #Array, #Dynamic Programming(DP), #Queue, #Sliding Window, #Heap(Priority Queue), #Monotonic Queue

링크

https://leetcode.com/problems/jump-game-vi/ 2169 회 연결

익명
작성일 2022.07.10 10:36


# deque dp solution
import collections
class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        deque = collections.deque([(nums[0],0)])
        dp = [0]*len(nums)
        dp[0] = nums[0]
        
        for i in range(1, len(nums)):
            while deque[0][1] < i-k : deque.popleft()
            
            dp[i] = nums[i] + deque[0][0]
            
            while deque and deque[-1][0] < dp[i]:
                deque.pop()
            deque.append((dp[i],i))
        
        return dp[-1]


# max heap dp solution
import heapq
class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        if len(nums) == 1: return nums[0]
        max_heap = [(-nums[0], 0)]
        heapq.heapify(max_heap)
        
        for i in range(1, len(nums)):
            while max_heap[0][1] < i-k: heapq.heappop(max_heap)
            
            curr_max = max_heap[0][0]
            heapq.heappush(max_heap,(curr_max-nums[i], i))
            if i == len(nums) -1: return -(curr_max - nums[i])

[code=Python 3]
# deque dp solution
import collections
class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        deque = collections.deque([(nums[0],0)])
        dp = [0]*len(nums)
        dp[0] = nums[0]
        
        for i in range(1, len(nums)):
            while deque[0][1] < i-k : deque.popleft()
            
            dp[i] = nums[i] + deque[0][0]
            
            while deque and deque[-1][0] < dp[i]:
                deque.pop()
            deque.append((dp[i],i))
        
        return dp[-1]

# max heap dp solution
import heapq
class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        if len(nums) == 1: return nums[0]
        max_heap = [(-nums[0], 0)]
        heapq.heapify(max_heap)
        
        for i in range(1, len(nums)):
            while max_heap[0][1] < i-k: heapq.heappop(max_heap)
            
            curr_max = max_heap[0][0]
            heapq.heappush(max_heap,(curr_max-nums[i], i))
            if i == len(nums) -1: return -(curr_max - nums[i])
        
        [/code]