Sponsored
Sponsored
This approach employs dynamic programming in combination with a deque to keep track of the best scores possible when jumping to each index within constraints. By using a decreasing deque, we efficiently maintain the maximum score in a window of size k.
Time Complexity: O(n), because each element is inserted and removed from the deque at most once. Space Complexity: O(k), to store the window of indices in the deque.
1from collections import deque
2
3def maxResult(nums, k):
4 n = len(nums)
5 dq = deque([0])
6 for i in range(1, n):
7 while dq and dq[0] < i - k:
8 dq.popleft()
9 nums[i] += nums[dq[0]]
10 while dq and nums[i] >= nums[dq[-1]]:
11 dq.pop()
12 dq.append(i)
13 return nums[-1]We initialize a deque and iterate over the array. For each index, we ensure that the deque only contains valid indices. We update the current index with the maximum score that can be achieved from the deque front, then maintain a decreasing order in the deque by popping elements smaller than the current score.
This approach involves using a dynamic programming solution where we keep track of the best scores using a max-heap (priority queue). By pushing elements onto the heap, we ensure that the maximum score is always accessible, facilitating quick updates for each step in our process.
Time Complexity: O(n log k) due to heap operations. Space Complexity: O(k), maintaining up to k elements in the heap.
1import
For this Java approach, the PriorityQueue is customized to function as a max-heap using negative values of scores. This allows accessing the highest scores quickly to progress through the array with maximized scores.