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Suppose we have a list of numbers nums and another value k. Here the items at nums[i] represents the costs of landing at index i. If we start from index 0 and end at the last index of nums. In each step we can jump from some position X to any position up to k steps away. We have to minimize the sum of costs to reach last index, so what will be the minimum sum?

So, if the input is like nums = [2, 3, 4, 5, 6] k = 2, then the output will be 12, as we can select 2 + 4 + 6 to get a minimum cost of 12.

To solve this, we will follow these steps −

- ans := 0
- h := an empty heap
- for i in range 0 to size of nums, do
- val := 0
- while h is not empty, do
- [val, index] := h[0]
- if index >= i - k, then
- come out from the loop

- otherwise,
- delete top from the heap h

- ans := nums[i] + val
- insert pair (ans, i) into heap h

- return ans

Let us see the following implementation to get better understanding −

from heapq import heappush, heappop class Solution: def solve(self, nums, k): ans = 0 h = [] for i in range(len(nums)): val = 0 while h: val, index = h[0] if index >= i - k: break else: heappop(h) ans = nums[i] + val heappush(h, (ans, i)) return ans ob = Solution() nums = [2, 3, 4, 5, 6] k = 2 print(ob.solve(nums, k))

[2, 3, 4, 5, 6], 2

12

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