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Suppose we have a list of numbers called nums, we have to find the length of the longest sublist where 2 * (minimum of sublist) > (maximum of sublist).

So, if the input is like nums = [10, 2, 6, 6, 4, 4], then the output will be 4, because the sublist [6, 6, 4, 4] is the longest sublist that meet the given criteria (2*4) > 6.

To solve this, we will follow these steps −

- ret := 0
- minq := an empty double ended queue
- maxq := an empty double ended queue
- l := 0
- r := 0
- while r < size of nums, do
- n := nums[r]
- while minq is not empty and n < nums[last element of minq], do
- delete last element from minq

- insert r at the end of minq
- while maxq is not empty and n > nums[last element of maxq], do
- delete last element from maxq

- insert r at the end of maxq
- r := r + 1
- while l < r and nums[minq[0]] * 2 <= nums[maxq[0]], do
- if minq[0] is same as l, then
- left left item from minq

- if maxq[0] is same as l, then
- delete last item of maxq

- l := l + 1

- if minq[0] is same as l, then
- ret := maximum of ret and (r - l)

- return ret

Let us see the following implementation to get better understanding −

from collections import deque def solve(nums): ret = 0 minq, maxq = deque(), deque() l, r = 0, 0 while r < len(nums): n = nums[r] while minq and n < nums[minq[-1]]: minq.pop() minq.append(r) while maxq and n > nums[maxq[-1]]: maxq.pop() maxq.append(r) r += 1 while l < r and nums[minq[0]] * 2 <= nums[maxq[0]]: if minq[0] == l: minq.popleft() if maxq[0] == l: maxq.popleft() l += 1 ret = max(ret, r - l) return ret nums = [10, 2, 6, 6, 4, 4] print(solve(nums))

[10, 2, 6, 6, 4, 4]

4

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