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Suppose we have a list of mumbers called nums, and find the length of the longest sublist in nums such that the equality relation between every consecutive numbers changes alternatively between less-than and greater-than operation. The first two numbers' inequality may be either less-than or greater-than.

So, if the input is like nums = [1, 2, 6, 4, 5], then the output will be 4, as the longest inequality alternating sublist is [2, 6, 4, 5] as 2 < 6 > 4 < 5.

To solve this, we will follow these steps −

Define a function get_direction(). This will take a, b

return 0 if a is same as b otherwise -1 if a < b otherwise 1

if size of nums < 2, then

return size of nums

max_length := 1, cur_length := 1, last_direction := 0

for i in range 0 to size of nums - 1, do

direction := get_direction(nums[i], nums[i + 1])

if direction is same as 0, then

cur_length := 1

otherwise when direction is same as last_direction, then

cur_length := 2

otherwise,

cur_length := cur_length + 1

max_length := maximum of max_length and cur_length

last_direction := direction

return max_length

Let us see the following implementation to get better understanding−

class Solution: def solve(self, nums): if len(nums) < 2: return len(nums) def get_direction(a, b): return 0 if a == b else -1 if a < b else 1 max_length = 1 cur_length = 1 last_direction = 0 for i in range(len(nums) - 1): direction = get_direction(nums[i], nums[i + 1]) if direction == 0: cur_length = 1 elif direction == last_direction: cur_length = 2 else: cur_length += 1 max_length = max(max_length, cur_length) last_direction = direction return max_length ob = Solution() nums = [1, 2, 6, 4, 5] print(ob.solve(nums))

[1, 2, 6, 4, 5]

4

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