# Program to find length of longest alternating inequality elements sublist in Python

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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−

## Example

Live Demo

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))

## Input

[1, 2, 6, 4, 5]

## Output

4