3105. Longest Strictly Increasing or Strictly Decreasing Subarray - Explanation

1. Brute Force

class Solution:
    def longestMonotonicSubarray(self, nums: List[int]) -> int:
        n = len(nums)
        res = 1

        for i in range(n - 1):
            curLen = 1
            for j in range(i + 1, n):
                if nums[j] == nums[j - 1] or ((nums[i] < nums[i + 1]) != (nums[j - 1] < nums[j])):
                    break
                curLen += 1

            res = max(res, curLen)

        return res

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(1)$

2. Iteration - I

class Solution:
    def longestMonotonicSubarray(self, nums: List[int]) -> int:
        cur = 1
        res = 1
        increasing = 0

        for i in range(1, len(nums)):
            if nums[i - 1] < nums[i]:
                if increasing > 0:
                    cur += 1
                else:
                    cur = 2
                    increasing = 1
            elif nums[i - 1] > nums[i]:
                if increasing < 0:
                    cur += 1
                else:
                    cur = 2
                    increasing = -1
            else:
                cur = 1
                increasing = 0
            res = max(res, cur)

        return res

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

3. Iteration - II

class Solution:
    def longestMonotonicSubarray(self, nums: List[int]) -> int:
        inc = dec = 1
        res = 1

        for i in range(1, len(nums)):
            if nums[i] == nums[i - 1]:
                inc = dec = 1
            elif nums[i] > nums[i - 1]:
                inc, dec = inc + 1, 1
            else:
                inc, dec = 1, dec + 1

            res = max(res, inc, dec)

        return res

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

4. Iteration - III

class Solution:
    def longestMonotonicSubarray(self, nums: List[int]) -> int:
        curLen = res = 1

        for i in range(1, len(nums)):
            if (nums[i] == nums[i - 1] or
                ((nums[i - curLen] < nums[i - curLen + 1]) != (nums[i - 1] < nums[i]))
            ):
                curLen = 1 if (nums[i] == nums[i - 1]) else 2
                continue

            curLen += 1
            res = max(res, curLen)

        return res

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$