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NeetCode #49LC-3105EasyArrays & Hashing
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#49 · #3105 · Longest Strictly Increasing or Strictly Decreasing Subarray(最长的严格递增或递减子数组)

📌 Problem Statement & Constraints

Given an array of integers nums, return the length of the longest subarray that is either strictly increasing or strictly decreasing. Constraints: 1 <= nums.length <= 50, 1 <= nums[i] <= 50.

💡 Core Algorithmic Approaches

  1. Track two running lengths: inc for the current strictly increasing run and dec for the strictly decreasing run.
  2. Compare each element with its predecessor. If nums[i] > nums[i-1], extend inc and reset dec to 1; if smaller, extend dec and reset inc; if equal, reset both to 1.
  3. The answer is the maximum of both counters over the whole scan.
  4. Because a subarray of length 1 is trivially both, both counters start at 1 and the answer is at least 1.

💻 Benchmark Python3 Implementation

class Solution:
    def longestMonotonicSubarray(self, nums: List[int]) -> int:
        inc = dec = 1                      # single-element runs
        best = 1
        for i in range(1, len(nums)):
            if nums[i] > nums[i - 1]:
                inc += 1
                dec = 1                    # the decreasing run is broken
            elif nums[i] < nums[i - 1]:
                dec += 1
                inc = 1
            else:
                inc = dec = 1              # equality breaks both
            best = max(best, inc, dec)
        return best

⚡ Complexity Deep Dive

⏱️ Time Complexity
O(n): one pass with constant work per element.
💾 Space Complexity
O(1): three counters.

⚠️ Interview Pitfalls & Follow-ups

  • Using >= / <= instead of strict comparisons: equal adjacent elements break monotonicity, so the comparison must be strict.
  • Forgetting to reset the opposite counter: [1, 2, 3] would then report a decreasing run that does not exist.
  • Initialising the counters to 0: a single element is a valid monotonic subarray of length 1.
  • Handling equality by only resetting one counter: equality breaks both directions.