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3105. Longest Strictly Increasing or Strictly Decreasing Subarray 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  // Similar to 978. Longest Turbulent Subarray
  int longestMonotonicSubarray(vector<int>& nums) {
    int ans = 1;
    int increasing = 1;
    int decreasing = 1;

    for (int i = 1; i < nums.size(); ++i) {
      if (nums[i] > nums[i - 1]) {
        increasing += 1;
        decreasing = 1;
      } else if (nums[i] < nums[i - 1]) {
        decreasing += 1;
        increasing = 1;
      } else {
        increasing = 1;
        decreasing = 1;
      }
      ans = max({ans, increasing, decreasing});
    }

    return ans;
  }
};

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class Solution {
  // Similar to 978. Longest Turbulent Subarray
  public int longestMonotonicSubarray(int[] nums) {
    int ans = 1;
    int increasing = 1;
    int decreasing = 1;

    for (int i = 1; i < nums.length; ++i) {
      if (nums[i] > nums[i - 1]) {
        increasing += 1;
        decreasing = 1;
      } else if (nums[i] < nums[i - 1]) {
        decreasing += 1;
        increasing = 1;
      } else {
        increasing = 1;
        decreasing = 1;
      }
      ans = Math.max(ans, Math.max(increasing, decreasing));
    }

    return ans;
  }
}

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class Solution:
  # Similar to 978. Longest Turbulent Subarray
  def longestMonotonicSubarray(self, nums: list[int]) -> int:
    ans = 1
    increasing = 1
    decreasing = 1

    for i in range(1, len(nums)):
      if nums[i] > nums[i - 1]:
        increasing += 1
        decreasing = 1
      elif nums[i] < nums[i - 1]:
        decreasing += 1
        increasing = 1
      else:
        increasing = 1
        decreasing = 1
      ans = max(ans, increasing, decreasing)

    return ans