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3349. Adjacent Increasing Subarrays Detection I 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  bool hasIncreasingSubarrays(vector<int>& nums, int k) {
    int increasing = 1;
    int prevIncreasing = 0;

    for (int i = 1; i < nums.size(); ++i) {
      if (nums[i] > nums[i - 1]) {
        ++increasing;
      } else {
        prevIncreasing = increasing;
        increasing = 1;
      }
      if (increasing / 2 >= k || min(prevIncreasing, increasing) >= k)
        return true;
    }

    return false;
  }
};

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class Solution {
  public boolean hasIncreasingSubarrays(List<Integer> nums, int k) {
    int increasing = 1;
    int prevIncreasing = 0;

    for (int i = 1; i < nums.size(); ++i) {
      if (nums.get(i) > nums.get(i - 1)) {
        ++increasing;
      } else {
        prevIncreasing = increasing;
        increasing = 1;
      }
      if (increasing / 2 >= k || Math.min(prevIncreasing, increasing) >= k)
        return true;
    }

    return false;
  }
}

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class Solution:
  def hasIncreasingSubarrays(self, nums: list[int], k: int) -> bool:
    increasing = 1
    prevIncreasing = 0

    for a, b in itertools.pairwise(nums):
      if b > a:
        increasing += 1
      else:
        prevIncreasing = increasing
        increasing = 1
      if increasing // 2 >= k or min(prevIncreasing, increasing) >= k:
        return True

    return False