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2393. Count Strictly Increasing Subarrays 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  long long countSubarrays(vector<int>& nums) {
    long long ans = 0;

    for (int i = 0, j = -1; i < nums.size(); ++i) {
      if (i > 0 && nums[i] <= nums[i - 1])
        j = i - 1;
      ans += i - j;
    }

    return ans;
  }
};

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class Solution {
  public long countSubarrays(int[] nums) {
    long ans = 0;

    for (int i = 0, j = -1; i < nums.length; ++i) {
      if (i > 0 && nums[i] <= nums[i - 1])
        j = i - 1;
      ans += i - j;
    }

    return ans;
  }
}

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class Solution:
  def countSubarrays(self, nums: list[int]) -> int:
    ans = 0

    j = -1
    for i, num in enumerate(nums):
      if i > 0 and num <= nums[i - 1]:
        j = i - 1
      ans += i - j

    return ans