Skip to content

3350. Adjacent Increasing Subarrays Detection II 👍

  • Time: $O(n)$
  • Space: $O(1)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution {
 public:
  // Similar to 3349. Adjacent Increasing Subarrays Detection I
  int maxIncreasingSubarrays(vector<int>& nums) {
    int ans = 0;
    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;
      }
      ans = max(ans, increasing / 2);
      ans = max(ans, min(prevIncreasing, increasing));
    }

    return ans;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
  // Similar to 3349. Adjacent Increasing Subarrays Detection I
  public int maxIncreasingSubarrays(List<Integer> nums) {
    int ans = 0;
    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;
      }
      ans = Math.max(ans, increasing / 2);
      ans = Math.max(ans, Math.min(prevIncreasing, increasing));
    }

    return ans;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution:
  # Similar to 3349. Adjacent Increasing Subarrays Detection I
  def maxIncreasingSubarrays(self, nums: list[int]) -> int:
    ans = 0
    increasing = 1
    prevIncreasing = 0

    for a, b in itertools.pairwise(nums):
      if b > a:
        increasing += 1
      else:
        prevIncreasing = increasing
        increasing = 1
      ans = max(ans, increasing // 2)
      ans = max(ans, min(prevIncreasing, increasing))

    return ans