Hard

Count the Number of Incremovable Subarrays IIPython

Full explanation · Time O(n) · Space O(1)

# Time:  O(n)
# Space: O(1)

# two pointers
class Solution(object):
    def incremovableSubarrayCount(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        for j in reversed(xrange(1, len(nums))):
            if not nums[j-1] < nums[j]:
                break
        else:
            return (len(nums)+1)*len(nums)//2
        result = len(nums)-j+1
        for i in xrange(len(nums)-1):
            while j < len(nums) and not (nums[i] < nums[j]):
                j += 1
            result += len(nums)-j+1
            if not (nums[i] < nums[i+1]):
                break
        return result