Python O(N) with O(1) space complexity. No sorting

  • 1

    Before, I explain anything this is my submitted solution is 96% and even though my code is not clean, I think it explains my thinking about this problem.

    First, we have two pointer that finds first non-ascending, l, and first non-descending, r in array.
    First if statement checks if l is greater than r which means the array is already sorted.
    Now, second set of while loop is for any numbers outside of violating zone that should be included.
    Such as following example that would not get caught by first pass:

    [1, 3, 7, 2, 5, 4, 6, 10] : left would catch at value 7, index: 2 and right would catch at value 4, index: 5

    But, looking at the array the function should also include 3 and 6. And this is why we do another round of while loop.
    Instead incrementing and decrementing respective, l and r, we decrement and increment respective l and r.
    Thinking about it as adjusting lower bound and upper bound is better.
    This adjustment depends on two things: Minimum and maximum values of array nums[l:r+1].
    Now, we decrement l value until we find something that is lower than minimum.
    Also, we increment r value until we find something that is greater than maximum.

    After all that you just need to do l + r - 1.

    I hope this helped everyone.

    At worst case, nums is already sorted: O(2n) = O(n)
    At worst case, nums is not sorted bound begins in the middle and two values are min and max of entire array: O(2n) = O(n)
    Space complexity: Using only 5 variables and it is constant: O(5) = O(1)

    class Solution(object):
        def findUnsortedSubarray(self, nums):
            :type nums: List[int]
            :rtype: int
            if len(nums) < 2: return 0
            l, r = 0, len(nums) - 1
            while l < len(nums) - 1 and nums[l] <= nums[l + 1]:
                l += 1
            while r > 0 and nums[r] >= nums[r -1]:
                r -= 1
            if l > r:
                return 0
            temp = nums[l:r+1]
            tempMin = min(temp)
            tempMax = max(temp)
            while l > 0 and tempMin < nums[l-1]:
                l -= 1
            while r < len(nums) - 1 and tempMax > nums[r+1]:
                r += 1
            return r - l + 1

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