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Problem solution · Python

Trionic Array II

Trionic Array II: a Python solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Sliding window or two pointers
Source
Kamyu LeetCode Solutions
Length
31 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Trionic Array II, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 31 lines of Python from the credited upstream file trionic-array-ii.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeTrionic Array II · PythonPython
Use this to learn the idea, then write your own version.
# Time:  O(n)# Space: O(1) # two pointers, greedyclass Solution(object):    def maxSumTrionic(self, nums):        """        :type nums: List[int]        :rtype: int        """        result = float("-inf")        left = p = q = 0        prefix = nums[0]        for right in xrange(1, len(nums)):            prefix += nums[right]            if nums[right-1] > nums[right]:                if right-2 >= 0 and nums[right-2] < nums[right-1]:                    p = right-1                    while left < q or (nums[left] < 0 and left+1 < p):                        prefix -= nums[left]                        left += 1            elif nums[right-1] < nums[right]:                if right-2 >= 0 and nums[right-2] > nums[right-1]:                    q = right-1                if left != p:                    result = max(result, prefix)            else:                left = p = q = right                prefix = nums[right]        return result

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