Maximum Container Area

1. Step 1: Initialize Solution.maxArea

   Replace the empty starter with the first real state owned by Solution.maxArea. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.

2. Step 2: Assemble the primary transition

   Extend the initialized state with the next contiguous part of the popular solution. The transition explains how one input element or operation changes the state; boundaries are easier to reason about after this invariant is visible.

3. Step 3: Pass the Two Walls case

   Complete the readable core algorithm for one representative Interview case. Use opposite-end pointers and prove why moving the shorter wall discards only non-improving pairs.

4. Step 4: Harden the Mixed Heights boundary

   Repair the reviewed boundary and pass the complete submission contract. For the current shorter wall, every pair with a smaller width has height capped by that wall and cannot improve the area; discarding it therefore preserves every potentially better pair.

• Moving the taller wall cannot improve the limiting height while width always shrinks.
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• Discard Pairs with Two Pointers(opens in a new tab)

  Discard Pairs with Two Pointers isolates every pair outside [left, right] has already been counted or proven too large, and no unresolved pair is skipped. That focused state discipline is required when implementing container with most water as a complete Interview Problem.

Start at both ends, record the current area, and move the pointer at the shorter wall inward.

Why it works: For the current shorter wall, every pair with a smaller width has height capped by that wall and cannot improve the area; discarding it therefore preserves every potentially better pair.

class Solution:
    def maxArea(self, heights):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        left = 0
        right = len(heights) - 1
        best = 0
        while left < right:
            best = max(best, (right - left) * min(heights[left], heights[right]))
            if heights[left] <= heights[right]:
                left += 1
            else:
                right -= 1
        return best

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