Q22: Container With Most Water. - DSA Patterns (Multi-Lang)

In Simple Words: Two pointers at extremes. Water capacity is bounded by the shorter line: min(h[l], h[r]) * (r - l). Always shift the pointer with the smaller height inward to potentially find taller vertical bars.

Time: $O(N)$ | Space: $O(1)$

Multi-Language SolutionsPython 3 • JavaScript • TypeScript • Java • C++

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def maxArea(height: list[int]) -> int:
    l, r = 0, len(height) - 1
    max_w = 0
    while l < r:
        area = min(height[l], height[r]) * (r - l)
        max_w = max(max_w, area)
        if height[l] < height[r]:
            l += 1
        else:
            r -= 1
    return max_w