Container With Most Water in Python

from bisect import bisect_left

# Container With Most Water in Python
#
# PURPOSE
# Two pointers maximize the area between a pair of height bars.
#
# METHOD
# Evaluate width times the smaller height. Move the smaller-height endpoint inward:
# retaining it with a shorter width cannot improve the area.
#
# COMPLEXITY
# O(n) time, O(1) auxiliary space.
#
# CONTRACT
# Nonnegative heights. Areas must fit signed 32-bit. Fewer than two bars returns 0.
# Public APIs infer array lengths except in C, which accepts pointer/length pairs. Vertex
# counts and matrix dimensions remain explicit. Arrays must contain signed 32-bit integers.
# Sorted order, correct tree topology, and caller-owned data-structure invariants are
# preconditions; checking them on every query would change query complexity.
# Scalar and shape checks are included. Value-range checks are included where needed to
# prevent invalid indexing or nontermination. Allocation sizes are capped at 4,000,000
# workspace entries. Arithmetic outside Python raises an error on 32-bit overflow; Python
# uses arbitrary precision. Native allocation failures propagate. In-place outputs may be
# partly updated after arithmetic failure. These are single-threaded, in-memory
# implementations.
#
# LANGUAGE DETAILS
# Python 3.9+: automatic lengths, native lists, arbitrary-precision arithmetic, and
# bisect_left for LIS. Invalid scalar/shape inputs raise ValueError. Integer element types
# and sorted/tree invariants remain caller preconditions.
#
# EXAMPLE
# Expected return: 49.
# Run: python example.py
# Tests for this exact source artifact run before publication. The accompanying source
# archive includes unit cases, runtime results, and source hashes.

def alg_require(condition):
    if not condition:
        raise ValueError("invalid algorithm input; see contract")
def alg_size(n):
    alg_require(0 <= n <= 4000000)
    return n

def max_container(a: list[int]) -> int:
    alg_require(isinstance(a, (list, tuple)) and len(a) <= 1000000)
    n: int = len(a)
    alg_require(n >= 0 and n <= 1000000)
    alg_require(len(a) == n)
    for value in a:
        alg_require(value >= 0)
    left: int = 0
    right: int = n - 1
    best: int = 0
    while left < right:
        best = max(best, (right - left) * min(a[left], a[right]))
        if a[left] <= a[right]:
            left += 1
        else:
            right -= 1
    return best

a = [1, 8, 6, 2, 5, 4, 8, 3, 7]
result = max_container(a)
print(result)