Trapping Rain Water in Python

from bisect import bisect_left

# Trapping Rain Water in Python
#
# PURPOSE
# Two boundary maxima determine trapped water above each bar.
#
# METHOD
# Process the side with the smaller boundary maximum, because the opposite side already
# provides a sufficiently high boundary. Accumulate that side's maximum minus its bar
# height.
#
# COMPLEXITY
# O(n) time, O(1) auxiliary space.
#
# CONTRACT
# Heights are nonnegative. Total volume must fit signed 32-bit.
# 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: 6.
# 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 trapped_water(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
    left_max: int = 0
    right_max: int = 0
    total: int = 0
    while left <= right:
        if left_max <= right_max:
            left_max = max(left_max, a[left])
            total += left_max - a[left]
            left += 1
        else:
            right_max = max(right_max, a[right])
            total += right_max - a[right]
            right -= 1
    return total

a = [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]
result = trapped_water(a)
print(result)