Implement a Lazy Tee Iterator in Python

Lazy Tee Iterator

Implement a function lazy_tee(iterable, n=2) that takes an input iterable and returns a tuple of n independent iterators. Each returned iterator must provide access to all items from the source iterable.

Your implementation must be lazy: if one consumer stops iterating, the source should not be advanced beyond what is required by the other consumers. Because iterators consume data linearly, you must implement a buffering mechanism to store items that have been read from the source but not yet consumed by all n iterators.

Requirements

• The source iterator should only be advanced when necessary.
• Memory usage should be proportional to the "lag" between the fastest and slowest consumer, not the total size of the iterable.
• Once an item has been consumed by all n iterators, it should be removed from the buffer to allow for garbage collection.

Function Signature

def lazy_tee(iterable, n=2):
    # Returns a tuple of n independent iterators
    pass

Examples

# Basic usage
it = iter([1, 2, 3])
a, b = lazy_tee(it)
print(list(a))  # [1, 2, 3]
print(list(b))  # [1, 2, 3]

# Uneven consumption
it = iter([1, 2, 3, 4])
a, b = lazy_tee(it)
next(a)
print(list(b))  # [1, 2, 3, 4]
print(next(a))  # 2

Edge Cases

• The input iterable might be empty.
• One of the returned iterators might never be used.
• One of the returned iterators might consume the entire stream before the others are even started.