First, you should never claim something is faster without actually benchmarking it. Python has many hidden optimizations in built-in functions.

Second, I guess my solution would be faster in most cases because any stops at the first True, and all stops at the first False, so it would only reach the end of the sequence if it consists of all equal elements. Your algorithm will reach the end in any case. Consider the situation [True, False, ...(10000 elements more)]. My algorithm will stop after checking the first two elements; yours will check all the sequence in any case.

On the second thought, it would be faster in any case except, maybe, some very short sequences, because the first value of predicate would be either True or False, so one of the any or all functions is guaranteed to stop after checking it.

For your solution, I missed the iterator part in map. yes, the space complexity will be O(1) but then it is a slower iteration over sequence. a simple loop should be faster. Mind me I understand that they both are O(n) time complexity but simple loop should be a faster O(n) out of those two. let me know what you think.

How exactly creating new structures in memory can be more efficient? This solution has O(1) space complexity.

i think counter method is more efficient space wise. however , it has same time complexity

pretty much the same thing as mine... not bad!

Nice solution :-)

(notification)

how did it work and what does it have to do with "key = str"

What's cheating about that solution? Also you are as able as anyone else to submit this solution no?

Bloody cheats LOL i feel so annoyed that i can't cheat

nice, and here I am hardcoded the whole thing xD

It's still O(n).

It's fine! Learning is what's important :)

Thanks @Kacarott, I really didn't know. Thanks for that lesson.

@geans You are overlooking one important thing. map does NOT go through the whole list. It creates a generator, which produces values one at a time, which feeds into the other functions, which short circuit.

Here is proof:

def infinite():
    while True: yield 1
    
a = map(lambda x: x*2, infinite()) # Works instantly
b = [x*2 for x in infinite()] # Loops forever (timeout)