India - This one is for you! Special Array, Leetcode 3151

Рет қаралды 53,750

Greg Hogg

Ай бұрын

FAANG Coding Interviews / Data Structures and Algorithms / Leetcode

Пікірлер: 87

@GregHogg Ай бұрын

If you're from India, please watch til the end - you will probably be surprised haha

@tejakovvuri3042 Ай бұрын

Surprised for real 😂

@user-qx4sk3ud6n Ай бұрын

Haha surprised😂

@mangoman-vx4pz Ай бұрын

The last part❤

@lakshyamathur8808 Ай бұрын

Surprised indeed😂😂

@harshnj 27 күн бұрын

that was really great!

@prathamyadav2422 Ай бұрын

In last ,You sang it really well 😂👍 Love from India ❤🇮🇳

@user-qx4sk3ud6n Ай бұрын

for i in range(0,♾️): print("Thank you. Love from India")

@yashwanthkumar2513 Ай бұрын

TLE

@shreehari2589 Ай бұрын

Memory be like : is it a free real estate for you?

@batteraquette5843 Ай бұрын

just do "while True:"

@pavliv 28 күн бұрын

You crashed by browser

@ARkhan-xw8ud 10 күн бұрын

it will throw error

@ngneerin Ай бұрын

That's not how Indians solve it. Here is how we do: def isArraySpecial(nums): for i in range(1, len(nums)) : if nums[i - 1]&1 != nums[i]&1: return False return True

@KonstantinUb Ай бұрын

Using len() inside of range() is a code smell; you should use zip() instead :)

@not_vinkami Ай бұрын

@@KonstantinUb I don't see how zip can make this code simpler tho

@KonstantinUb Ай бұрын

@@not_vinkami return all( x & 1 != y & 1 for x, y in zip(nums, nums[1:]) )

@_greysama_ Ай бұрын

Welldone. But as a computer science major, we use if ((nums[i] ^ nums[i-1]) & 1): return False return True Ps: Doing this just for fun, there is much better way to do it by reducing calculations..like just taking the parity of first element and comparing it with others..that way is faster

@KonstantinUb Ай бұрын

@@_greysama_ As a computer science major, you should be aware that `x & 1 != y & 1` and `(x ^ y) & 1` are completely equivalent as boolean expressions. In fact, in a compiled language like C/C++, they produce the exact same machine instructions (check with Godbolt). In Python, they generate equivalent bytecode (check with `import dis`). So there is no difference, other than the first being a tad more readable / communicating the intent a bit more clearly, in my opinion. Again, using indices like that *in Python specifically* is considered non-Pythonic and a code smell. A pairwise iterator is recommended instead.

@davidgillies620 Ай бұрын

You can take the bitwise XOR of adjacent elements and look at the LSB. def pairwiseparity(nums: list) -> bool: pparity = True for i in range(1, len(nums)): pparity = pparity and bool((nums[i] ^ nums[i - 1]) % 2) return pparity

@DavideCanton 25 күн бұрын

True , but you can avoid traversing the entire list if pparity becomes false

@JohnDoe-sp3dc Ай бұрын

Why would you use string comparison for the parity check instead of a Boolean?

@joaooliveira9852 Ай бұрын

I suppose it was just to be more intuitive for the viewer

@michaelvankley862 Ай бұрын

Because why not? It’s readable and won’t make a performance difference. To be fair, I probably would have used a function called is_even() and returned a Boolean, but there’s a lot of ways to express the same reasoning.

@xingyuxiang1637 Ай бұрын

One can also use pairwise from Python for more general cases.

@tommasochiti4237 Ай бұрын

@@michaelvankley862Won’t make a performance difference? A string, on a modern PC is at least 8bytes (the pointer to the heap allocated region) and usually you also have a capacity and a size (which are two integers). A boolean is as simple as a bit.

@Kahoneki Ай бұрын

@@tommasochiti4237 i agree wholeheartedly with what you're saying. Minor correction though - a boolean is still stored as a byte, not a bit. This is because a byte is the smallest addressable size in memory :]

@nagasaijairam2343 Ай бұрын

Love from India 🎉

@user-lz9ld5tj6i 4 күн бұрын

it’s O(2n), instead you should store compiled parity in a memo for O(n)

@tranhung5493 Ай бұрын

I think we can do this way: if (nums[i-1] + nums[i])%2==0 then False

@victormoisa6935 Ай бұрын

You might run into some integer overflows with this one, I think it would still work just keep an eye on that. Also not a problem on Python

@asagiai4965 Ай бұрын

That actually could work, just put some extra conditions.

@edudictivecoder4644 Ай бұрын

Chak de India 😍😍😍😍 I follow your shorts to improve my last submitted codes on leet code if you post those problem shorts otherwise will take your shrot as motivation for my next problem on leetcode

@jbparkes190 28 күн бұрын

You could also check whether the sum arr[i] + arr[i+1] is always odd. This is probably computationally cheaper.

@arsheyajain7055 Ай бұрын

Very good 😊

@lakshyamathur8808 Ай бұрын

Very well done Greg 🎉🎉

@hlubradio2318 25 күн бұрын

Damn Greg! You've done it again. I gotta find some time.

@kartekjadhav8481 Ай бұрын

Love from India ❤

@abhi.4164 Ай бұрын

Thanks buddy

@AbhishekRai-qp7ww Ай бұрын

Loved it Greg❤

@crissdell Ай бұрын

Love from Indiaaa

@amarjeetusnale4699 Ай бұрын

love from INDIA

@indyplaygames3066 Ай бұрын

for in range(1, len(nums)): if nums[i] % 2 == nums[i - 1] % 2: return False return True

@augustinradjou3909 Ай бұрын

After 2 months of break I am again starting with leetcode..thanks man...this video Kickstart me

@parthrathod2607 Ай бұрын

This was an array now think in terms of multi dimensional Matrix 😅

@youthpost_34 Ай бұрын

Got the logic of algo ❤

@denysdanov88 Ай бұрын

Parity function is overkill and makes it slower, just use if nums[i] % 2 == nums[i - 1] % 2 And avoid string comparison at any cost especially in this question

@nexushare8105 Ай бұрын

why?

@KonstantinUb Ай бұрын

The performance impact is negligible, and the function makes the code a bit more readable. But string comparisons are definitely unnecessary, agreed on that one.

@pulkitjain8135 28 күн бұрын

Didn't expect last part, chak de india

@user-uj2mt6nv2g Ай бұрын

You can speed this up to be twice as fast on average with a simple check. If p(num[0]) = p(num[last odd index]) return false. For a random data there is a 50% chance that the last element is of the wrong parity and since you can do this before the cycle you dont add any complexity.

@not_vinkami Ай бұрын

We can simplify the condition to just num[i] ^ num[i+1] & 1 == 0 because the xor operation can be interpreted as returning 1 if the inputs are different. I'm not sure if this is faster tho

@kotXbit Ай бұрын

Why not use a bitwise op for the parity check

@danieledispirito2692 Ай бұрын

the best (elegant) solution is the bitwise one: if (first & 1) ^ (second & 1) == 0 return False

@user-uj2mt6nv2g Ай бұрын

Another thing - you dont have to compare elements at all. Just use (if nums[i] % 2 == A) where A is redefined as A = A XOR 1). This will work like 10 times faster compared to authors solution

@naivedyam2675 Ай бұрын

A better solution would be adding the two numbers. Odd+Odd gives even and even+even also gives even. So we just have to check if the sum is even or not and if it is then we return false

@user-oj9iz4vb4q Ай бұрын

set(map(lambda (i,x): i%2 == x%2, enumerate(array))).size()==1

@KatariaDeepak Ай бұрын

🇮🇳🇮🇳🇮🇳

@albin_joby Ай бұрын

love from India

@blackHeretic00 Ай бұрын

haha didn't expect this from you

@lakku153 Ай бұрын

Very 👍

@TheCrazyCoder-tb3ux Ай бұрын

What coding language do you use ?

@laytonjr6601 Ай бұрын

You're calculating the parity of each number twice (except for the edges)

@Destimate Ай бұрын

W song

@shreehari2589 Ай бұрын

I believe 30% to 40% of your audience are from India 😂

@nigh_anxiety 29 күн бұрын

This solution is checking every index twice. Instead, you can set a boolean flag based on whether index 0 is odd or even. Then make sure the flag flips at each index. Or you could use bitwise XOR for every pair of indexes and just verify the XOR result is true.

@sanscipher9166 Ай бұрын

for i in range is an unpythonic code smell.

@dakshbhatnagar Ай бұрын

NPCI asked this? 😳

@f.herumusu8341 Ай бұрын

Some people consider multiple returns as bad style and each elements parity is calculated two times. If you have a picky interviewer ...

@mytube3486 Ай бұрын

😂😂

@joyxprogamers2005 29 күн бұрын

So Indian companies do ask leetcode pdoblems 😂😂😂

@cstates94 27 күн бұрын

This is taking 'readability over performance' waaaay too far. At least put a temporary bool to cut calculations in half.

@Sehyo Ай бұрын

Man is doing excessive string comparisons

@creativeusername4400 Күн бұрын

Lol

@itsmaxim01 Ай бұрын

so you do useless string comparison operations (“even” and “odd”, O(n+1) complexity) instead of comparing booleans (O(1)). sounds like something the average indian uber eats driver would do.

@technicallytechnical1 Ай бұрын

lil bro stop hating

@sachinpangal2194 26 күн бұрын

Bro give me a leetcode premium 😅

@GregHogg 26 күн бұрын

Noooo

@sachinpangal2194 26 күн бұрын

@@GregHogg bro replied legit to defend himself 🥲😭

@GregHogg 26 күн бұрын

@@sachinpangal2194 I genuinely have no idea what you're talking about

@sachinpangal2194 26 күн бұрын

@@GregHogg nothing bro u just motivate me to code 😤

@KonstantinUb Ай бұрын

Using indices to iterate pairwise through a list is not Pythonic. Try this instead: def parity(x): return x % 2 == 1 return all( parity(x) != parity(y) for x, y in itertools.pairwise(nums) ) And if you want to do it using only builtins, you can implement `pairwise` yourself: def pairwise(nums): return zip(nums, nums[1:])