Junior Software Dev vs Senior Dev solving Valid Anagram

FAANG Coding Interviews / Data Structures and Algorithms / Leetcode

[GregHogg]

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[alxjones]

This is less rookie vs. senior and more unprepared rookie vs. prepared rookie.

[Taurdil]

problems with strings such as this one that solved by hash maps could usually be optimised further by counting in array instead of hash map (lower constant in O(1)). Hint is "ascii only characters". This is how you get to top 5% in both speed and memory

So iterate s, incrementing stats[s_char], then iterate t, decrementing. Return true of stats has all zeroes

[Basta]

In practical code, sorting an array can be faster and more memory efficient than creating objects and hashing if we don't expect huge n's. In many cases, we don't.

[Leafycoke]

The point of leetcode is to try the brute force way first then learn the optimised method and apply that to new problems.

[hlubradio]

Oh anagrams are words using the same letters?

[astral]

I could easily solve this in C, but not in Python.

In C, I'd just make an int array of size 26, iterate through the string while doing +/- 1 on the corresponding array elements, and check if the array has a non-zero.

Easiest O(n) ever.

[nk]

I'd rather go forward through one and backward through another instead of using the hashmap, more clear, more fun, less space, and makes more sense to anyone who sees it

[MrBranh]

Senior devs are actually solving real problems, and are usually annoyed when they have to solve problems like this

[lOmaine]

This would require a space of o(m+n) which can be reduced to o(n) by only storing count of s in hashmap and decrementing count while traversing t. If -1 os even reached, return false.

[ZeruelB]

from experience, this comes up like never and most stuff thats breaking or causing issues is the architecture and/or used frameworks and/or libraries

[ShinAkuma]

Still a bad solution.
take an integer array of size 26.
increment (s1.charAt(i) - 'a') in a loop.
Then decrement (s2.charAt(i) - 'a') in a loop.
Check if the array has any non-zero value.

The complexity is O(s1Lenth + s2Length + 26) = O(n)

[HhddGufif]

I wonder if you used bit packing or flags on the most common letters to build a checksum value. Then compare on both to break early for 99.99% of cases. The. For the remaining do any slower verification, doesnt matter for performance

[bart]

"Sort is n log n!"
Dude, you're sorting about 10 items, so the O of the algorithm is irrelevant.

[hmkl]

Lol, coding is a really cool thing for me. I love maths, but not only to have to find a solution, but also optimize it is really challenging

[Nellak]

😂

const solution = (s, t) => Array.from(t).every(el => Set(s).has(el))

stupid solution but correct

[numbdigger]

Lmao, if you wanted to over-optimize your code like that:
1. You wouldn't use hash tables. A single array indexed from 0 to 25 (or from 0 to 255 if we allow all ASCII characters) is faster.
2. You wouldn't use Python, lol.

[rishujeetrai]

O(n) with space complexity of O(n) as well
Or
O(nlogn) with space complexity of O(n)

Make sure to tell both approaches the interviewer

PS- Since python doesn't let you sort string in-place both approaches are O(n) space for python developers

[thegoodhen]

Me: Eh, convert both to lists, go over the first list and if you don't find the letter from the first list in the second, they aren't anagrams; if you find it, remove it. It's O(n^2), but hey, that's still gonna be fast for most reasonable inputs... if not, you can always use 2 arrays for the numbers of occurances of each letter and compare them, I guess...

[ungrim]

The expert dev is the one that realises that your prod code probably wouldn't notice the difference between the two solutions but broke because someone asked if 🤢💛🍇🍊🍌🍍🍎 is an anagram of 🤢🍎💛🍇🍊🍌🍍