Can you solve this Easy Google Interview Question? Ransom Note - Leetcode 383

Can you solve this Easy Google Interview Question? Ransom Note - Leetcode 383

[GregHogg]

Master Data Structures & Algorithms for FREE at AlgoMap.io!

[ShinAkuma]

This question was asked for the position of the janitor.

You can just use a integer frequency array. Increase the frequency of character from string 1 and decrease the frequency of characters from string 2 and check if the array has any non-zero elements left.

[llumnate]

A way faster approach would be to build frequency arrays on both and make sure each element in the note is smaller than the one in the magazine

[TheOmegaAlfa]

I like your shorts because they show how ignorant am I and cool stuff I could learn. I only do simple coding for work but like the theory pretty much.

[ZaeemMZia]

Can we just sort both of them and compare them

[BovineFreedo]

How about a bucket sort, have an array of how many a's, b's and so forth. Then you see if you have more or equal in the mag array than the note array for each letter. It is false if any letter has more in the note array. This is also an O(n) solution.

[nihalbhamrah]

Whats the difference between hashmap and dictionary i built in python?

[abdulsaboor]

I tried a different approch

def check (st1, st2):
st1= st1.lower();
st2= st2.lower();
list =[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
for char in st1:
index=ord(char)-ord('a')
list[index]+=1
for char in st2:
index=ord(char)-ord('a')
list[index]-=1

for x in list:
if x!=0:
return False
return True


st1="abc"
st2="cba"
print(check(st1, st2))


st1="aaabd"
st2="bdaa"
print(check(st1, st2))

[troys]

how do you do something like the minimum number of magazine "strips" where strips is a contiguous substring?

[guilhermewxs]

Use a frequency array with each letter being represented by its ASCII value subtracted by the ASCII value of 'a'. Do this for both the Note and the magazine and then loop through each one. That is way better than using a dictoonary, even though both solutions are technically O(n).

[rdubb]

Two frequency counter maps. Then loop the ransom map: if ransom key not in magazine map or ransom counter value > magazine counter value: return False

[obgbbyn]

Greg with another banger for noob coders like me

[saikiranmajji]

Brute force would be sort(a), sort(b) and return a==b

[arinkulkarni]

We don't need to decrement the counter.
Just store elements of ransomnote & magazine in map.
Now for every element of ransomnote, make sure that the element u r checking in ransomnote, the frequency is greater or equal to in map of magazine than in ransomnote

[paulsingh]

What’s the point.

Going to get my Elevator Certification

[MadChrisp]

Sort both. Iterate and return false if a char doesnt match. O(n)

[architmishra]

Lol the butt hurted people in the comments

[BloodyIron]

I'd rather put my efforts into running my own business (tech) than join Google and eventually join their graveyard. This isn't 2001, Google lost its shine long ago.