Single HashMap is enough.

Space complexity is wrong -> O(1)

Simpler solution.

Mismatch between constraint and test cases

The classic and magic solution!

if the character going out was in hashmap, put it back in the hashmap??

wrong space complexity analysis

Hashmap additions???

Alternate C++ solution -- 2 maps

"if window_end >= len(pattern) - 1" could be replaced by "if window_end == len(p...

Can someone help me with the time/space complexity of this solution?

Actually, no windows shrinking is required to solve the task. We jus go forwar...

Hey, I don't understand how "if window_end >= len(pattern) - 1" check works for ...

why do we need to compare the len(matches) to the len(freq) why not something el...

Isn't the time complexity just O(n)? since it will always be the case that n >= ...

When I enter this solution on leetcode it is giving me an error. https://leetcod...

I personally like going without the matched variable and instead checking if the...

I've done it like this: using 2 hashMaps, one of them to "reset" the original ha...

"If at any time, the number of characters matched is equal to the number of dist...

How about bit masking?function isPermuationInString(str, pattern) { const bit...

Shouldn't the condition for shrinking the window be windowLength > patternLength...

Go based solutionfunc findPermutation(input, pattern string) bool { winStart := ...

Should we be able to answer the problem challenges on our own after going throug...