TOP K FREQUENT WORDS| LEETCODE 692 | PYTHON CUSTOM HEAP SOLUTION

[@2sourcerer]

You can use a tuple instead of HeapItem. Tuple is set as (-count, word). Tuple comparison is resolved on the first element. If first element is equal, then it'll starts to compare the second element, and it happens that it's a mean heap so it's already in lexicographical order.

[@seant11]

Can you explain why in your __lt__ the comparison is self.word > to_compare.word? My assumption is that if you're trying to compare if the current word is lexicographically smaller than the "to_compare" word then it would be self.word < to_compare.word?

[@keithang9335]

think about it this way: for 2 strings a and b, a>b if a comes after b in the dictionary. e.g “leetcode” > “idiom” # True you want the 0th element of your min heap to be the one with smallest count and shows up last in the dictionary, so you want (self.count==other.count and self.word > other.word) or (self.count < other.count)

[@pranjalgupta6995]

I think its because we are reversing the result at the end.

[@satwiktatikonda764]

you channel is very underrated

[@crackfaang]

I agree haha

[@prodbyCheez]

Is the to_compare the item that is already on the heap? Also is __lt__ called when you when you first call HeapItem(word, count)?

[@kamila5207]

so beautiful solution!! thank you for explaining

[@bigboyshassen59]

`return list(reversed(res))`