Counter(arr).items() returns a list of [(key1, value1), (key2, value2), ...] where the keys are the distinct numbers in arr and the values are the occurrences of that key in arr. You can pass this into the max function with a custom key to retrieve the max value. The key takes in a (key, value) instance and compares it against other elements by first comparing x[1] then x[0] where x[1] is the occurrence of that number and x[0] is the number as stated before. You need to apply [0] to the outputted (key,value) of the max function to obtain the distinct number.

I discovered this too, good to know

your post helped me discover decorators, thanks

can you please explain your solution, thanks

Wasn't familar with the formatting So looked for it in the stackoverflow and realised your wonderfull simple solution..

one of the main reason that keeps me in codewar is to see codes like this.

I tried to do something similar,

normally efficiency. But here, it's done in a pretty stupid way, which makes this solution even slower than the O(n²) solution without the set... XD

=> See the fork.

clever for using the slicing at the beginning

This is really similar to my idea. but I need to learn more to have this kind of reflexes.
good job

Clever using list.index, didn't thought of that