my difficulty: easy — sol1) brute force This was accepted but runtime was also 4000ms(not acceptable for me). because its time complexity is O(n²) sol2) slicing by slicing the substring, we can reduce time complexity to O(n)

my difficulty: easy — sol) implementation algorithm was not complex. but creating new node was a little bit annoying. I create new node when carry has been generated or nodes are remained to perform addition

my difficulty: super easy — sol) if statements nothing to explain.

my difficulty: super easy — sol1) dictionary nothing to explain

my difficulty: super easy — sol1) python’s library nothing to explain. but its time complexity is O(n²) sol2) indexing as Follow-up 1’s asked, What if the given array is already sorted? I think the time complexity can be optimized by O(n)

my difficulty: normal — sol) indexing solving in an in-place way was not easy. Key idea was swapping all the non-zero numbers to the first zero

my difficulty: super easy — sol) math fortunately, there is only one missing number, difference between sum of 0~n and sum of nums will be answer

my difficulty: super easy — sol1) sorting anagram means each character’s number of appearance is same. so if some words are anagram, sorting their characters will be same sol2) iteration as I said above, to check whether some two words are anagram, we can count their characters. and then compare each character’s number of appearance is same or not.

my difficulty: super easy — sol) intuition nothing to explain