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class Solution: # time O(N+2K^2) space(N) def toGoatLatin(self, S: str) -> str: if not S: return [] sen = S.split(' ') # O(N) vo = {'a','e','i','o','u'} res = list() for idx, word in enumerate(sen): if word[0].lower() not in vo: # edge case: Capical vowel word = word[1:] + word[0] #... [Read More]

ituition solution class Solution: # time O(lgN) space O(N) # main idea: Hash Table + Sort def frequencySort(self, s: str) -> str: d = dict() for i in range(len(s)): if s[i] not in d: d[s[i]] = 1 else: d[s[i]] += 1 tups = sorted(d.items(), key=lambda k : k[1], reverse= True... [Read More]

intuition solution class Solution: # time (N/M * M) space O(1) if not consider output space def distributeCandies(self, candies: int, num_people: int) -> List[int]: res = [0] * num_people ptr = 0 round = 0 gives = 0 cnt = 0#round * num_people + index # termination: candies - cnt... [Read More]

class Solution: def merge(self, intervals: List[List[int]]) -> List[List[int]]: """ tc O(NlgN) sc O(1) main idea: find overlapping pair ==> previous end is bigger than current start time 1. sort by start 2. for each time pair, check if overlap with stack top 3. if overlap, update stack top, otherwise push... [Read More]

class Solution: def eraseOverlapIntervals(self, A: List[List[int]]) -> int: """ tc O(NlgN) sc O(1) greedy,get earliest end time as possible to spare more space to get more potential non-overlaps 1. sort by end 2. loop intervals, check if current start >= prev end time. if yes, update previous valid end time... [Read More]