Lists

Challenge Given a integer converts it to it’s Roman notation. E.g 14 becomes XIV Solution First I was thinking about adding all conversion in a hash map but it not that will be ask in a real interview scenario. class Solution: def find_representation(self, num, one, five, ten): # if less than 3 rep = "" if 0: return "" if num <= 3: for i in range(num): rep += one # four is five "minus" one if num == 4: rep += one + five # five use five if num == 5: return five # between five and nine use five + one(s) if num > 5 and num < 9: rep = five for i in range(num - 5): rep += one if num == 9: rep += one + ten if num == 10: rep += ten return rep def intToRoman(self, num: int) -> str: thousand = self.find_representation((num % 10000) // 1000, "M","","") hundred = self.find_representation((num % 1000) // 100, "C","D","M") ten = self.find_representation((num % 100) // 10 , "X", "L", "C") ones = self.find_representation(num % 10, "I", "V", "X") return thousand + hundred + ten + ones

Challenge Given two lists of intervals, return the intersection of these two interval lists. Example: list1 = [[0, 2], [5,6], [24, 25]] list2 = [[1,5],[8,13], [25, 26]] OUTPUT = [[1,2],[5,5],[25,25]] Solution #1 - Memory Limit Exceeded class Solution: def interval_intersection(self, firstList: List[List[int]], secondList: List[List[int]]) -> List[List[int]]: if not firstList or not secondList: return [] max_val = max(firstList[-1][1],secondList[-1][1]) first = [0 for i in range(0, max_val+1)] second = [0 for i in range(0, max_val+1)] for f in firstList: for i in range(f[0],f[1]): first[i] = 1 for s in secondList: for i in range(s[0],s[1]): second[i] = 1 i = 0 output = [] while i < max_val + 1: if first[i] == second[i] == 1: start_index = i while first[i] == second[i] == 1: i += 1 else: end_index = i output.append([start_index, end_index]) i += 1 i = 0 while i < max_val + 1: if first[i-1] == 0 and first[i] == 1 and second[i-1] == 1 and second[i] == 0: output.append([i,i]) if second[i-1] == 0 and second[i] == 1 and first[i-1] == 1 and first[i] == 0: output.append([i,i]) i += 1 return sorted(output) Memory inefficient because the biggest value of the two lists is used to crate first and second. ...