class Solution(object): def merge(self, intervals): intervals.sort(key=lambda x: x[0]) merged = [] for interval in intervals: if not merged or merged[-1][1] < interval[0]: merged.append(interval) else: merged[-1][1] = max(merged[-1][1], interval[1]) return merged give a detailed 2-3 sentence analysis of the time complexity and space complexity of this algorithm in big O notation

class Solution(object): def merge(self, intervals): intervals.sort(key=lambda x: x[0]) merged = [] for interval in intervals: if not merged or merged[-1][1] < interval[0]: merged.append(interval) else: merged[-1][1] = max(merged[-1][1], interval[1]) return merged give a detailed 2-3 sentence analysis of the time complexity and space complexity of this algorithm in big O notation

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Racket: Execute these lists: ( (lambda (x) (* x x)) 1) ( (lambda (x) (* x x)) 2) ( (lambda (x) (+ x x)) 3) ( (lambda (x) (- x x)) 4) ( (lambda (x) (/ x x)) 5) ( (lambda (x) (modulo x x)) 6) ( (lambda (x) (modulo (+ x 2) x)) 6) Explain clearly what these expressions seem to mean
Write the program that allows the user to sort using the Bubble Sort, Selection Sort, Insertion Sort and Shell Short The program should be able to read in data from a binary file. The first element of the binary file will be used to tell how many elements to read in. Once all the data has been read in, the program should sort the data. The user should be able to choose which algorithm to use to sort the data. The program should print the time before and after the sort - be sure to not print the start time until after the algorithm has been chosen from your menue. The last part of the program should print out the value and location of three random positions in your array The name of each algorithm:Insertion Sort A description of the elapsed time found for each input file: 10numbers; 12 seconds 100number: 30 seconds ... A screenshot of the output of your program showing the start time and stop time of each algorithm running on the largest file (1000000numbers) as well as the…
What is the value of the median-of-3 for the following list. [80, 47, 41, 21, 40, 68, 78, 18, 75, 46, 30] Notes: • Your answer should be a single valid, non-negative, literal Python int value. For example, 123 is a valid int literal. • This question is asking about before any partition operation is carried out. • This is not asking for the median-of-3 index. • You can pre-check your answer (to check it's a valid, non-negative, literal int).
def large_matrix(matrix: list[list[int]]) -> int:"""Returns the area of the rectangle in the area of a rectangle is defined by number of 1's that it contains.The matrix will only contain the integers 1 and 0.>>> case = [[1, 0, 1, 0, 0],... [1, 0, 1, 1, 1],... [1, 1, 1, 1, 1],... [1, 0, 0, 1, 0]]>>> largest_in_matrix(case1)6"" You must use this helper code: def large_position(matrix: list[list[int]], row: int, col: int) -> int: a = rowb = colmax = 0temp = 0rows = len(matrix)column = len(matrix[a])while a < rows and matrix[a][col] == 1:temp = 0while b < column and matrix[a][b] == 1:temp = temp + 1b = b + 1column = ba = a + 1if (a != row+1):temp2 = temp * (a - row)else:temp2 = tempif max < temp2:max = temp2b = colreturn max """ remember: Please do this on python you should not use any of the following: dictionaries or dictionary methods try-except break and continue statements recursion map / filter import""'
1. Write an algorithm to determine whether a given element x belongs to a set S := {s1, . . . , sn}.
Solve C, please. Solve C, please. Solve C, please.
joinList(n) (2 pts) Takes an integer number n. This functions returns the list [1,2,...,n,n,...2,1] for every n>0. Preconditions: While the function will receive integer values, you should not make any assumptions about the value of n. Input An integer value Output list Starting at 1 moving upwards, a list with the sequence 1 to n, followed the sequence that starts at n moving downwards to 1 None Argument do not satisfy the conditions of the function Example: >>> joinList(5) [1, 2, 3, 4, 5, 5, 4, 3, 2, 1] >>> joinList(1) [1, 1] >>> joinlist(-3) is None True
251e def intersection (self, source): 252e 253 254 Efficiency: 255 Returns a new set with only the values that appear in both self and source. 256 257 258 Maximum size of target Set is the smaller of the maximum sizes of 259 self and source. 260 Use: target = self.intersection(source) 261 262 Parameters: 263 source - a set (Set) 264 Returns: 265 target - a set containing one copy of all the values that appear in both self and source (Set) 266 267 268 Examples: (1,2,3) intersection (3,2,1) is (1,2,3) (1,2,3) intersection (4,5,6) is () (1,2,3,4,5,6) intersection (0,2,4,6,8) is (2,4,6) 269 270 271 272 273 274