Python: Find the starting and ending position of a given value in a given array of integers, sorted in ascending order
Python Basic - 1: Exercise-76 with Solution
Write a Python program to find the starting and ending position of a given value in a given array of integers, sorted in ascending order.
If the target is not found in the array, return [0, 0].
Input: [5, 7, 7, 8, 8, 8] target value = 8
Output: [0, 5]
Input: [1, 3, 6, 9, 13, 14] target value = 4
Output: [0, 0]
def search_Range(array_nums, target_val): result_arra =  start_pos = 0 end_pos = 0 for i in range(len(array_nums)): if target_val == array_nums[i] and start_pos == -1: start_pos = i end_pos = i elif target_val == array_nums[i] and start_pos != -1: end_pos = i result_arra.append(start_pos) result_arra.append(end_pos) return result_arra print(search_Range([5, 7, 7, 8, 8, 8], 8)) print(search_Range([1, 3, 6, 9, 13, 14], 4)) print(search_Range([5, 7, 7, 8, 10], 8))
[0, 5] [0, 0] [0, 3]
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Python: Tips of the Day
How to make a flat list out of list of lists?
Given a list of lists l
flat_list = [item for sublist in l for item in sublist]
flat_list =  for sublist in l: for item in sublist: flat_list.append(item)
is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:
flatten = lambda l: [item for sublist in l for item in sublist]
As evidence, you can use the timeit module in the standard library:
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' '[item for sublist in l for item in sublist]' 10000 loops, best of 3: 143 usec per loop $ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' 'sum(l, )' 1000 loops, best of 3: 969 usec per loop $ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' 'reduce(lambda x,y: x+y,l)' 1000 loops, best of 3: 1.1 msec per loop
Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.
The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.
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