Content-Length: 7176 | pFad | http://github.com/codebasics/data-structures-algorithms-python/pull/90.diff

thub.com diff --git a/1.py b/1.py new file mode 100644 index 0000000..860e313 --- /dev/null +++ b/1.py @@ -0,0 +1,105 @@ +# ### Binary Search Exercise +# 1. When I try to find number 5 in below list using binary search, it doesn't work and returns me -1 index. Why is that? + +# ```numbers = [1,4,6,9,10,5,7]``` + +# This is because the array is not sorted in order from lowest to highest. +# Once it splits the first time, it starts looking in the [1,4,6] range and doesn't find 5 + +# 1. Find index of all the occurances of a number from sorted list + +# ``` +# numbers = [1,4,6,9,11,15,15,15,17,21,34,34,56] +# number_to_find = 15 +# ``` +# This should return 5,6,7 as indices containing number 15 in the array + +from util import time_it + +@time_it +def linear_search(numbers_list, number_to_find): + for index, element in enumerate(numbers_list): + if element == number_to_find: + return index + return -1 + +@time_it +def binary_search(numbers_list, number_to_find): + left_index = 0 + right_index = len(numbers_list) - 1 + mid_index = 0 + + while left_index <= right_index: + mid_index = (left_index + right_index) // 2 + mid_number = numbers_list[mid_index] + + if mid_number == number_to_find: + return mid_index + + if mid_number < number_to_find: + left_index = mid_index + 1 + else: + right_index = mid_index - 1 + + return -1 + +def binary_search_recursive(numbers_list, number_to_find, left_index, right_index): + if right_index < left_index: + return -1 + + mid_index = (left_index + right_index) // 2 + if mid_index >= len(numbers_list) or mid_index < 0: + return -1 + + mid_number = numbers_list[mid_index] + + if mid_number == number_to_find: + return mid_index + + if mid_number < number_to_find: + left_index = mid_index + 1 + else: + right_index = mid_index - 1 + + return binary_search_recursive(numbers_list, number_to_find, left_index, right_index) + +#this should run the binary search, find the index, and then recursively run the search on both the right and left side +def binary_search_multiple(numbers_list, number_to_find): + + index = binary_search(numbers_list,number_to_find) + result_indices = [index] + + # find all indices on the left + i = index - 1 + while i>=0: + if numbers_list[i] == numbers_list[index]: + result_indices.append(i) + else: + break + i = i-1 + + # find all indices on the right + i = index + 1 + while i elements[j+1]: + tmp = elements[j] + elements[j] = elements[j+1] + elements[j+1] = tmp + swapped = True + + if not swapped: + break + +def bubble_sort_by_key(elements, key): + size = len(elements) + + for i in range(size-1): + swapped = False + for j in range(size-1-i): + if elements[j][key] > elements[j+1][key]: + tmp = elements[j] + elements[j] = elements[j+1] + elements[j+1] = tmp + swapped = True + + if not swapped: + break + + +elements = [5,9,2,1,67,34,88,34] +elements = [1,2,3,4,2] +elements = ["mona", "dhaval", "aamir", "tina", "chang"] + +bubble_sort(elements) +print(elements) + +elements2 = [ { 'name': 'kathy', 'transaction_amount': 200, 'device': 'vivo'}, + { 'name': 'dhaval', 'transaction_amount': 400, 'device': 'google pixel'}, + { 'name': 'aamir', 'transaction_amount': 800, 'device': 'iphone-8'}, + { 'name': 'mona', 'transaction_amount': 1000, 'device': 'iphone-10'}, + ] +bubble_sort_by_key(elements2,key='transaction_amount') +print(elements2) diff --git a/util.py b/util.py new file mode 100644 index 0000000..2a7d2d5 --- /dev/null +++ b/util.py @@ -0,0 +1,9 @@ +import time +def time_it(func): + def wrapper(*args, **kwargs): + start = time.time() + result = func(*args,**kwargs) + end = time.time() + print(func.__name__ +" took " + str((end-start)*1000) + " mil sec") + return result + return wrapper \ No newline at end of file

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