• Nov 19, 2022 •CodeCatch
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""" Binary Search Algorithm ---------------------------------------- """ #iterative implementation of binary search in Python def binary_search(a_list, item): """Performs iterative binary search to find the position of an integer in a given, sorted, list. a_list -- sorted list of integers item -- integer you are searching for the position of """ first = 0 last = len(a_list) - 1 while first <= last: i = (first + last) / 2 if a_list[i] == item: return ' found at position '.format(item=item, i=i) elif a_list[i] > item: last = i - 1 elif a_list[i] < item: first = i + 1 else: return ' not found in the list'.format(item=item) #recursive implementation of binary search in Python def binary_search_recursive(a_list, item): """Performs recursive binary search of an integer in a given, sorted, list. a_list -- sorted list of integers item -- integer you are searching for the position of """ first = 0 last = len(a_list) - 1 if len(a_list) == 0: return ' was not found in the list'.format(item=item) else: i = (first + last) // 2 if item == a_list[i]: return ' found'.format(item=item) else: if a_list[i] < item: return binary_search_recursive(a_list[i+1:], item) else: return binary_search_recursive(a_list[:i], item)
• May 31, 2023 •CodeCatch
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my_list = [1, 2, 3, 4, 5] removed_element = my_list.pop(2) # Remove and return element at index 2 print(removed_element) # 3 print(my_list) # [1, 2, 4, 5] last_element = my_list.pop() # Remove and return the last element print(last_element) # 5 print(my_list) # [1, 2, 4]
• Nov 18, 2022 •AustinLeath
primes=[] products=[] def prime(num): if num > 1: for i in range(2,num): if (num % i) == 0: return False else: primes.append(num) return True for n in range(30,1000): if len(primes) >= 20: break; else: prime(n) for previous, current in zip(primes[::2], primes[1::2]): products.append(previous * current) print (products)
• Apr 21, 2023 •sebastianagauyao2002-61a8
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print("hellur")
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from collections import defaultdict def combine_values(*dicts): res = defaultdict(list) for d in dicts: for key in d: res[key].append(d[key]) return dict(res) d1 = {'a': 1, 'b': 'foo', 'c': 400} d2 = {'a': 3, 'b': 200, 'd': 400} combine_values(d1, d2) # {'a': [1, 3], 'b': ['foo', 200], 'c': [400], 'd': [400]}
• May 5, 2026 •CodeCatch
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{"output":{"name":"format_final_json_response","arguments":{"output":{}}}}