<jupyter_start><jupyter_text>For the given list of strings, return common letters for the strings that starts with letter 'A'- Implement this in Filter + Reduce way -> The reason: imagine, the fruit list is very large<jupyter_code>## for example:

fruit = ["Apple", "Banana", "Pear", "Apricot", "Orange"]

# common letters for the strings that starts with letter 'A' are: 'A', 'p'

## Hint:
set("Apple")
set("Apricot")
set("Apple").intersection(set("Apricot"))<jupyter_output><empty_output><jupyter_text>Obtain the largest element of a list (without using max)- Implement it in reduce way<jupyter_code>reduce(lambda x, y: x if x > y else y, [1, 5, 2, 10, 13, 2])
reduce(lambda x, y: max(x, y), [1, 5, 2, 10, 13, 2])
# the above operations is different in terms of computation with the following if ls (input list argument) is too large
max([1, 5, 2, 10, 13, 2])<jupyter_output><empty_output>
{'fraction_non_alphanumeric': 0.13580246913580246, 'fraction_numerical': 0.026936026936026935, 'mean_word_length': 5.5588235294117645, 'pattern_counts': {'":': 0, '<': 9, '<?xml version=': 0, '>': 11, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 0, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
{'chain_length': '2', 'dir': 'jupyter-structured-clean-dedup', 'id': '292926', 'license': 'MIT', 'n_tokens_mistral': 326, 'n_tokens_neox': 299, 'n_words': 122, 'path': 'Activities/in_class_activity.ipynb', 'repo_name': 'UPstartDeveloper/DS-2.3-Data-Science-in-Production', 'size': '891'}