grokking-algorithms/chapter11/substring.py

58 lines
1.3 KiB
Python

import logging
from tabulate import tabulate
logging.basicConfig(
level=logging.DEBUG,
format="%(asctime)s %(levelname)s\n\r%(message)s",
datefmt="%H:%M:%S",
)
logger = logging.getLogger(__name__)
def longest_common_substring(X, Y, len_x, len_y):
# create table and zero fill it
table = [[0 for _ in range(len_x + 1)] for _ in range(len_y + 1)]
longest = 0
for i in range(1, len_y + 1):
for j in range(1, len_x + 1):
if X[j - 1] == Y[i - 1]:
table[i][j] = table[i - 1][j - 1] + 1
longest = max(longest, table[i][j])
else:
table[i][j] = 0
format_and_print(X, Y, table)
return longest
def format_and_print(X, Y, table):
# enumerate first row (headings) + label each row in column0
for i, row in enumerate(table):
if i == 0:
continue
row[0] = Y[i - 1]
table[0] = [None, *X]
# print tabularised 2D array
logger.info(tabulate(table, tablefmt="pretty"))
X = "fish"
Y = "vish"
logger.info(f"words: {X} {Y}")
longest_substring = longest_common_substring(X, Y, len(X), len(Y))
print(f"Longest common substring: {longest_substring}")
X = "vista"
Y = "hish"
logger.info(f"words: {X} {Y}")
longest_substring = longest_common_substring(X, Y, len(X), len(Y))
print(f"Longest common substring: {longest_substring}")