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recursion solution in Speedy category for Count Substring Occurrences by zshlab
def count_occurrences(main_str: str, sub_str: str) -> int:
try:
first = main_str.lower().index(sub_str.lower())
return 1 + count_occurrences(main_str[first + 1:], sub_str)
except:
return 0
print("Example:")
print(count_occurrences("hello world hello", "hello"))
# These "asserts" are used for self-checking
assert count_occurrences("hello world hello", "hello") == 2
assert count_occurrences("Hello World hello", "hello") == 2
assert count_occurrences("hello", "world") == 0
assert count_occurrences("hello world hello world hello", "world") == 2
assert count_occurrences("HELLO", "hello") == 1
assert count_occurrences("appleappleapple", "appleapple") == 2
assert count_occurrences("HELLO WORLD", "WORLD") == 1
assert count_occurrences("hello world hello", "o w") == 1
assert count_occurrences("apple apple apple", "apple") == 3
assert count_occurrences("apple Apple apple", "apple") == 3
assert count_occurrences("apple", "APPLE") == 1
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Feb. 10, 2025
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