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recursion solution in Clear category for Count Substring Occurrences by kdim
def count_occurrences(main_str: str, sub_str: str) -> int:
m, s = main_str.lower(), sub_str.lower()
return 0 if m.find(s) < 0 else 1 + count_occurrences(m[m.index(s) + 1:], s)
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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Jan. 5, 2024
