rfind(), rindex()를 이용하여 위 문제를 풀었다.
a = ['a', 'b', 'c', 'd', 'e']
print(a.index('c'))
# 결과: 2
b = 'hello world'
print(b.find('o'))
# 결과: 4
a = 'banana'
print(a.rindex('a'))
# 결과: 5
b = 'hello world'
print(b.rfind('o'))
# 결과: 7
def solution(s):
already = []
result = []
for i in range(len(s)):
if s[i] not in already:
already.append(s[i])
result.append(-1)
else:
a_index = ''.join(already).rindex(s[i])
already.append(s[i])
result.append(i - a_index)
return result