# My accepted Python solution

• ``````class Solution:
# @param s, a string
# @return an integer
def lengthOfLastWord(self, s):
count = 0
for i in range(len(s)-1,-1,-1):
if s[i] != ' ':
while s[i] != ' ' and i >= 0:
count += 1
i -= 1
break
return count``````

• I went with the more lazy option. Probably executes slower, but much simpler to write.

``````class Solution:
# @param s, a string
# @return an integer
def lengthOfLastWord(self, s):
if len(s) == 0:
return 0

return len(s.strip().split(' ')[-1])``````

• Neat solution !

But I am afraid if using Python functions like strip() and split() , which would reduce the LOC significantly, will go well with the interviewer.

• Mine has just one loop, if you could have a look:

``````class Solution:
# @param s, a string
# @return an integer
def lengthOfLastWord(self, s):
count = 0
lock = False
if len(s) == 0:
return 0
for i in range(len(s)-1,-1,-1):
if s[i] != " ":
count += 1
lock = True
if (s[i] == " " and lock) or (i == 0):
return count
``````

I was using a lock to see if this the second times to see the space

and BTW, if I used the xrange instead of range, this could be better, the runtime is 38ms

• Hi, I am new to LeetCode. I am wondering is it ok to use build-in function during interviews?

• Can someone take a look at my solution?

``````class Solution:
# @param {string} s
# @return {integer}
def lengthOfLastWord(self, s):
for i in range(len(s)):
length = 0
if s[i] == ' ' and s[i+1] != ' ':
length = 1
while s[i+2] and s[i+2] != ' ':
length += 1
return length
``````

I think it should be accepted, but the system hints " Line 7: IndexError: string index out of range " .....

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