Given a string of words, reverse all the words. For example:
Given:
‘This is the best’
Return:
‘best the is This’
As part of this exercise you should remove all leading and trailing whitespace. So that inputs such as:
‘ space here’ and ‘space here ‘
both become:
We could take advantage of Python’s abilities and solve the problem with the use of split() and some slicing or use of reversed:
def rev_word1(s): return " ".join(reversed(s.split())) #Or def rev_word2(s): return " ".join(s.split()[::-1])
rev_word1('Hi John, are you ready to go?')
‘go? to ready you are John, Hi’
rev_word2('Hi John, are you ready to go?')
‘go? to ready you are John, Hi’
While these are valid solutions, in an interview setting you’ll have to work out the basic algorithm that is used. In this case what we want to do is loop over the text and extract words form the string ourselves. Then we can push the words to a “stack” and in the end opo them all to reverse. Let’s see what this actually looks like:
def rev_word3(s): """ Manually doing the splits on the spaces. """ words = [] length = len(s) spaces = [' '] # Index Tracker i = 0 # While index is less than length of string while i < length: # If element isn't a space if s[i] not in spaces: # The word starts at this index word_start = i while i < length and s[i] not in spaces: # Get index where word ends i += 1 # Append that word to the list words.append(s[word_start:i]) # Add to index i += 1 # Join the reversed words return " ".join(reversed(words))
rev_word3(' Hello John how are you ')
‘you are how John Hello’
rev_word3(' space before')
‘before space’
""" RUN THIS CELL TO TEST YOUR SOLUTION """ from nose.tools import assert_equal class ReversalTest(object): def test(self,sol): assert_equal(sol(' space before'),'before space') assert_equal(sol('space after '),'after space') assert_equal(sol(' Hello John how are you '),'you are how John Hello') assert_equal(sol('1'),'1') print("ALL TEST CASES PASSED") # Run and test t = ReversalTest() t.test(rev_word)
ALL TEST CASES PASSED
Recommended: Understand Big-O Notation Complexity Of Algorithm
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