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reverse words #22

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58 changes: 57 additions & 1 deletion lib/reverse_words.rb
Original file line number Diff line number Diff line change
@@ -1,4 +1,60 @@
# A method to reverse each word in a sentence, in place.
require 'pry'
def string_reverse(my_string)
return if my_string == nil || my_string.length < 1
i = 0 #first ele
j = my_string.length - 1 #last ele
while i < j # until position of first ele meets last, so there is the center element in between
temp = my_string[i] #stores frist ele in temp
my_string[i] = my_string[j] #sets first ele to the last ele's position
my_string[j] = temp #sets last element to first ele's position
i += 1 #increment
j -= 1
end
return my_string
end

# " evol "

def reverse_words(my_words)
raise NotImplementedError
# puts my_words
unless my_words == nil || my_words.empty?
last = 0
first = 0
until last == my_words.length - 1
while my_words[first] == " " && first != my_words.length - 1
first += 1
end
last = first
while my_words[last] != " " && last != my_words.length - 1
last += 1
end
if first == last || last == my_words.length - 1
my_words[first..last] = string_reverse(my_words[first..last])

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I didn't see any comments on time and space complexity. As we saw in the first few slides of https://drive.google.com/file/d/0B__DV26QHsH4bWJmS1A0QXBad1U/view?usp=sharing because of using the substring method e.g. my_words[first..last], your space complexity will be O(n).

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Can you think of a solution by updating your string_reverse method to take two additional parameters first and last along with the complete my_words and then only reversing the characters in my_words from index first to index last?

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Once you've given it some thought, take a look at the solution at https://github.com/Ada-C10/reverse_words/blob/solution/lib/reverse_words.rb

else
my_words[first..last - 1] = string_reverse(my_words[first..last - 1])
end
first = last
end
end
# puts "finished #{my_words}"
end
#1, check if first character is " "
#check if the index to the right of it (the next highest index) is a space
#if true, that section is one word
# (the beginning and end of each word needs an index)
#once i find a since word section call string_reverse helper method on it
#else, continue incrementation +1

# "En gineer"
# "reenig nE"
#letter at index0 must swap with final index
#letter at index0 decrements index[0] -1
# so lets assign i = index[0] which is E
#so i = E
# index[i - 1] = -1
# j = index[length -1]
# so j = r
# the value in j should be assigned to the first position
# storage = index[0]
# j = storage