Finding longest substring in alphabetical order

Question

EDIT: I am aware that a question with similar task was already asked in SO but I\'m interested to find out the problem in this specific piece of code. I am

Answer 1

a lot more looping, but it gets the job done

s = raw_input("Enter string")
fin=""
s_pos =0
while s_pos < len(s):
    n=1
    lng=" "
    for c in s[s_pos:]:
        if c >= lng[n-1]:
            lng+=c
            n+=1
        else :
            break
    if len(lng) > len(fin):
        fin= lng`enter code here`
    s_pos+=1    
print "Longest string: " + fin

Answer 2

s = 'azcbobobegghakl' 

i=1
subs=s[0]
subs2=s[0]
while(i<len(s)):
    j=i
    while(j<len(s)):
        if(s[j]>=s[j-1]):
            subs+=s[j]
            j+=1 
        else:
            subs=subs.replace(subs[:len(subs)],s[i])   
            break

        if(len(subs)>len(subs2)):
            subs2=subs2.replace(subs2[:len(subs2)], subs[:len(subs)])
    subs=subs.replace(subs[:len(subs)],s[i])
    i+=1
print("Longest substring in alphabetical order is:",subs2)

Answer 3

s = input("insert some string: ")
start = 0
end = 0
temp = ""
while end+1 <len(s):
    while end+1 <len(s) and s[end+1] >= s[end]:
        end += 1
    if len(s[start:end+1]) > len(temp):
        temp = s[start:end+1]
    end +=1
    start = end
print("longest ordered part is: "+temp)

Answer 4

There are many things to improve in your code but making minimum changes so as to make it work. The problem is you should have if last_pos(i) != None: in your for loop (i instead of i+1) and you should compare diff (not diff - 1) against maxLen. Please read other answers to learn how to do it better.

for i in range(len(s)):
    if last_pos(i) != None:
        diff = last_pos(i) - i + 1
    if diff > maxLen:
        maxLen = diff
        startPos = i
        endPos = startPos + diff - 1

Answer 5

Simple and easy.

Code :

s = 'hixwluvyhzzzdgd' 
r,p,t = '','',''
for c in s:
    if p <= c:
        t += c
        p = c
    else:
        if len(t) > len(r):
            r = t
        t,p = c,c
if len(t) > len(r):
    r = t
print 'Longest substring in alphabetical order is: ' + r

Output :

Longest substring in alphabetical order which appeared first: luvy

Answer 6

Here is a single pass solution with a fast loop. It reads each character only once. Inside the loop operations are limited to

• 1 string comparison (1 char x 1 char)
• 1 integer increment
• 2 integer subtractions
• 1 integer comparison
• 1 to 3 integer assignments
• 1 string assignment

No containers are used. No function calls are made. The empty string is handled without special-case code. All character codes, including chr(0), are properly handled. If there is a tie for the longest alphabetical substring, the function returns the first winning substring it encountered. Case is ignored for purposes of alphabetization, but case is preserved in the output substring.

def longest_alphabetical_substring(string):
    start, end = 0, 0     # range of current alphabetical string
    START, END = 0, 0     # range of longest alphabetical string yet found
    prev = chr(0)         # previous character

    for char in string.lower():   # scan string ignoring case
        if char < prev:       # is character out of alphabetical order?
            start = end       #     if so, start a new substring 
        end += 1              # either way, increment substring length 

        if end - start > END - START:  # found new longest?  
            START, END = start, end    #     if so, update longest 
        prev = char                    # remember previous character

    return string[START : END]   # return longest alphabetical substring 

Result

>>> longest_alphabetical_substring('drurotsxjehlwfwgygygxz')
'ehlw'
>>> longest_alphabetical_substring('eseoojlsuai')
'jlsu'
>>> longest_alphabetical_substring('hixwluvyhzzzdgd')
'luvy'
>>>