Python code

s = 'woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz'
count = 1
max = 1
for i in range(99):
    if s[i+1] > s[i]:
        count += 1
        if count > max:
            max = count
    else:
        count = 1
max

public class sequ

{

void main()

{                                                                                                                                                                                                                                                                                                                      String x="woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz"   ;    

    int max=0;


    for(int i=0;i<x.length();i++)
    {
       String y="";
        for(int j=i;j<x.length();j++)
        {
            char ch = x.charAt(j);
            y=y+ch;
            String z="";
            for(char k='a';k<='z';k++)
            {
                for(int l=0;l<y.length();l++)
                {

                char c = y.charAt(l);
                if(c==k)
                z=z+c;
            }


        if(y.equalsIgnoreCase(z))
        {

            if(y.length()>max)
            max=y.length();
        }

    }
}
}

System.out.println(max);

}

}

It's pretty trivial to do this in Javascript. Essentially, I converted each character to its corresponding number (ie. a → 1, b → 2, c → 3, ... , z → 26) and then found the length of the longest subsequence for which those numbers were increasing:

var string = 'woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz'.toUpperCase();
var characterNumbers = [];
var increasingLengths = [];

for (var i = 0; i < string.length; i++) {
    // Find the corresponding number for this character
    characterNumbers[i] = string.charCodeAt(i) - 64;

    if (characterNumbers[i] > characterNumbers[i - 1] && i > 0) {
        // If this character is alphabetically "greater" than the one before it, add to the length of the current sequence
        // Also, if we're just starting out a sequence, mark this number and the one before it
        increasingLengths[increasingLengths.length - 1] += (increasingLengths[increasingLengths.length - 1] == 0) ? 2 : 1
    } else if (increasingLengths[increasingLengths.length - 1] != 0) {
        // Otherwise, if a sequence just ended, create a new sequence
        increasingLengths.push(0);
    }
}

// Log the length of the longest sequence of characters in alphabetical order
console.log(Math.max.apply(Math, increasingLengths));

Java Solution :::

import java.io.*;

public class StringSting

{

public static void main(String args[])

{

    String

s="woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmi gz";

    char c='w';

    int i=0,j=s.length(),k=0,l=0;

    for(i=1;i<j;i++)

    {

        c=s.charAt(i);

        if(c>s.charAt(i-1))

        {

            k++;

        }

        else

        {

            if(k>l)

                l=k;

            k=0;

        }

    }

    System.out.println(l+1);

}

}

OUTPUT :: 4

Python solution:

s = 'woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz'
ans = []

past = 'z'
x = 1
for index in xrange(len(s)):
    if index:
        past = current
    current = s[index]

    if ord(current) >= ord(past):
        x += 1
    else:
        ans.append(x)
        x = 1

Here is a solution in C language with logic explained: /* Take three variables : strLength: length of current found sub-string endIndex: array index of the sub-string with maximum length previousLength: length of sub-string with maximum length found last time **/

int strLength = 1;
int endIndex = 0;
int previousLength = 0;
char *randomString = (char *)"woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz";

/* 
Finds the length of longest sub-string with letters in alphabetical order and 
stores the end index of sub-string with maximum length and length of the sub-string.
**/
for (int i = 0; i < strlen(randomString); i++)
{
    if (randomString[i] <= randomString[i+1])
    {
        strLength += 1;
    }
    else
    {
        if (strLength > previousLength)
        {
            endIndex = i+1;
            previousLength = strLength;
        }
        strLength = 1;
    }
}

//prints the found sub-string and its length. 
printf("Length of sub string is %d\n", previousLength);
for (int  i = endIndex - previousLength; i < endIndex; i++)
{
    printf("%c",randomString[i]);
}

Solution printed:

Length of sub string is 4.

sub-string is: bjvx

This is the C program I wrote:

char s[] = "woi........migz";
int index1, index2, count = 0, count2, temp, result = 0;

index1 = 0;
temp = 'a';
while(s[index1] != '\0'){

    for(index2 = temp; index2 <= 'z'; index2++) {

        count2 = count;
        if(s[index1] == index2) {

            count++;
            temp = index2;
            break;
        }
    }
    if(count == count2) {

        if(count > result) {

            result = count;
        }
        count = 0;
        temp = 'a';
    }
    else {

        index1++;
    }
}
printf("%d", result);

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

wtf = "woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz"
wtf_mark = ""

for i in range(len(wtf)-1):
    if wtf[i] > wtf[i+1]:
        wtf_mark += "0"
    else:
        wtf_mark += "1"

print wtf
print wtf_mark

Use your eyes haha

In Python (with explanation)

def checkString( s ):
    #to put running streak
    idx1 = 1
    #to put max running streak
    idx2 = 1
    #start loop from second character
    for aLoop in range(1,len(s)):
        if s[aaa] >= s[aaa-1]:
            #same character or in alphabetical order, add the streak
            idx1 += 1
        else:
            #this streak ends here
            if idx1 > idx2:
                #save if this is the longest streak
                idx2 = idx1
            #reset the running streak
            idx1 = 1
    #show the result
    print idx2

#put any string to check here
s= 'woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz'
checkString(s)

C Code

main() {

int i, x=1, bigger=0, p1, p2;

char s[] = "woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz";

for(i=0; i<100; i++)
   {
      p1=s[i];
      p2=s[i+1];
      if( p1 > p2 )
      x=1;
      else x++;
      if(x>bigger)
      {
         bigger=x;
      }    
   } 
   printf("%d\n", bigger);

include<iostream>

#include<stdio.h>
using namespace std;

int main() 
{
    char A[200];
    gets(A);
    char last=A[0];
    int length=1,maxlength=1;
    for(int i=1;A[i]!='\0';i++)
    {
        if(A[i]>=last)
        {
            length++;
            if(length>maxlength)
            maxlength=length;
            last=A[i];
        }
        else
        {
            last=A[i];
            length=1;
        }
    } 
    cout<<maxlength;
}

I just saw the string ...wrote it on the paper and found out all the strings which are in alphabetical order in string s ..... :P

C++ Code

#include <iostream>
#include <string>
using namespace std;

int main() {
    string str;
    getline(cin, str);
    int longest=0;
    int length=0;
    string maxword, word;
    char x;
    for(int i=0; i<str.length(); i++)
    {

        x=str[i+1];
        if(str[i]<=x)
        {
            length++;
            word+=str[i];
        }
        else
        {
            word+=str[i];
            if(length>longest)
            {
                longest=length;
                maxword=word;
            }
            length=0;
            word="";
        }
    }
    cout << maxword;
    return 0;
}

Run it here

I was able to find the solution by writing the following python code:

s = '....'
n, i, p = 1, 1, 0
#n stores the current highest sequence
#i is the counter
#p is the ascii value of the previous character

for c in s:
  if ord(c) >= p:
    i += 1 #if c follows p, increase the counter
    n = 1 if i > n else n
  else:
    i = 1 #reset the counter to one if it doesn't follow p
  p = ord(c) #set p to the ascii value of c

print n

This is a solution using the borland c++ compiler turboc 3.0,though this should run on other compilers by just changing the headers

#include<iostream.h>
#include<conio.h>
#include<string.h>

char s[100]="woimoepzbjvxfafpyfpzgmxugjodtemcjcpoxobfgbsmokkmcdpmawcwwaxhqwabzdlplvteszqgtkamxjkswkpnzpxpudxcmigz";

int i,j;

void main()
{
clrscr();
cout<<"s="<<s<<endl<<endl;
int k,c=0,l; //c stores the length of the longest substring in alphabetical order //l stores the no of characters which are in alphabetical order in each iteration
k=strlen(s);
for(i=0;i<k;i++)
{
l=1;
for(j=i+1;j<k;j++)
{
if(s[j]>=s[j-1])
{
l++;
}
else break;
}
if(l>=c)
{
c=l;
}
}
cout<<"The highest substring of s in alphabetical order is"<<c<<"\n";
getch();
}