CODEWITHSUNDEEP
javastring
Visaali Sundaram
Visaali Sundaram

Reputation: 35

How to specify the number of alphabets (and not blank spaces) to include in a substring?

I am trying to print a substring using index value. I need to exclude the blank space while counting but it should print the output along with blank space. I want to display, say, n alphabets from the main string. The blank spaces will be as they are but the number of alphabets from the lower bound to upper bound index should be n. My code is

public class Test {
    public static void main(String args[])
    {
        String Str=new String("Welcome to the class");
        System.out.println("\nReturn value is:");
        System.out.println(Str.substring(2,9));
    }
}

Output: lcome t

In the above mentioned code, it counts the space between the "Welcome" and "to". i need not want to count the space between them. My expected output is lcome to

Upvotes: 4

Views: 251

Answers (5)

progyammer
progyammer

Reputation: 1508

What you want to do is display n number of characters from the string including the spaces but n doesn't include the no. of blank spaces. For that, you could simply be using a loop instead of a library function.

The Logic: Keep displaying characters of the String str from index = 2 to index = 9-1 in a while loop. If the current character is a blank space, then increase the value of n, which is the upper bound of the string index for the sub string, by 1, i.e., the program will now display an extra character beyond the upper bound for each blank space encountered.

Consider the code below. 

String str = "Welcome to the class";
int index = 2, n = 9;
while(index < n){
    char c = str.charAt(index);
    System.out.print(c);
    if(c==' ')
    n++;
    index++;
}

Output: lcome to
Hope you can understand this code.

EDIT
As @Finbarr O'B said, a check to prevent StringIndexOutOfBoundsException would be necessary for the program for which, the loop will have to be defined as: 

while(index < n && index < str.length()){
    ...
}

Upvotes: 3

majk
majk

Reputation: 857

One way would be to extract it to using a regex ^.{2}([^ ] *){7}.

Another option is to use a simple for loop to traverse the string and calculate the end point to use for substring.

int non_whitespace = 0; int i;
for(i = 2; non_whitespace < 7; ++non_whitespace, ++i) { 
    while (str.charAt(i) == ' ') ++i;
}
return str.substring(2, i);

It is up to you which method do you consider more readable, and assess which one leads to better performance if speed is a concern.

Upvotes: 4

SomeJavaGuy
SomeJavaGuy

Reputation: 7357

You could use simple mathematics. Just substring it, remove all whitespaces and compare the original length to the String without whitespaces. Afterwards add the difference in size to your end index for the substring.

public static void main(String args[]) {
    String Str = "Welcome to the class";
    System.out.println("\nReturn value is:");
    String sub = Str.substring(2, 9);
    String wsRemoved = sub.replaceAll(" ", "");
    String wsBegginingRemoved = sub.replaceAll("^ *", "");
    String outputSub = Str.substring(2+(sub.length()-wsBegginingRemoved.length()), 9+(sub.length()-wsRemoved.length()+(sub.length() - wsBegginingRemoved.length())));
    System.out.println(outputSub);
}

Edit: not ignoring leading whitespaces anymore

O/P

lcome to

O/P "My name is Earl"

name is E

Upvotes: 6

gotnull
gotnull

Reputation: 27224

You could use replaceFirst for this:

String Str = "Welcome to the class"; // remove new String()
Str = Str.replaceFirst("^ *", "");
System.out.println("\nReturn value is:");
System.out.println(Str.substring(2, 10)); // increment end index by 1

Output:

lcome to

Upvotes: -2

Maroun
Maroun

Reputation: 95998

If you don't want to use regex, you can implement your own version of substring. The straightforward solution:

private static String substring(int begin, int end, String str) {
    StringBuilder res = new StringBuilder();
    while (begin < end) {
        if (str.charAt(begin) == ' ') {
            end++;
        }
        res.append(str.charAt(begin));
        begin++;
    }
    return res.toString();
}

The trick here is to ignore the "count" of a space, by incrementing end when it's encountered, forcing the loop to make one extra iteration.

The code complexity is O(n).

System.out.println(substring(2, 9, "Welcome to the class"));
>> lcome to

Upvotes: 0

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