"number of segments in a string" not working for a particular input

I have to find "contiguous sequence of non-space characters" in a string.
My output is coming wrong for the input

Input=", , , ,        a, eaefa"

My answer is coming as 13 instead of 6.Though I have only counted words except for spaces.

class Solution {
    public int countSegments(String s) 
    {
        if(s.isEmpty()){
            return 0;
        }
        else
        {
            int count=0;
            String s1[]=s.split(" ");
            for(int i=0;i<s1.length;i++)
            {
                if(s1[i]!=" ")
                    count++;
            }
            return count;
        }
    }
}

Others have suggesting using:

s.split("\\s+").length

However, there are complications in using split. Specifically, the above will give incorrect answers for strings with leading spaces. Even if these issues are fixed it's still overly expensive as we're creating count new strings, and an array to hold them.

We can implement countSegments directly by iterating through the string and counting the number of times we go from a non-space character to a space character, or the end of the string:

public static int countSegments(String s)
{
	int count = 0;
	for(int i=1; i<=s.length(); i++)
	{
		if((s.charAt(i-1) != ' ') && (i == s.length() || s.charAt(i) == ' ')) count++;
	}
	return count;
}

Test:

for(String s : new String[] {"", " ", "a", " a", "a ", " a ", ", , , ,        a, eaefa"})
	System.out.format("<%s> : %d%n", s, countSegments(s));

Output:

<> : 0
< > : 0
<a> : 1
< a> : 1
<a > : 1
< a > : 1
<, , , ,        a, eaefa> : 6

You should use split on multiple spaces, and then you have the segments already divided up for you, so you don't need to make a for-loop or anything.

//The trim is because split gets messed up with leading spaces, as SirRaffleBuffle said
s = s.trim();
if (s.isEmpty()) return 0;
return s.split("\\s+").length;

• If you want only sequences of alphanumeric characters, you can try this regex instead: "\\W+"
• If you want only sequences of English letters, you can do the same thing but with the regex "[^A-Za-z]+".

Here, it splits on multiple spaces instead of just one.

The way you're currently doing it, you count every single letter that's not a whitespace instead of "contiguous sequences of no-space characters". That's why you're getting 13 instead of 6.

Notice that count is incremented anytime it finds something that isn't a space, but if you do want to do this with a for-loop, you should have a boolean flag telling you that you've entered a sequence, so you only increment count when that flat was previously false (you were outside a sequence) and then you find a space.

Also, using != for String comparison is wrong, you should use the equals method.

> “number of segments in a string” not working for a particular input

You can do it easily by using the regex, \\s+ as follows:

public class Main {
	public static void main(String[] args) {
		String str = ", , , ,        a, eaefa";
		str = str.trim();// Remove the leading and trailing space
		System.out.println(str.isEmpty() ? 0 : str.split("\\s+").length);
	}
}

Output:

6

The regex, \\s+ matches on one or more consecutive spaces.

On a side note, you are using != to compare strings, which is not correct. Note that == and != are used to compare the references, not the values.