Question: Remove Zeroes (need a java approach)

[Bhakti Khedkar]

You are given a string s. Each character is either 0 or 1.

You want all 1s in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all ones (1) form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.

You may erase some (possibly none) zeroes(0) from the string. What is the minimum number of zeros (0) that you have to erase?
Input
The first line contains one integer t (1≤t≤100) — the number of test cases.

Then t lines follow, each representing a test case. Each line contains one string s (1≤|s|≤100); each character of s is either 0 or 1.

Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of zeros(0) that you have to erase from s).

Example
Input
3 010011 0 1111000

Output
2 0 0

Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).

[Piet Souris]

hi Bhakti,

welcome to the Ranch and enjoy the stay!

Do you have any question about this exercise?

[Campbell Ritchie]

Welcome to the Ranch

I shall give you a hint: Strings are immutable objects, but they have a mutable counterpart, called StringBuilder.
Another hint: The question doesn't ask you to convert the String, “0010101101001010101011010101011100110100,” to, “001111111111111111111100.” It asks for a count of 0s.
What does the following mean?

s (1≤|s|≤100)

Neither the absolute value operator |x| nor inequality ≤ can be applied to text.
There is no absolute value operator in Java®.

[Bhakti Khedkar]

its the number of characters we can input,ie ,the maximum length of string.

[Campbell Ritchie]

In that case it should read s (1≤|s.length()|≤100).
There should be no need for a limit to the number of Strings nor their individual length; a correctly written program should be able to handle any number. The absolute value operator is redundant since Strings never have a negative length.

[Campbell Ritchie]

Well done getting a solution, but please don't edit old posts. That can cause confusion and can also cause you solution to be ignored. I am refusing the edit; please post the solution in a new post. Please use the code button; I shall use it on your original post so you can see how much better it looks

[Bhakti Khedkar]

OK SIR.....I wont edit the original post.
Also, sir I am new to coderanch . Hence, trying to get familiar with it.

[Bhakti Khedkar]

/////SOLUTION FOR THE QUESTION:- REMOVING ZEROES.
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
	public static void main (String[] args) throws java.lang.Exception
	{
		//your code here
      Scanner sc=new Scanner(System.in);
        int t=sc.nextInt(); //number of testcases.
        String s=" ";
        for(int i=1;i<=t;i++){
          s=sc.next();
          char[] ch=s.toCharArray();
          int count=0;
          int pt1=0;
          int pt2=0;
          for(int j=0;j<ch.length;j++){ ///searching for 1st "1" from the beginning 
            if(ch[j]=='1'){
              pt1=j;
              break;
            } 
          }
          for(int j=ch.length-1;j>pt1;j--){ //searching for 1st "1" from the end
            if(ch[j]=='1'){
              pt2=j;
              break;
            }
          }
          ///after searching for "1" from beginnig and end ,
          ///then search for zeros in between them.
          for(int j=pt1;j<pt2;j++){
            if(ch[j]=='0'){
              count++;
            }
          }
          System.out.println(count);
      }
}}

[Campbell Ritchie]

Thank you for showing us your solution; it works nicely (), butI think I can improve it. I would create a class to do the calculations, which I shall show you last.
I am not convinced that for loops are the best loops for all cases. You can change the loop in line 21 to a while loop:-int last1 = s.length() - 1; while (s.charAt(last1) == '0') {   last1--; }You can do the same with the loop in line 20, but that may be slightly easier.
There is a very naughty way to reduce those loops to a single line each
Please reserve // comments for commenting‑out and short things; if you need to write anything long use /* comments */ instead.
Don't declare that method throws Exception because it doesn't throw any checked exceptions.
Don't declare import java.lang.Anything; because that is always unnecessary.
Don't declare import java.io.Anything; because you are using nothing from that package. Only declare import java.util.Scanner; because that is the only import you are using.
Make sure your variable names are obvious; I can't tell what pt1 and pt2 mean from simply reading them.
Make sure to indent your code, which you have done quite well except for lines 11 and 41, but you are mixing K&R and Allman indentation. Use one convention and stick to it throughout.
Make sure to space your code out with spaces around binary operators./* Before */ for(int i=1;i<=t;i++){/* After */ for (int i = 1; i <= t; i++) {Regard the following as the standard version of the for loop. It uses < and 0 rather than <= and 1:-for (int i = 0; i < t; i++) {

class ZeroCounter
{
    private final String s;
     
    ZeroCounter(String s)
    {
        if (!s.matches("[01]*"))
        {
            throw new IllegalArgumentException(
                    "Text may only contain 1s and 0s.");
        }
        this.s = s;
    }
 
    public int countIncluded0s()
    {
        int count = 0;
        int first1 = s.indexOf("1");
        int last1 = s.lastIndexOf("1");
        for (int i = first1; i < last1; i++)
        {
            if (s.charAt(i) == '0')
            {
                count++;
            }
        }
        return count;
    }
}

The bottom quote shows a suggestion of how ZeroCounter can be used. It may be possible to “lose” the local variables in lines 18‑19.

JShell on my computer wrote:|  created class ZeroCounter
. . .
jshell> new ZeroCounter("0010101101001010101011010101011100110100").countIncluded0s()
$4 ==> 16

[Campbell Ritchie]

[edit]There is a mistake in this post, corrected later.
Can I do it with a Stream? See Java™ Tutorials (parts of each section in that link), Stream interface, and java.util.stream package./* Following code not tested */ System.out.println(s.chars() // Stream of ints, one for each char                    .limit(s.lastIndexOf("1")) // stop where last 1 was                    .filter(j -> j == '0') // keep only 0s                    .count()); // Should be obvious what count does.

[Campbell Ritchie]

Afraid I missed out a line from the Stream code. Sorry. Corrected version follows, with line 3 added./* Following code not tested */ System.out.println(s.chars() // Stream of ints, one for each char                    .limit(s.lastIndexOf("1")) // stop where last 1 was                    .dropWhile(j -> j == '0') // lose leading 0s                    .filter(j -> j == '0') // keep only 0s                    .count()); // Should be obvious what count does.

[Bhakti Khedkar]

Thank you sir for giving me optimized solution for the question.

[Junilu Lacar]

Here's a way without using Streams:

  public static int includedZeroes(String s) {
    if (s.indexOf("01") == -1) return 0;
    return s.substring(s.indexOf("1"), s.lastIndexOf("1") )
            .replaceAll("1", "")
            .length();
  }

[Campbell Ritchie]

This morning, I wrote:. . . I am not convinced that for loops are the best loops for all cases. . . .

When I changed it to a while loop, I was using sentinel‑controlled repetition; a for loop works best with counter‑controlled repetition.