Pattern Count | Editorial by aksingh007

We need to just find the '1' in the string and then just keep moving forward until a character (other than 0 or 1) is met or '1' is found. In the case of '1', 
if the previous number was '0', increment the count.

```
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

static int patternCount(String s){
       int count=0;int i,j;
        int[] arr=new int[50000];
        int pos=0;
        for(i=0;i<s.length();i++){
            if(s.charAt(i)=='1'){
                arr[pos]=i;
                pos++;
            }
        }
       
        int k=0;
        while(k<pos-1){
            boolean flag=true;
            i=arr[k]+1;j=arr[k+1];
             int tot=0;
            
            for(;i<j;i++){
                //System.out.println(i+" "+j+" "+s.charAt(i));
                if(s.charAt(i)!='0'){
                    flag=false;
                    break;
                }
                else
                    if(s.charAt(i)=='0')
                         tot++;
            }
            if(flag==true && tot>0)
                count++;
            k++;
        }
        return count;
    }

public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int q = in.nextInt();
        for(int a0 = 0; a0 < q; a0++){
            String s = in.next();
            int result = patternCount(s);
            System.out.println(result);
        }
    }
}

```

Pattern Count | Editorial by aksingh007

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