### Problem Statement

You are given two Strings: A and B. Each character in A is either '0' or '1'. Each character in B is '0', '1', or '?'.

A string C matches B if we can change B into C by changing each '?' in B either to '0' or to '1'. Different occurrences of '?' may be changed to different digits. For example, C = "0101" matches B = "01??". Note that each character in C must be either '0' or '1', there cannot be any '?' remaining.

Consider all possible strings that match B. How many of these strings occur as a (contiguous) substring in A? Compute and return their number. Note that we only count each good string once, even if it has multiple occurrences in A.

### Definition

 Class: NonDeterministicSubstring Method: ways Parameters: String, String Returns: long Method signature: long ways(String A, String B) (be sure your method is public)

### Limits

 Time limit (s): 2 Memory limit (MB): 256 Stack limit (MB): 256

### Constraints

- A will contain between 1 and 50 characters, inclusive.
- B will contain between 1 and 50 characters, inclusive.
- Each character in A will be '0' or '1'.
- Each character in B will be '0', '1' or '?'.

### Examples

0)

 `"00010001"` `"??"`
`Returns: 3`
 There are four strings that match B: the strings "00", "01", "10", and "11". Out of these, the string "11" does not occur in A as a substring. The other three do occur, hence the answer is 3.
1)

 `"00000000"` `"??0??0"`
`Returns: 1`
 There are 16 different strings that match B, but out of those only the string "000000" is a substring of A.
2)

 `"001010101100010111010"` `"???"`
`Returns: 8`
 Each of the 8 strings that match B occurs somewhere in A.
3)

 `"00101"` `"??????"`
`Returns: 0`
 Here, the length of B is greater than the length of A. Clearly, a string that matches this B cannot be a substring of this A.
4)

 `"1101010101111010101011111111110001010101"` `"???????????????????????????????????"`
`Returns: 6`

=========================================================
My approach

1. A>B and
2. Find B length
3. Make all possible String with length of B and push HashSet
4. HashSet will filter duplication.
5. and finally only unique strings remain in hashset
6. compare all string with B if index of character is not ?
7. count all if it is match

my code is below
```public long ways(String A, String B){
HashSet<String> set = new HashSet<String>();

int blen = B.length();
for (int i = 0; i <= A.length()-blen; i++) {
}

Iterator<String> iter =  set.iterator();
int count =0;
while (iter.hasNext()) {
String str = iter.next();
boolean found = true;
for (int i = 0; i < blen; i++) {
if(B.charAt(i)!= '?'){
if(A.charAt(i) != B.charAt(i)){
found =false;
break;
}
}
}
if(found){
count++;
}
}
return count;
}```