> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/hashmap-and-hashset-and-sliding-window/check-if-a-string-contains-all-binary-codes-of-size-k.md).

# Check If a String Contains All Binary Codes of Size K

Given a binary string `s` and an integer `k`.

Return *True* if every binary code of length `k` is a substring of `s`. Otherwise, return *False*.

**Example 1:**

```
Input: s = "00110110", k = 2
Output: true
Explanation: The binary codes of length 2 are "00", "01", "10" and "11". They can be all found as substrings at indicies 0, 1, 3 and 2 respectively.
```

**Example 2:**

```
Input: s = "00110", k = 2
Output: true
```

**Example 3:**

```
Input: s = "0110", k = 1
Output: true
Explanation: The binary codes of length 1 are "0" and "1", it is clear that both exist as a substring. 
```

**Example 4:**

```
Input: s = "0110", k = 2
Output: false
Explanation: The binary code "00" is of length 2 and doesn't exist in the array.
```

**Example 5:**

```
Input: s = "0000000001011100", k = 4
Output: false
```

**Constraints:**

* `1 <= s.length <= 5 * 10^5`
* `s` consists of 0's and 1's only.
* `1 <= k <= 20`

```java
class Solution {
    public boolean hasAllCodes(String s, int k) {
        Set<String> seen = new HashSet<>();
        // Adding all substrings of size -> K
        for (int i = k; i <= s.length(); ++i)
            seen.add(s.substring(i - k, i));
        return seen.size() == 1 << k;
    }
}
```