# 1461. Check If a String Contains All Binary Codes of Size K

### Description

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*.

### Constraints

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

### Approach

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/check-if-a-string-contains-all-binary-codes-of-size-k/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="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.
{% endtab %}

{% tab title="Example 2" %}
**Input:** s = "00110", k = 2

**Output:** true
{% endtab %}

{% tab title="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.
{% endtab %}

{% tab title="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.
{% endtab %}

{% tab title="Example 5" %}
**Input:** s = "0000000001011100", k = 4

**Output:** false
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(N*K). Where N is length of s. We need to iterate the 
 *    string, and use O(K) to calculate the hash of each substring.
 * Space complexity : O(N*K). There are at most N strings with length K in the set.
 */

class Solution {
    public boolean hasAllCodes(String s, int k) {
        int n = s.length();
        int requiredNoOfCodes = 1 << k;
        Set<String> codes = new HashSet();
        
        for(int i = 0; i+k <= n; i++) {
            codes.add(s.substring(i, i+k));
            if(codes.size() == requiredNoOfCodes) {
                return true;
            }
        }
        
        return false;
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*