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

Suppose we have a binary string s and an integer k. We have to check whether every binary code of length k is a substring of s. Otherwise, return False.

So, if the input is like S = "00110110", k = 2, then the output will be true. The binary codes of length 2 are "00", "01", "10" and "11". These are present at indices 0, 1, 3, and 2 respectively.

To solve this, we will follow these steps −

• Define one set v

• temp := blank string

• req := 2^k

• for initialize i := 0, when i < size of s, update (increase i by 1), do −

• temp := temp + s[i]

• if i >= k, then −

• delete one character from the first index of temp

• if i >= k - 1, then −

• insert temp into v

• if size of v is same as req, then −

• return true

• return false

## Example

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
lli fastPow(lli b, lli p){
lli ret = 1;
while (p) {
if (p & 1) {
ret *= b;
}
b *= b;
p >>= 1;
}
return ret;
}
bool hasAllCodes(string s, int k) {
unordered_set<string> v;
string temp = "";
lli req = fastPow(2, k);
for (lli i = 0; i < s.size(); i++) {
temp += s[i];
if (i >= k) {
temp.erase(0, 1);
}
if (i >= k - 1) {
v.insert(temp);
}
if ((lli)v.size() == req)
return true;
}
return false;
}
};
main(){
Solution ob;
cout << (ob.hasAllCodes("00110110",2));
}

## Input

"00110110",2

## Output

1