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