Check if a binary string contains all permutations of length k

Difficulty Level : Easy
Last Updated : 06 Jan, 2021

Given a binary string and k, to check whether it’s contains all permutations of length k or not.
Examples:

Input : Binary string 11001
        k : 2
Output : Yes
11001 contains all possibilities of 
binary sequences with k = 2, 00, 01, 
10, 11

Input : Binary string: 1001
        k : 2
Output: No
1001 does not contain all possibilities of
binary sequences with k = 2. Here 11 
sequence is missing

Method 1:
Explanation:
In this example one binary sequence of length k is not found it is 0110.
So all binary sequences with k=4 will be 2⁴=16. they are following
0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111,
1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111
All should be sub string of given binary string then print Yes otherwise No
Algorithm
Taking binary string and the size k. In binary string we check binary sequences are matched or not. Binary sequence is made of size k, as we know that in binary using 0 and 1 digit so to generate total binary subsets is 2^k element. The main idea behind it, to store all binary value in list as string and then compare list all item to given binary string as subset. If all are occur inside the binary string then print “Yes” otherwise print “No”.

JAVA

// Java program to Check a binary string
// contains all permutations of length k.
 
import java.util.*;
public class Checkbinary {
 
    // to check all Permutation in given String
    public static boolean tocheck(String s, int k)
    {
        List<String> list = BinarySubsets(k);
 
        // to check binary sequences are available 
        // in string or not
        for (String b : list) 
            if (s.indexOf(b) == -1)
                return false;        
 
        return true;
    }
 
    // to generate all binary subsets of given length k
    public static List<String> BinarySubsets(int k)
    {
        // Declare the list as String
        List<String> list = new ArrayList<>();
 
        // to define the format of binary of 
        // given length k
        String format = "%0" + k + "d";
 
        // returns the string representation of the
        // unsigned integer value represented by
        // the argument in binary (base 2)  using
        // Integer.toBinaryString and convert it
        // into integer using Integer.valueOf.
        // Loop for 2<sup>k</sup> elements
        for (int i = 0; i < Math.pow(2, k); i++) 
        {
            // To add in the list all possible 
            // binary sequence of given length
            list.add(String.format(format,
                Integer.valueOf(Integer.toBinaryString(i))));
 
            /* To Show all binary sequence of given 
               length k 
            System.out.println(String.format(format,
            Integer.valueOf(Integer.toBinaryString(i))));*/
        }
        return list;
    }
 
    // drive main
    public static void main(String[] args)
    {
        String str = "11001";
        int num = 2;
        if (tocheck(str, num))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Output

Yes

Method 2:
Algorithm
Taking binary string and the size k. In binary string we check binary sequences are matched or not. Binary sequence is made of size k, as we know that in binary using 0 and 1 digit so to generate total binary subsets is 2^k element. The main idea behind it, to store all the substring of size k of the given string to the set i.e. storing the distinct substring of size k. If the size of the set is equal to 2^k than print “YES” otherwise print “NO”.

C++

// C++ Program to Check If a 
// String Contains All Binary
// Codes of Size K
#include <bits/stdc++.h>
using namespace std;
#define int long long
 
bool hasAllcodes(string s, int k)
{
     
    // Unordered map of type string
    unordered_set<string> us;
   
    for (int i = 0; i + k <= s.size(); i++) 
    {
        us.insert(s.substr(i, k));
    }
    return us.size() == 1 << k;
}
 
// Driver Code
signed main()
{
    string s = "00110110";
    int k = 2;
    if (hasAllcodes) 
    {
        cout << "YES\n";
    }
    else
    {
        cout << "NO\n";
    }
}

Java

// Java Program to Check If a 
// String Contains All Binary
// Codes of Size K
import java.io.*;
import java.util.*;
class GFG
{
    static boolean hasAllcodes(String s, int k)
    {
       
        // Unordered map of type string
        Set<String> us= new HashSet<String>();
        for(int i = 0; i + k <= s.length(); i++)
        {
            us.add(s.substring(i, i + k));
        }
        return (us.size() == (1 << k));
    }
   
    // Driver code
    public static void main (String[] args) 
    {
        String s = "00110110";
        int k = 2;
        if(hasAllcodes(s, k))
        {
            System.out.println("YES");
        }
        else
        {
            System.out.println("NO");
        }
    }
}
 
//  This code is contributed by avanitrachhadiya2155

Python3

# Python3 Program to Check If a 
# String Contains All Binary
# Codes of Size K
def hasAllcodes(s, k) :
     
    # Unordered map of type string
    us = set()
    for i in range(len(s) + 1) :    
        us.add(s[i : k])    
    return len(us) == 1 << k
 
# Driver code
s = "00110110"
k = 2
if (hasAllcodes) :
    print("YES")
else :
    print("NO")
 
    # This code is contributed by divyeshrabadiya07

C#

// C# Program to Check If a 
// String Contains All Binary
// Codes of Size K
using System;
using System.Collections.Generic;
class GFG {
 
  static bool hasAllcodes(string s, int k)
  {
 
    // Unordered map of type string
    HashSet<string> us = new HashSet<string>(); 
 
    for (int i = 0; i + k <= s.Length; i++) 
    {
      us.Add(s.Substring(i, k));
    }
 
    return us.Count == 1 << k;
  }
 
  // Driver code
  static void Main() 
  {
    string s = "00110110";
    int k = 2;
    if(hasAllcodes(s, k)) 
    {
      Console.WriteLine("YES");
    }
    else
    {
      Console.WriteLine("NO");
    }
  }
}
 
// This code is contributed by divyesh072019

Output

YES

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