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Check if a binary string contains all permutations of length k

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  • Difficulty Level : Easy
  • Last Updated : 01 Feb, 2023
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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 2k 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”. 

Implementation:

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");
    }
}

Python3




# Python3 program to Check a binary string
# contains all permutations of length k.
def tocheck(s, k):
    list1 = binary_subsets(k)
     
    # to check binary sequences are available
    # in string or not
    for b in list1:
        if s.find(b) == -1:
            return False
    return True
     
# to generate all binary subsets of given length k
def binary_subsets(k):
    # Declare the list as array
    list1 = []
     
    # to define the format of binary of
    # given length k
    format = '0' * k
     
    # returns the string representation of the
    # unsigned integer value represented by
    # the argument in binary (base 2)
    # Loop for 2^k elements
    for i in range(2 ** k):
        # To add in the list all possible
        # binary sequence of given length
        list1.append(format[0:k - len(bin(i)[2:])] + bin(i)[2:])
    return list1
     
# driver code
string = "11001"
num = 2
 
# function call
if tocheck(string, num):
    print("Yes")
else:
    print("No")
 
# This code is contributed by phasing17

C#




// C# code to implement the approach
 
using System;
using System.Collections.Generic;
 
class GFG {
    // to check all Permutation in given String
    public static bool ToCheck(string s, int k)
    {
        var list = BinarySubsets(k);
 
        // to check binary sequences are available
        // in string or not
        foreach(string b in 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
        var list = new List<string>();
 
        // to define the format of binary of
        // given length k
        var format = "{0:D" + k + "}";
 
        // Loop for 2^k 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,
                                   Convert.ToString(i, 2)));
 
            /* To Show all binary sequence of given
               length k
            Console.WriteLine(string.Format(format,
            Convert.ToString(i, 2)));*/
        }
        return list;
    }
 
    // drive main
    public static void Main(string[] args)
    {
        var str = "11001";
        var num = 2;
        if (ToCheck(str, num))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by phasing17

Javascript




// JavaScript program to Check a binary string
// contains all permutations of length k.
 
function tocheck(s, k) {
  let list = binarySubsets(k);
 
  // to check binary sequences are available
  // in string or not
  for (let b of list) {
    if (s.indexOf(b) === -1) {
      return false;
    }
  }
  return true;
}
 
// to generate all binary subsets of given length k
function binarySubsets(k) {
  // Declare the list as array
  let list = [];
 
  // to define the format of binary of
  // given length k
  let format = '0'.repeat(k);
 
  // returns the string representation of the
  // unsigned integer value represented by
  // the argument in binary (base 2)  using
  // parseInt(n,2) and convert it
  // into integer using toString(2).
  // Loop for 2^k elements
  for (let i = 0; i < Math.pow(2, k); i++) {
    // To add in the list all possible
    // binary sequence of given length
    list.push(i.toString(2).padStart(k, '0'));
  }
  return list;
}
 
// driver code
let str = "11001";
let num = 2;
 
// function call
if (tocheck(str, num)) {
  console.log("Yes");
} else {
  console.log("No");
}
 
 
// This code is contributed by phasing17

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 2k 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 2k then print “YES” otherwise print “NO”.

Implementation:

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

Javascript




<script>
 
// Javascript program to Check If a
// String Contains All Binary
// Codes of Size K
function hasAllcodes(s,k)
{
     
    // Unordered map of type string
    let us = new Set();
    for(let i = 0; i + k <= s.length; i++)
    {
        us.add(s.substring(i, i + k));
    }
    return (us.size == (1 << k));
}
 
// Driver code
let s = "00110110";
let k = 2;   
 
if (hasAllcodes(s, k))
{
    document.write("YES");
}
else
{
    document.write("NO");
}
 
// This code is contributed by ab2127
 
</script>

Output

YES

 Method : 3 (Two Pointer Based)

In this method, we will take a window of size k and move that window to the end, and mark all possible permutations in the visited array. then check if there is any value that is not marked as visited then return false, otherwise return true.

Implementation:

C++




#include <bits/stdc++.h>
using namespace std;
 
bool hasAllCodes(string s, int k)
{
    int n = s.size();
    if (n < k)
        return false;
    int size = pow(2, k);
    vector<bool> visited(size, false);
    int i = 0;
    int j = 0;
    int val = 0;
    while (j < k - 1) {
        val = 2 * val + (s[j] - '0');
        j++;
    }
    while (j < n) {
        val = 2 * val + (s[j] - '0');
        visited[val] = true;
        if (s[i] == '1')
            val -= pow(2, k - 1);
        j++;
        i++;
    }
    for (int i = 0; i < size; i++) {
        if (!visited[i])
            return false;
    }
    return true;
}
// Driver Code
int main()
{
    string s = "00110110";
    int k = 2;
    if (hasAllCodes(s, k)) {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
}

Java




// Java program to implement the approach
import java.util.*;
 
class GFG {
 
  // Method to check if the string contnains
  // all the binary codes of size k
  static boolean hasAllCodes(String s, int k)
  {
    int n = s.length();
    if (n < k)
      return false;
    int size = (int)Math.pow(2, k);
 
    // ArrayList that tracks if a certain code
    // has been found
    ArrayList<Boolean> visited
      = new ArrayList<Boolean>();
    for (int i = 0; i < size; i++)
      visited.add(false);
 
    int i = 0;
    int j = 0;
    int val = 0;
    while (j < k - 1) {
      val = 2 * val + (s.charAt(j) - '0');
      j++;
    }
    while (j < n) {
      val = 2 * val + (s.charAt(j) - '0');
      visited.set(val, true);
      if (s.charAt(i) == '1')
        val -= (int)Math.pow(2, k - 1);
      j++;
      i++;
    }
 
    // Checking if all the codes have been found
    for (i = 0; i < size; i++) {
      if (!visited.get(i))
        return false;
    }
    return true;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    String s = "00110110";
    int k = 2;
 
    // Function call
    if (hasAllCodes(s, k)) {
      System.out.print("YES\n");
    }
    else {
      System.out.print("NO\n");
    }
  }
}
 
// This code is contributed by phasing17

Python3




# Python3 Program to Check If a String Contains All Binary Codes of Size K
def hasAllcodes(s, k):
    n = s.len()
    size = pow(2, k)
    visited = [True for i in range(size)]
    i = 0
    j = 0
    val = 0
    while(j < k-1):
        val = (2 * val) + (s[j] - '0')
        j += 1
    while(j < size):
        val = (2 * val) + (s[j] - '0')
        visited[val] = True
        if(s[i] == '1'):
            val = val - pow(2,k-1)
        j += 1
        i += 1
    for i in range(size):
        if(visited[i] == False):
            return False
    return True
 
 # Driver code
s = "00110110"
k = 2
if (hasAllcodes):
    print("YES")
else:
    print("NO")
     
# This code is contributed by Ajay Makvana

Javascript




// JavaScript Program to Check If a String Contains All Binary Codes of Size K
function hasAllcodes(s, k)
{
    var n = s.length;
    var size = Math.pow(2, k);
    var visited = new Array(size).fill(true);
    var i = 0;
    var j = 0;
    var val = 0;
    while (j < k-1)
    {
        val = (2 * val) + (s[j] - '0');
        j += 1;
    }
    while (j < size)
    {
        val = (2 * val) + (s[j] - '0');
        visited[val] = true;
        if(s[i] == '1')
            val = val - Math.pow(2, k - 1);
        j += 1;
        i += 1;
    }
     
    for (var i = 0; i < size; i++)
    {
        if(visited[i] == false)
            return false;
    }
    return true;
}
 
// Driver code
var s = "00110110";
var k = 2;
if (hasAllcodes)
    console.log("YES");
else
    console.log("NO");
     
// This code is contributed by phasing17

Output

YES

Time Complexity : O(n)


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