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:
#include <bits/stdc++.h> using namespace std;
// Generate all binary subsets of given length k. vector<string> binarySubsets( int k)
{ vector<string> list;
// Loop for 2^k elements
for ( int i = 0; i < pow (2, k); i++)
{
// Add all possible binary sequence of given length
// to the list
list.push_back(
bitset<32>(i).to_string().substr(32 - k));
}
return list;
} // Check if binary string contains all permutations of // length k. bool tocheck( const string& s, int k)
{ vector<string> list = binarySubsets(k);
// Check if binary sequences are available in string or
// not
for ( const auto & b : list) {
if (s.find(b) == string::npos) {
return false ;
}
}
return true ;
} // Driver code int main()
{ string str = "11001" ;
int num = 2;
// Function call
if (tocheck(str, num)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
} // This code is contributed by phasing17. |
// 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 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# 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 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 |
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++ 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 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 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# 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 |
<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> |
YES
Time Complexity: O(n) we iterating over string from index 0 to n-k, so O(n) time complexity
Space Complexity: O(2k) as here, in the above solution we are creating an unordered set which stores the all possible binary substring of size k so space complexity will be 2k
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:
#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 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 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 |
// C# code for the above approach using System;
using System.Collections.Generic;
class GFG
{ // Method to check if the string contnains
// all the binary codes of size k
static bool hasAllCodes( string s, int k)
{
int n = s.Length;
if (n < k)
return false ;
int size = ( int )Math.Pow(2, k);
bool [] visited= new bool [size];
int i = 0;
// ArrayList that tracks if a certain code
// has been found
for (i=0; i<size; i++)
visited[i] = false ;
i=0;
int j = 0;
int val = 0;
while (j < k - 1) {
val = 2 * val + (( int )s[j]- '0' );
j++;
}
while (j < n) {
int x= (( int )s[j]- '0' );
val = 2 * val + x;
visited[val] = true ;
if (s[i] == '1' )
val -= (( int )Math.Pow(2, k - 1));
j++;
i++;
}
for (i = 0; i < size; i++) {
if (!visited[i])
return false ;
}
return true ;
}
// Driver Code
static void Main( string [] args)
{
string s = "00110110" ;
int k = 2;
if (hasAllCodes(s, k)) {
Console.WriteLine( "YES" );
}
else {
Console.WriteLine( "NO" );
}
}
} |
// 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 |
YES
Time Complexity : O(n)
Space Complexity: O(2k) as here, in the above solution we are creating a visited array of size pow(2,k) so we need that extra space