Check if a binary string contains all permutations of length k
Last Updated :
20 Feb, 2023
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:
C++
#include <bits/stdc++.h>
using namespace std;
vector<string> binarySubsets(int k)
{
vector<string> list;
for (int i = 0; i < pow(2, k); i++)
{
list.push_back(
bitset<32>(i).to_string().substr(32 - k));
}
return list;
}
bool tocheck(const string& s, int k)
{
vector<string> list = binarySubsets(k);
for (const auto& b : list) {
if (s.find(b) == string::npos) {
return false;
}
}
return true;
}
int main()
{
string str = "11001";
int num = 2;
if (tocheck(str, num)) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
return 0;
}
|
Java
import java.util.*;
public class Checkbinary {
public static boolean tocheck(String s, int k)
{
List<String> list = BinarySubsets(k);
for (String b : list)
if (s.indexOf(b) == -1)
return false;
return true;
}
public static List<String> BinarySubsets(int k)
{
List<String> list = new ArrayList<>();
String format = "%0" + k + "d";
for (int i = 0; i < Math.pow(2, k); i++)
{
list.add(String.format(format,
Integer.valueOf(Integer.toBinaryString(i))));
}
return list;
}
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
def tocheck(s, k):
list1 = binary_subsets(k)
for b in list1:
if s.find(b) == -1:
return False
return True
def binary_subsets(k):
list1 = []
format = '0' * k
for i in range(2 ** k):
list1.append(format[0:k - len(bin(i)[2:])] + bin(i)[2:])
return list1
string = "11001"
num = 2
if tocheck(string, num):
print("Yes")
else:
print("No")
|
C#
using System;
using System.Collections.Generic;
class GFG {
public static bool ToCheck(string s, int k)
{
var list = BinarySubsets(k);
foreach(string b in list) if (s.IndexOf(b)
== -1) return false;
return true;
}
public static List<string> BinarySubsets(int k)
{
var list = new List<string>();
var format = "{0:D" + k + "}";
for (int i = 0; i < Math.Pow(2, k); i++) {
list.Add(string.Format(format,
Convert.ToString(i, 2)));
}
return list;
}
public static void Main(string[] args)
{
var str = "11001";
var num = 2;
if (ToCheck(str, num))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
|
Javascript
function tocheck(s, k) {
let list = binarySubsets(k);
for (let b of list) {
if (s.indexOf(b) === -1) {
return false;
}
}
return true;
}
function binarySubsets(k) {
let list = [];
let format = '0'.repeat(k);
for (let i = 0; i < Math.pow(2, k); i++) {
list.push(i.toString(2).padStart(k, '0'));
}
return list;
}
let str = "11001";
let num = 2;
if (tocheck(str, num)) {
console.log("Yes");
} else {
console.log("No");
}
|
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++
#include <bits/stdc++.h>
using namespace std;
#define int long long
bool hasAllcodes(string s, int k)
{
unordered_set<string> us;
for (int i = 0; i + k <= s.size(); i++)
{
us.insert(s.substr(i, k));
}
return us.size() == 1 << k;
}
signed main()
{
string s = "00110110";
int k = 2;
if (hasAllcodes)
{
cout << "YES\n";
}
else
{
cout << "NO\n";
}
}
|
Java
import java.io.*;
import java.util.*;
class GFG
{
static boolean hasAllcodes(String s, int k)
{
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));
}
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");
}
}
}
|
Python3
def hasAllcodes(s, k) :
us = set()
for i in range(len(s) + 1) :
us.add(s[i : k])
return len(us) == 1 << k
s = "00110110"
k = 2
if (hasAllcodes) :
print("YES")
else :
print("NO")
|
C#
using System;
using System.Collections.Generic;
class GFG {
static bool hasAllcodes(string s, int k)
{
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;
}
static void Main()
{
string s = "00110110";
int k = 2;
if(hasAllcodes(s, k))
{
Console.WriteLine("YES");
}
else
{
Console.WriteLine("NO");
}
}
}
|
Javascript
<script>
function hasAllcodes(s,k)
{
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));
}
let s = "00110110";
let k = 2;
if (hasAllcodes(s, k))
{
document.write("YES");
}
else
{
document.write("NO");
}
</script>
|
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:
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;
}
int main()
{
string s = "00110110";
int k = 2;
if (hasAllCodes(s, k)) {
cout << "YES\n";
}
else {
cout << "NO\n";
}
}
|
Java
import java.util.*;
class GFG {
static boolean hasAllCodes(String s, int k)
{
int n = s.length();
if (n < k)
return false;
int size = (int)Math.pow(2, k);
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++;
}
for (i = 0; i < size; i++) {
if (!visited.get(i))
return false;
}
return true;
}
public static void main(String[] args)
{
String s = "00110110";
int k = 2;
if (hasAllCodes(s, k)) {
System.out.print("YES\n");
}
else {
System.out.print("NO\n");
}
}
}
|
Python3
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
s = "00110110"
k = 2
if (hasAllcodes):
print("YES")
else:
print("NO")
|
C#
using System;
using System.Collections.Generic;
class GFG
{
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;
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;
}
static void Main(string[] args)
{
string s = "00110110";
int k = 2;
if (hasAllCodes(s, k)) {
Console.WriteLine("YES");
}
else {
Console.WriteLine("NO");
}
}
}
|
Javascript
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;
}
var s = "00110110";
var k = 2;
if (hasAllcodes)
console.log("YES");
else
console.log("NO");
|
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
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