Check if binary representations of 0 to N are present as substrings in given binary string

Give binary string str and an integer N, the task is to check if the substrings of the string contain all binary representations of non-negative integers less than or equal to the given integer N.

Examples:

Input: str = “0110″, N = 3 
Output: True 
Explanation:
Since substrings “0″, “1″, “10″, and “11″ can be formed from given string. Hence all binary representations of 0 to 3 are present as substrings in given binary string.

Input: str = “0110”, N = 4 
Output: False 
Explanation:
Since substrings “0″, “1″, “10″, and “11″ can be formed from given string, but not “100”. Hence the answer is False

Approach: 
The above problem can be solved using BitSet and HashMap. Follow the steps given below to solve the problem

• Initialize a map[] to mark the strings and take a bit-set variable ans to convert the number from decimal to binary.
• Take one more variable count as zero.
• run the loop from N to 1 using the variable i and check the corresponding numbers are marked in a map or not.
• if number i is not marked in a map[] then convert the current number into binary using the bit-set variable ans.
• then check if converted binary string is substring of the given string or not.
• if it is not a substring then
• run while loop unless i is not marked and binary number becomes zero
  • mark the i in a map
  • increment the count
  • do the right shift of converted number. This is done because if any string x is converted into binary (say 111001) and this substring is already marked in map, then 11100 will already be marked automatically. 
    This is based on the fact that if i exists, i>>1 also exists.
• Finally check if count ? N + 1, then print True 
  Else print False

Below is the implementation of above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to convert decimal to binary
// representation
string decimalToBinary(int N)
{

    string ans = "";

    // Iterate over all bits of N
    while (N > 0) {

        // If bit is 1
        if (N & 1) {
            ans = '1' + ans;
        }
        else {
            ans = '0' + ans;
        }

        N /= 2;
    }

    // Return binary representation
    return ans;
}

// Function to check if binary conversion
// of numbers from N to 1 exists in the
// string as a substring or not
string checkBinaryString(string& str, int N)
{

    // To store the count of number
    // exists as a substring
    int map[N + 10], cnt = 0;

    memset(map, 0, sizeof(map));

    // Traverse from N to 1
    for (int i = N; i > 0; i--) {

        // If current number is not
        // present in map
        if (!map[i]) {

            // Store current number
            int t = i;

            // Find binary of t
            string s = decimalToBinary(t);

            // If the string s is a
            // substring of str
            if (str.find(s) != str.npos) {

                while (t && !map[t]) {

                    // Mark t as true
                    map[t] = 1;

                    // Increment the count
                    cnt++;

                    // Update for t/2
                    t >>= 1;
                }
            }
        }
    }

    // Special judgement '0'
    for (int i = 0; i < str.length(); i++) {
        if (str[i] == '0') {
            cnt++;
            break;
        }
    }
    // If the count is N+1, return "yes"
    if (cnt == N + 1)
        return "True";
    else
        return "False";
}

// Driver Code
int main()
{
    // Given String
    string str = "0110";

    // Given Number
    int N = 3;

    // Function Call
    cout << checkBinaryString(str, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to convert decimal to binary
// representation
static String decimalToBinary(int N)
{
    String ans = "";

    // Iterate over all bits of N
    while (N > 0)
    {

        // If bit is 1
        if (N % 2 == 1)
        {
            ans = '1' + ans;
        }
        else
        {
            ans = '0' + ans;
        }
        N /= 2;
    }

    // Return binary representation
    return ans;
}

// Function to check if binary conversion
// of numbers from N to 1 exists in the
// String as a subString or not
static String checkBinaryString(String str, int N)
{

    // To store the count of number
    // exists as a subString
    int []map = new int[N + 10];
    int cnt = 0;

    // Traverse from N to 1
    for(int i = N; i > 0; i--)
    {

        // If current number is not
        // present in map
        if (map[i] == 0)
        {

            // Store current number
            int t = i;

            // Find binary of t
            String s = decimalToBinary(t);

            // If the String s is a
            // subString of str
            if (str.contains(s))
            {
                while (t > 0 && map[t] == 0)
                {

                    // Mark t as true
                    map[t] = 1;

                    // Increment the count
                    cnt++;

                    // Update for t/2
                    t >>= 1;
                }
            }
        }
    }

    // Special judgement '0'
    for(int i = 0; i < str.length(); i++)
    {
        if (str.charAt(i) == '0')
        {
            cnt++;
            break;
        }
    }

    // If the count is N+1, return "yes"
    if (cnt == N + 1)
        return "True";
    else
        return "False";
}

// Driver Code
public static void main(String[] args)
{

    // Given String
    String str = "0110";

    // Given number
    int N = 3;

    // Function call
    System.out.print(checkBinaryString(str, N));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of
# the above approach

# Function to convert decimal to
# binary representation
def decimalToBinary(N):

    ans = ""

    # Iterate over all bits of N
    while(N > 0):

        # If bit is 1
        if(N & 1):
            ans = '1' + ans
        else:
            ans = '0' + ans

        N //= 2

    # Return binary representation
    return ans

# Function to check if binary conversion
# of numbers from N to 1 exists in the
# string as a substring or not
def checkBinaryString(str, N):

    # To store the count of number
    # exists as a substring
    map = [0] * (N + 10)
    cnt = 0

    # Traverse from N to 1
    for i in range(N, -1, -1):

        # If current number is not
        # present in map
        if(not map[i]):

            # Store current number
            t = i

            # Find binary of t
            s = decimalToBinary(t)

            # If the string s is a
            # substring of str
            if(s in str):
                while(t and not map[t]):

                    # Mark t as true
                    map[t] = 1

                    # Increment the count
                    cnt += 1

                    # Update for t/2
                    t >>= 1

    # Special judgement '0'
    for i in range(len(str)):
        if(str[i] == '0'):
            cnt += 1
            break

    # If the count is N+1, return "yes"
    if(cnt == N + 1):
        return "True"
    else:
        return "False"

# Driver Code
if __name__ == '__main__':

    # Given String
    str = "0110"

    # Given Number
    N = 3

    # Function Call
    print(checkBinaryString(str, N))

# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;

class GFG{

// Function to convert decimal to binary
// representation
static String decimalToBinary(int N)
{
    String ans = "";

    // Iterate over all bits of N
    while (N > 0)
    {

        // If bit is 1
        if (N % 2 == 1)
        {
            ans = '1' + ans;
        }
        else
        {
            ans = '0' + ans;
        }
        N /= 2;
    }

    // Return binary representation
    return ans;
}

// Function to check if binary conversion
// of numbers from N to 1 exists in the
// String as a subString or not
static String checkBinaryString(String str, int N)
{

    // To store the count of number
    // exists as a subString
    int []map = new int[N + 10];
    int cnt = 0;

    // Traverse from N to 1
    for(int i = N; i > 0; i--)
    {

        // If current number is not
        // present in map
        if (map[i] == 0)
        {

            // Store current number
            int t = i;

            // Find binary of t
            String s = decimalToBinary(t);

            // If the String s is a
            // subString of str
            if (str.Contains(s))
            {
                while (t > 0 && map[t] == 0)
                {

                    // Mark t as true
                    map[t] = 1;

                    // Increment the count
                    cnt++;

                    // Update for t/2
                    t >>= 1;
                }
            }
        }
    }

    // Special judgement '0'
    for(int i = 0; i < str.Length; i++)
    {
        if (str[i] == '0')
        {
            cnt++;
            break;
        }
    }

    // If the count is N+1, return "yes"
    if (cnt == N + 1)
        return "True";
    else
        return "False";
}

// Driver Code
public static void Main(String[] args)
{

    // Given String
    String str = "0110";

    // Given number
    int N = 3;

    // Function call
    Console.Write(checkBinaryString(str, N));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program for the above approach

// Function to convert decimal to binary
// representation
function decimalToBinary(N)
{

    var ans = "";

    // Iterate over all bits of N
    while (N > 0) {

        // If bit is 1
        if (N % 2 == 1){
            ans = '1' + ans;
        }
        else {
            ans = '0' + ans;
        }

        N = parseInt(N/2);
    }

    // Return binary representation
    return ans;
}

// Function to check if binary conversion
// of numbers from N to 1 exists in the
// string as a substring or not
function checkBinaryString(str, N)
{

    // To store the count of number
    // exists as a substring
    var map = Array(N+10).fill(0), cnt = 0;

    // Traverse from N to 1
    for (var i = N; i > 0; i--) {

        // If current number is not
        // present in map
        if (!map[i]) {

            // Store current number
            var t = i;

            // Find binary of t
            var s = decimalToBinary(t);

            // If the string s is a
            // substring of str
            if (str.includes(s)) {

                while (t>0 && map[t] == 0) {

                    // Mark t as true
                    map[t] = 1;

                    // Increment the count
                    cnt++;

                    // Update for t/2
                    t >>= 1;
                }
            }
        }
    }

    // Special judgement '0'
    for (var i = 0; i < str.length; i++) {
        if (str[i] == '0') {
            cnt++;
            break;
        }
    }
    // If the count is N+1, return "yes"
    if (cnt == N + 1)
        return "True";
    else
        return "False";
}

// Driver Code
// Given String
var str = "0110";
// Given Number
var N = 3;
// Function Call
document.write( checkBinaryString(str, N));

</script>

Output:

True

Time Complexity: O(N logN)
Auxiliary Space: O(N), as extra space of size N is used to make an array