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

Last Updated : 15 Jul, 2025

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# 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

Analysis of Algorithms

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