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

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  • Last Updated : 18 Jun, 2022

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


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