# 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   ` `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 judgment '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 judgment '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 judgment '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 judgment '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 `

Output:

```True
```

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