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++
- Java
- Python3
- C#
- Javascript
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
Python3
C#
Javascript
True
Time Complexity: O(N logN)
Auxiliary Space: O(N), as extra space of size N is used to make an array
