Generate all binary strings of length n with sub-string “01” appearing exactly twice
Given an integer N, the task is to generate all possible binary strings of length N which contain “01” as the sub-string exactly twice.
Examples:
Input: N = 4
Output:
0101
“0101” is the only binary string of length 4
that contains “01” exactly twice as the sub-string.Input: N = 5
Output:
00101
01001
01010
01011
01101
10101
Approach: This problem can solved using backtracking. To generate a binary string, we implement a function that generate each bit at a time, update the state of the binary string (current length, number of occurrences of the pattern). Then call the function recursively, and according to the current state of the binary string, the function will decide how to generate the next bit or print out the binary string (if the problem’s requirement is met).
For this problem, backtracking strategy looks like we generate a binary tree with each node can have either value 0 or 1.
For example, with N = 4, the tree will look like:
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> #include <stdlib.h> using namespace std; // Utility function to print the given binary string void printBinStr(int* str, int len) { for (int i = 0; i < len; i++) { cout << str[i]; } cout << endl; } // This function will be called recursively // to generate the next bit for given // binary string according to its current state void generateBinStr(int* str, int len, int currlen, int occur, int nextbit) { // Base-case: if the generated binary string // meets the required length and the pattern "01" // appears twice if (currlen == len) { // nextbit needs to be 0 because each time // we call the function recursively, // we call 2 times for 2 cases: // next bit is 0 or 1 // The is to assure that the binary // string is printed one time only if (occur == 2 && nextbit == 0) printBinStr(str, len); return; } // Generate the next bit for str // and call recursive if (currlen == 0) { // Assign first bit str[0] = nextbit; // The next generated bit will wither be 0 or 1 generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); } else { // If pattern "01" occurrence is < 2 if (occur < 2) { // Set next bit str[currlen] = nextbit; // If pattern "01" appears then // increase the occurrence of pattern if (str[currlen - 1] == 0 && nextbit == 1) { occur += 1; } generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); // Else pattern "01" occurrence equals 2 } else { // If previous bit is 0 then next bit cannot be 1 if (str[currlen - 1] == 0 && nextbit == 1) { return; // Otherwise } else { str[currlen] = nextbit; generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); } } } } // Driver code int main() { int n = 5; // Length of the resulting strings // must be at least 4 if (n < 4) cout << -1; else { int* str = new int[n]; // Generate all binary strings of length n // with sub-string "01" appearing twice generateBinStr(str, n, 0, 0, 0); generateBinStr(str, n, 0, 0, 1); } return 0; } |
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Java
// Java implementation of the above approach class GFG { // Utility function to print the given binary string static void printBinStr(int[] str, int len) { for (int i = 0; i < len; i++) { System.out.print(str[i]); } System.out.println(); } // This function will be called recursively // to generate the next bit for given // binary string according to its current state static void generateBinStr(int[] str, int len, int currlen, int occur, int nextbit) { // Base-case: if the generated binary string // meets the required length and the pattern "01" // appears twice if (currlen == len) { // nextbit needs to be 0 because each time // we call the function recursively, // we call 2 times for 2 cases: // next bit is 0 or 1 // The is to assure that the binary // string is printed one time only if (occur == 2 && nextbit == 0) { printBinStr(str, len); } return; } // Generate the next bit for str // and call recursive if (currlen == 0) { // Assign first bit str[0] = nextbit; // The next generated bit will wither be 0 or 1 generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); } else // If pattern "01" occurrence is < 2 if (occur < 2) { // Set next bit str[currlen] = nextbit; // If pattern "01" appears then // increase the occurrence of pattern if (str[currlen - 1] == 0 && nextbit == 1) { occur += 1; } generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); // Else pattern "01" occurrence equals 2 } else // If previous bit is 0 then next bit cannot be 1 if (str[currlen - 1] == 0 && nextbit == 1) { return; // Otherwise } else { str[currlen] = nextbit; generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); } } // Driver code public static void main(String[] args) { int n = 5; // Length of the resulting strings // must be at least 4 if (n < 4) { System.out.print(-1); } else { int[] str = new int[n]; // Generate all binary strings of length n // with sub-string "01" appearing twice generateBinStr(str, n, 0, 0, 0); generateBinStr(str, n, 0, 0, 1); } } } // This code has been contributed by 29AjayKumar |
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Python3
# Python3 implementation of the approach # Utility function to print the # given binary string def printBinStr(string, length): for i in range(0, length): print(string[i], end = "") print() # This function will be called recursively # to generate the next bit for given # binary string according to its current state def generateBinStr(string, length, currlen, occur, nextbit): # Base-case: if the generated binary # string meets the required length and # the pattern "01" appears twice if currlen == length: # nextbit needs to be 0 because each # time we call the function recursively, # we call 2 times for 2 cases: # next bit is 0 or 1 # The is to assure that the binary # string is printed one time only if occur == 2 and nextbit == 0: printBinStr(string, length) return # Generate the next bit for # str and call recursive if currlen == 0: # Assign first bit string[0] = nextbit # The next generated bit will # either be 0 or 1 generateBinStr(string, length, currlen + 1, occur, 0) generateBinStr(string, length, currlen + 1, occur, 1) else: # If pattern "01" occurrence is < 2 if occur < 2: # Set next bit string[currlen] = nextbit # If pattern "01" appears then # increase the occurrence of pattern if string[currlen - 1] == 0 and nextbit == 1: occur += 1 generateBinStr(string, length, currlen + 1, occur, 0) generateBinStr(string, length, currlen + 1, occur, 1) # Else pattern "01" occurrence equals 2 else: # If previous bit is 0 then next bit cannot be 1 if string[currlen - 1] == 0 and nextbit == 1: return # Otherwise else: string[currlen] = nextbit generateBinStr(string, length, currlen + 1, occur, 0) generateBinStr(string, length, currlen + 1, occur, 1) # Driver code if __name__ == "__main__": n = 5 # Length of the resulting strings # must be at least 4 if n < 4: print(-1) else: string = [None] * n # Generate all binary strings of length n # with sub-string "01" appearing twice generateBinStr(string, n, 0, 0, 0) generateBinStr(string, n, 0, 0, 1) # This code is contributed by Rituraj Jain |
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C#
// C# implementation of the above approach using System; class GFG { // Utility function to print the given binary string static void printBinStr(int[] str, int len) { for (int i = 0; i < len; i++) { Console.Write(str[i]); } Console.Write("\n"); } // This function will be called recursively // to generate the next bit for given // binary string according to its current state static void generateBinStr(int[] str, int len, int currlen, int occur, int nextbit) { // Base-case: if the generated binary string // meets the required length and the pattern "01" // appears twice if (currlen == len) { // nextbit needs to be 0 because each time // we call the function recursively, // we call 2 times for 2 cases: // next bit is 0 or 1 // The is to assure that the binary // string is printed one time only if (occur == 2 && nextbit == 0) { printBinStr(str, len); } return; } // Generate the next bit for str // and call recursive if (currlen == 0) { // Assign first bit str[0] = nextbit; // The next generated bit will wither be 0 or 1 generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); } else // If pattern "01" occurrence is < 2 if (occur < 2) { // Set next bit str[currlen] = nextbit; // If pattern "01" appears then // increase the occurrence of pattern if (str[currlen - 1] == 0 && nextbit == 1) { occur += 1; } generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); // Else pattern "01" occurrence equals 2 } else // If previous bit is 0 then next bit cannot be 1 if (str[currlen - 1] == 0 && nextbit == 1) { return; // Otherwise } else { str[currlen] = nextbit; generateBinStr(str, len, currlen + 1, occur, 0); generateBinStr(str, len, currlen + 1, occur, 1); } } // Driver code public static void Main(String[] args) { int n = 5; // Length of the resulting strings // must be at least 4 if (n < 4) { Console.Write(-1); } else { int[] str = new int[n]; // Generate all binary strings of length n // with sub-string "01" appearing twice generateBinStr(str, n, 0, 0, 0); generateBinStr(str, n, 0, 0, 1); } } } // This code is contributed by Princi Singh |
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Output:
00101 01001 01010 01011 01101 10101
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