Count of non-overlapping sub-strings “101” and “010” in the given binary string
Given binary string str, the task is to find the count of non-overlapping sub-strings of either the form “010” or “101”.
Examples:
Input: str = “10101010101”
Output: 3
str[0..2] = “101”
str[3..5] = “010”
str[6..8] = “101”
Input: str = “111111111111110”
Output: 0
Approach: Initialize count = 0 and for every index i in the given string check whether the sub-string of size 3 starting at the current index i matches either with “010” or “101”. If it’s a match then update count = count + 1 and i = i + 3 (to avoid overlapping of sub-strings) else increment i by 1.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> using namespace std; // Function to return the count of // required non-overlapping sub-strings int countSubStr(string &s, int n) { // To store the required count int count = 0; for ( int i = 0; i < n - 2;) { // If "010" matches the sub-string // starting at current index i if (s[i] == '0' && s[i + 1] == '1' && s[i + 2] == '0' ) { count++; i += 3; } // If "101" matches the sub-string // starting at current index i else if (s[i] == '1' && s[i + 1] == '0' && s[i + 2] == '1' ) { count++; i += 3; } else { i++; } } return count; } // Driver code int main() { string s = "10101010101" ; int n = s.length(); cout << countSubStr(s, n); return 0; } |
Java
// Java implementation of the approach class GFG { // Function to return the count of // required non-overlapping sub-strings static int countSubStr( char [] s, int n) { // To store the required count int count = 0 ; for ( int i = 0 ; i < n - 2 ? { // If "010" matches the sub-string // starting at current index i if (s[i] == '0' && s[i + 1 ] == '1' && s[i + 2 ] == '0' ) { count++; i += 3 ; } // If "101" matches the sub-string // starting at current index i else if (s[i] == '1' && s[i + 1 ] == '0' && s[i + 2 ] == '1' ) { count++; i += 3 ; } else { i++; } } return count; } // Driver code public static void main(String[] args) { char [] s = "10101010101" .toCharArray(); int n = s.length; System.out.println(countSubStr(s, n)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # Function to return the count of # required non-overlapping sub-strings def countSubStr(s, n) : # To store the required count count = 0 ; i = 0 while i < (n - 2 ) : # If "010" matches the sub-string # starting at current index i if (s[i] = = '0' and s[i + 1 ] = = '1' and s[i + 2 ] = = '0' ) : count + = 1 ; i + = 3 ; # If "101" matches the sub-string # starting at current index i elif (s[i] = = '1' and s[i + 1 ] = = '0' and s[i + 2 ] = = '1' ) : count + = 1 ; i + = 3 ; else : i + = 1 ; return count; # Driver code if __name__ = = "__main__" : s = "10101010101" ; n = len (s); print (countSubStr(s, n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approach using System; class GFG { // Function to return the count of // required non-overlapping sub-strings static int countSubStr( char [] s, int n) { // To store the required count int count = 0; for ( int i = 0; i < n - 2;) { // If "010" matches the sub-string // starting at current index i if (s[i] == '0' && s[i + 1] == '1' && s[i + 2] == '0' ) { count++; i += 3; } // If "101" matches the sub-string // starting at current index i else if (s[i] == '1' && s[i + 1] == '0' && s[i + 2] == '1' ) { count++; i += 3; } else { i++; } } return count; } // Driver code public static void Main(String[] args) { char [] s = "10101010101" .ToCharArray(); int n = s.Length; Console.WriteLine(countSubStr(s, n)); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // javascript implementation of the approach // Function to return the count of // required non-overlapping sub-strings function countSubStr( s , n) { // To store the required count var count = 0; for (i = 0; i < n - 2;) { // If "010" matches the sub-string // starting at current index i if (s[i] == '0' && s[i + 1] == '1' && s[i + 2] == '0' ) { count++; i += 3; } // If "101" matches the sub-string // starting at current index i else if (s[i] == '1' && s[i + 1] == '0' && s[i + 2] == '1' ) { count++; i += 3; } else { i++; } } return count; } // Driver code var s = "10101010101" ; var n = s.length; document.write(countSubStr(s, n)); // This code contributed by Rajput-Ji </script> |
Output:
3
Time Complexity: O(n), where n is the length of the string.
Auxiliary Space: O(1) as constant extra space is used