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# Count of non-overlapping sub-strings “101” and “010” in the given binary string

• Last Updated : 03 May, 2021

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:
str[0..2] = “101”
str[3..5] = “010”
str[6..8] = “101”
Input: str = “111111111111110”
Output:

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 ``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

 ``
Output:
`3`

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