# Count of occurrences of a “1(0+)1” pattern in a string

Given an alphanumeric string, find the number of times a pattern 1(0+)1 occurs in the given string. Here, (0+) signifies the presence of non empty sequence of consecutive 0’s.

Examples:

```Input : 1001010001
Output : 3
First sequence is in between 0th and 3rd index.
Second sequence is in between 3rd and 5th index.
Third sequence is in between 5th and 9th index.
So total number of sequences comes out to be 3.

Input : 1001ab010abc01001
Output : 2
First sequence is in between 0th and 3rd index.
Second valid sequence is in between 13th and 16th
index. So total number of sequences comes out to
be 2.```

The idea to solve this problem is to first find a ‘1’ and keep moving forward in the string and check as mentioned below:

1. If any character other than ‘0’ and ‘1’ is obtained then it means pattern is not valid. So we go on in the search of next ‘1’ from this index and repeat these steps again.
2. If a ‘1’ is seen, then check for the presence of ‘0’ at previous position to check the validity of sequence.

Below is the implementation of above idea:

## C++

 `// C++ program to calculate number of times` `// the pattern occurred in given string` `#include` `using` `namespace` `std;`   `// Returns count of occurrences of "1(0+)1"` `// int str.` `int` `countPattern(string str)` `{` `    ``int` `len = str.size();` `    ``bool` `oneSeen = 0;`   `    ``int` `count = 0;  ``// Initialize result` `    ``for` `(``int` `i = 0; i < len ; i++)` `    ``{`   `        ``// Check if encountered '1' forms a valid` `        ``// pattern as specified` `        ``if` `(str[i] == ``'1'` `&& oneSeen == 1)` `            ``if` `(str[i - 1] == ``'0'``)` `                ``count++;`   `        ``// if 1 encountered for first time` `        ``// set oneSeen to 1` `        ``if` `(str[i] == ``'1'` `&& oneSeen == 0)` `           ``{` `            ``oneSeen = 1;` `            ``continue``;` `           ``}` `        ``// Check if there is any other character` `        ``// other than '0' or '1'. If so then set` `        ``// oneSeen to 0 to search again for new` `        ``// pattern` `        ``if` `(str[i] != ``'0'` `&& str[i] != ``'1'``)` `            ``oneSeen = 0;`     `    ``}`   `    ``return` `count;` `}`   `// Driver program to test above function` `int` `main()` `{` `    ``string str = ``"100001abc101"``;` `    ``cout << countPattern(str);` `    ``return` `0;` `}`

## Java

 `//Java program to calculate number of times` `//the pattern occurred in given string` `public` `class` `GFG` `{` `    ``// Returns count of occurrences of "1(0+)1"` `    ``// int str.` `    ``static` `int` `countPattern(String str)` `    ``{` `        ``int` `len = str.length();` `        ``boolean` `oneSeen = ``false``;` `        `  `        ``int` `count = ``0``;  ``// Initialize result` `        ``for``(``int` `i = ``0``; i < len ; i++)` `        ``{` `            ``char` `getChar = str.charAt(i);` `            `  `            ``// Check if encountered '1' forms a valid` `            ``// pattern as specified` `            ``if` `(getChar == ``'1'` `&& oneSeen == ``true``){` `                ``if` `(str.charAt(i - ``1``) == ``'0'``)` `                    ``count++;` `            ``}`   `            ``// if 1 encountered for first time` `            ``// set oneSeen to 1` `            ``if``(getChar == ``'1'` `&& oneSeen == ``false``)` `                ``oneSeen = ``true``;` `            `  `            ``// Check if there is any other character` `            ``// other than '0' or '1'. If so then set` `            ``// oneSeen to 0 to search again for new` `            ``// pattern` `            ``if``(getChar != ``'0'` `&& str.charAt(i) != ``'1'``)` `                ``oneSeen = ``false``;` `            `    `        ``}` `        ``return` `count;` `    ``}`   `    ``// Driver program to test above function` `    ``public` `static` `void` `main(String[] args)` `    ``{` `         ``String str = ``"100001abc101"``;` `         ``System.out.println(countPattern(str));` `    ``}`   `}` `// This code is contributed by Sumit Ghosh`

## Python3

 `# Python program to calculate number of times` `# the pattern occurred in given string`   `# Returns count of occurrences of "1(0+)1"` `def` `countPattern(s):` `    ``length ``=` `len``(s)` `    ``oneSeen ``=` `False` `    `  `    ``count ``=` `0`   `# Initialize result` `    ``for` `i ``in` `range``(length):`   `        ``# Check if encountered '1' forms a valid` `        ``# pattern as specified` `        ``if` `(s[i] ``=``=` `'1'` `and` `oneSeen):` `            ``if` `(s[i ``-` `1``] ``=``=` `'0'``):` `                ``count ``+``=` `1`   `        ``# if 1 encountered for first time` `        ``# set oneSeen to 1` `        ``if` `(s[i] ``=``=` `'1'` `and` `oneSeen ``=``=` `0``):` `            ``oneSeen ``=` `True` ` `  `        ``# Check if there is any other character` `        ``# other than '0' or '1'. If so then set` `        ``# oneSeen to 0 to search again for new` `        ``# pattern` `        ``if` `(s[i] !``=` `'0'` `and` `s[i] !``=` `'1'``):` `            ``oneSeen ``=` `False` ` `    ` `  `    ``return` `count`   `# Driver code` `s ``=` `"100001abc101"` `print``(countPattern(s))`   `# This code is contributed by Sachin Bisht`

## C#

 `// C# program to calculate number` `// of times the pattern occurred ` `// in given string ` `using` `System;`   `class` `GFG` `{` `// Returns count of occurrences ` `// of "1(0+)1" int str. ` `public` `static` `int` `countPattern(``string` `str)` `{` `    ``int` `len = str.Length;` `    ``bool` `oneSeen = ``false``;`   `    ``int` `count = 0; ``// Initialize result` `    ``for` `(``int` `i = 0; i < len ; i++)` `    ``{` `        ``char` `getChar = str[i];`   `        ``// Check if encountered '1' forms ` `        ``// a valid pattern as specified ` `        ``if` `(getChar == ``'1'` `&& ` `                 ``oneSeen == ``true``)` `        ``{` `            ``if` `(str[i - 1] == ``'0'``)` `            ``{` `                ``count++;` `            ``}` `        ``}`   `        ``// if 1 encountered for first` `        ``// time set oneSeen to 1 ` `        ``if` `(getChar == ``'1'` `&& ` `            ``oneSeen == ``false``)` `        ``{` `            ``oneSeen = ``true``;` `        ``}`   `        ``// Check if there is any other character ` `        ``// other than '0' or '1'. If so then set ` `        ``// oneSeen to 0 to search again for new ` `        ``// pattern ` `        ``if` `(getChar != ``'0'` `&& ` `                 ``str[i] != ``'1'``)` `        ``{` `            ``oneSeen = ``false``;` `        ``}`     `    ``}` `    ``return` `count;` `}`   `// Driver Code` `public` `static` `void` `Main(``string``[] args)` `{` `    ``string` `str = ``"100001abc101"``;` `    ``Console.WriteLine(countPattern(str));` `}`   `}`   `// This code is contributed ` `// by Shrikant13`

## PHP

 ``

## Javascript

 ``

Output

`2`

Time Complexity: O( N ), where N is the length of input string.
Auxiliary Space: O(1), as extra space is not required

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