# Count of substrings of a given Binary string with all characters same

Given binary string str containing only 0 and 1, the task is to find the number of sub-strings containing only 1s and 0s respectively, i.e all characters same.

Examples:

Input: str = “011”
Output: 4
Explanation:
Three sub-strings are “1, “1”, “11” which have only 1 in them, and one substring is there which contains only “0”.

Input: str = “0000”
Output: 10
Explanation:
There are no sub-strings having all ones in it.

Naive Approach: The idea is to generate all possible sub-string of the given string. For each sub-string check if the string contains all 1’s or all 0’s. If yes then count that sub-string. Print the count of sub-string after the above operations.

Time Complexity: O(N3)
Auxiliary Space: O(N)

Efficient Approach: The idea is to use the concept of Sliding window and Two pointers Approach. Below are the steps to find the count of the substring that contains only 1s:

1. Initialize two pointers say L and R and initialize them to 0.
2. Now iterate in the given string and check if the current character is equal to 1 or not. If it is, then extend the window by incrementing the value of R.
3. If the current character is 0, then the window L to R – 1 contains all ones.
4. Add the number of sub-strings of L to R – 1 to the answer that is ((R – L) * (R – L + 1)) / 2 and increment R and reinitialize L as R.
5. Repeat the process until L and R cross each other.
6. Print the count of all the substring in step 4.
7. To count the number of substring with all 0s flip the given string i.e., all 0s will be converted into 1s and vice-versa.
8. Repeat the above steps from step 1 to step 4 for character 1 in the flipped string to get a count of the substring that contains only 0s in it and print the count.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to count number of` `// sub-strings of a given binary` `// string that contains only 1` `int` `countSubAllOnes(string s)` `{` `    ``int` `l = 0, r = 0, ans = 0;`   `    ``// Iterate untill L and R cross` `    ``// each other` `    ``while` `(l <= r) {`   `        ``// Check if reached the end` `        ``// of string` `        ``if` `(r == s.length()) {` `            ``ans += ((r - l) * (r - l + 1)) / 2;` `            ``break``;` `        ``}`   `        ``// Check if encountered '1'` `        ``// then extend window` `        ``if` `(s[r] == ``'1'``)` `            ``r++;`   `        ``// Check if encountered '0' then` `        ``// add number of strings of` `        ``// current window and change the` `        ``// values for both l and r` `        ``else` `{`   `            ``ans += ((r - l) * (r - l + 1)) / 2;` `            ``l = r + 1;` `            ``r++;` `        ``}` `    ``}`   `    ``// Return the answer` `    ``return` `ans;` `}`   `// Function to flip the bits of string` `void` `flip(string& s)` `{`   `    ``for` `(``int` `i = 0; s[i]; i++) {` `        ``if` `(s[i] == ``'1'``)` `            ``s[i] = ``'0'``;` `        ``else` `            ``s[i] = ``'1'``;` `    ``}` `  ``cout<

## Java

 `// Java program for the above approach` `import` `java.util.*;`   `class` `GFG{`   `// Function to count number of` `// sub-Strings of a given binary` `// String that contains only 1` `static` `int` `countSubAllOnes(String s)` `{` `    ``int` `l = ``0``, r = ``0``, ans = ``0``;`   `    ``// Iterate untill L and R cross` `    ``// each other` `    ``while` `(l <= r)` `    ``{` `        `  `        ``// Check if reached the end` `        ``// of String` `        ``if` `(r == s.length()) ` `        ``{` `            ``ans += ((r - l) * (r - l + ``1``)) / ``2``;` `            ``break``;` `        ``}`   `        ``// Check if encountered '1'` `        ``// then extend window` `        ``if` `(s.charAt(r) == ``'1'``)` `            ``r++;`   `        ``// Check if encountered '0' then` `        ``// add number of Strings of` `        ``// current window and change the` `        ``// values for both l and r` `        ``else` `        ``{` `            ``ans += ((r - l) * (r - l + ``1``)) / ``2``;` `            ``l = r + ``1``;` `            ``r++;` `        ``}` `    ``}`   `    ``// Return the answer` `    ``return` `ans;` `}`   `// Function to flip the bits of String` `static` `String flip(``char` `[]s)` `{` `    ``for``(``int` `i = ``0``; i < s.length; i++)` `    ``{` `        ``if` `(s[i] == ``'1'``)` `            ``s[i] = ``'0'``;` `        ``else` `            ``s[i] = ``'1'``;` `    ``}` `    ``return` `String.valueOf(s);` `}`   `// Function to count number of` `// sub-Strings of a given binary` `// String that contains only 0s & 1s` `static` `int` `countSubAllZerosOnes(String s)` `{`   `    ``// count of subString` `    ``// which contains only 1s` `    ``int` `only_1s = countSubAllOnes(s);`   `    ``// Flip the character of String s` `    ``// 0 to 1 and 1 to 0 to count the` `    ``// subString with consecutive 0s` `    ``s = flip(s.toCharArray());`   `    ``// count of subString` `    ``// which contains only 0s` `    ``int` `only_0s = countSubAllOnes(s);`   `    ``return` `only_0s + only_1s;` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    `  `    ``// Given String str` `    ``String s = ``"011"``;`   `    ``// Function call` `    ``System.out.print(countSubAllZerosOnes(s) + ``"\n"``);` `}` `}`   `// This code is contributed by Rohit_ranjan`

## Python3

 `# Python3 program for ` `# the above approach`   `# Function to count number of` `# sub-strings of a given binary` `# string that contains only 1` `def` `countSubAllOnes(s):` `  `  `    ``l, r, ans ``=` `0``, ``0``, ``0`   `    ``# Iterate untill L and R cross` `    ``# each other` `    ``while` `(l <``=` `r):`   `        ``# Check if reached the end` `        ``# of string` `        ``if` `(r ``=``=` `len``(s)):` `            ``ans ``+``=` `((r ``-` `l) ``*` `                    ``(r ``-` `l ``+` `1``)) ``/``/` `2` `            ``break` `       `  `        ``# Check if encountered '1'` `        ``# then extend window` `        ``if` `(s[r] ``=``=` `'1'``):` `            ``r ``+``=` `1`   `        ``# Check if encountered '0' then` `        ``# add number of strings of` `        ``# current window and change the` `        ``# values for both l and r` `        ``else` `:` `            ``ans ``+``=` `((r ``-` `l) ``*` `                    ``(r ``-` `l ``+` `1``)) ``/``/` `2` `            ``l ``=` `r ``+` `1` `            ``r ``+``=` `1`   `    ``# Return the answer` `    ``return` `ans`   `# Function to flip the bits of string` `def` `flip(s):` `     `  `    ``arr ``=` `list``(s)` `    ``for` `i ``in` `range` `(``len``(s)):` `        ``if` `(arr[i] ``=``=` `'1'``):` `            ``arr[i] ``=` `'0'` `        ``else``:` `            ``arr[i] ``=` `'1'` `    ``s ``=` `''.join(arr)` `    ``return` `s`   `# Function to count number of` `# sub-strings of a given binary` `# string that contains only 0s & 1s` `def` `countSubAllZerosOnes(s):`   `    ``# count of substring` `    ``# which contains only 1s` `    ``only_1s ``=` `countSubAllOnes(s)`   `    ``# Flip the character of string s` `    ``# 0 to 1 and 1 to 0 to count the` `    ``# substring with consecutive 0s` `    ``s ``=` `flip(s)`   `    ``# count of substring` `    ``# which contains only 0s` `    ``only_0s ``=` `countSubAllOnes(s)`   `    ``return` `only_0s ``+` `only_1s`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `  `  `    ``# Given string str` `    ``s  ``=` `"011"`   `    ``# Function Call` `    ``print` `(countSubAllZerosOnes(s))` `   `  `# This code is contributed by Chitranayal`

## C#

 `// C# program for the above approach` `using` `System;` `class` `GFG{`   `// Function to count number of` `// sub-Strings of a given binary` `// String that contains only 1` `static` `int` `countSubAllOnes(String s)` `{` `    ``int` `l = 0, r = 0, ans = 0;`   `    ``// Iterate untill L and R cross` `    ``// each other` `    ``while` `(l <= r)` `    ``{` `        `  `        ``// Check if reached the end` `        ``// of String` `        ``if` `(r == s.Length) ` `        ``{` `            ``ans += ((r - l) * (r - l + 1)) / 2;` `            ``break``;` `        ``}`   `        ``// Check if encountered '1'` `        ``// then extend window` `        ``if` `(s[r] == ``'1'``)` `            ``r++;`   `        ``// Check if encountered '0' then` `        ``// add number of Strings of` `        ``// current window and change the` `        ``// values for both l and r` `        ``else` `        ``{` `            ``ans += ((r - l) * (r - l + 1)) / 2;` `            ``l = r + 1;` `            ``r++;` `        ``}` `    ``}`   `    ``// Return the answer` `    ``return` `ans;` `}`   `// Function to flip the bits of String` `static` `String flip(``char` `[]s)` `{` `    ``for``(``int` `i = 0; i < s.Length; i++)` `    ``{` `        ``if` `(s[i] == ``'1'``)` `            ``s[i] = ``'0'``;` `        ``else` `            ``s[i] = ``'1'``;` `    ``}` `    ``return` `String.Join(``""``,s);` `}`   `// Function to count number of` `// sub-Strings of a given binary` `// String that contains only 0s & 1s` `static` `int` `countSubAllZerosOnes(String s)` `{`   `    ``// count of subString` `    ``// which contains only 1s` `    ``int` `only_1s = countSubAllOnes(s);`   `    ``// Flip the character of String s` `    ``// 0 to 1 and 1 to 0 to count the` `    ``// subString with consecutive 0s` `    ``s = flip(s.ToCharArray());`   `    ``// count of subString` `    ``// which contains only 0s` `    ``int` `only_0s = countSubAllOnes(s);`   `    ``return` `only_0s + only_1s;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    `  `    ``// Given String str` `    ``String s = ``"011"``;`   `    ``// Function call` `    ``Console.Write(countSubAllZerosOnes(s) + ``"\n"``);` `}` `}`   `// This code is contributed by Rohit_ranjan`

Output:

```4

```

Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(1)

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