# 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:4Explanation:

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:10Explanation:

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(N ^{3})*

*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**:

- Initialize two pointers say
and**L**and initialize them to**R****0**. - 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**. - If the current character is
**0**, then the window**L to R – 1**contains all ones. - 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**. - Repeat the process until
**L and R**cross each other. - Print the count of all the substring in
**step 4**. - 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. - 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 <iostream>` `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<<s<<endl;` `}` `// Function to count number of` `// sub-strings of a given binary` `// string that contains only 0s & 1s` `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` ` ` `flip(s);` ` ` `cout<<s<<endl;` ` ` `// count of substring` ` ` `// which contains only 0s` ` ` `int` `only_0s = countSubAllOnes(s);` ` ` `return` `only_0s + only_1s;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given string str` ` ` `string s = ` `"011"` `;` ` ` `// Function Call` ` ` `cout << countSubAllZerosOnes(s) << endl;` ` ` `return` `0;` `}` |

## 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` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to count number of` `// sub-strings of a given binary` `// string that contains only 1` `function` `countSubAllOnes(s)` `{` ` ` `var` `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` `function` `flip(s)` `{` ` ` `for` `(` `var` `i = 0; s[i]; i++) {` ` ` `if` `(s[i] == ` `'1'` `)` ` ` `s[i] = ` `'0'` `;` ` ` `else` ` ` `s[i] = ` `'1'` `;` ` ` `}` ` ` ` ` `return` `s;` `}` `// Function to count number of` `// sub-strings of a given binary` `// string that contains only 0s & 1s` `function` `countSubAllZerosOnes(s)` `{` ` ` `// count of substring` ` ` `// which contains only 1s` ` ` `var` `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.split(` `''` `));` ` ` `// count of substring` ` ` `// which contains only 0s` ` ` `var` `only_0s = countSubAllOnes(s);` ` ` `return` `only_0s + only_1s;` `}` `// Driver Code` `// Given string str` `var` `s = ` `"011"` `;` `// Function Call` `document.write( countSubAllZerosOnes(s));` `// This code is contributed by famously.` `</script>` |

**Output:**

4

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

**Auxiliary Space:**

*O(1)*