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# Count number of substrings of a string consisting of same characters

Given a string. The task is to find out the number of substrings consisting of the same characters.

Examples:

Input: abba
Output:
The desired substrings are {a}, {b}, {b}, {a}, {bb}

Input: bbbcbb
Output: 10

Approach: It is known for a string of length n, there are a total of n*(n+1)/2 number of substrings.
Let’s initialize the result to 0. Traverse the string and find the number of consecutive element(let’s say count) of same characters. Whenever we find another character, increment the result by count*(count+1)/2, set count to 1, and from that index, repeat the above process.
Remember, for each different character, the number of our desired substring is 1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation``// of the above approach``#include ``using` `namespace` `std;` `// Function to return the``// number of substrings of``// same characters``void` `findNumbers(string s)``{``    ``if` `(s.empty()) ``return` `;``    ``// Size of the string``    ``int` `n = s.size();` `    ``// Initialize count to 1``    ``int` `count = 1;``    ``int` `result = 0;` `    ``// Initialize left to 0 and``    ``// right to 1 to traverse the``    ``// string``    ``int` `left = 0;``    ``int` `right = 1;` `    ``while` `(right < n) {` `        ``// Checking if consecutive``        ``// characters are same and``        ``// increment the count``        ``if` `(s[left] == s[right]) {``            ``count++;``        ``}` `        ``// When we encounter a``        ``// different characters``        ``else` `{` `            ``// Increment the result``            ``result += count * (count + 1) / 2;` `            ``// To repeat the whole``            ``// process set left equals``            ``// right and count variable to 1``            ``left = right;``            ``count = 1;``        ``}` `        ``right++;``    ``}` `    ``// Store the final``    ``// value of result``    ``result += count * (count + 1) / 2;` `    ``cout << result << endl;``}` `// Driver code``int` `main()``{``    ``string s = ``"bbbcbb"``;` `    ``findNumbers(s);``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `    ``// Function to return the``    ``// number of substrings of``    ``// same characters``    ``static` `void` `findNumbers(String s)``    ``{``        ``// Size of the string``        ``int` `n = s.length();``    ` `        ``// Initialize count to 1``        ``int` `count = ``1``;``        ``int` `result = ``0``;``    ` `        ``// Initialize left to 0 and``        ``// right to 1 to traverse the``        ``// string``        ``int` `left = ``0``;``        ``int` `right = ``1``;``    ` `        ``while` `(right < n)``        ``{``    ` `            ``// Checking if consecutive``            ``// characters are same and``            ``// increment the count``            ``if` `(s.charAt(left) == s.charAt(right))``            ``{``                ``count++;``            ``}``    ` `            ``// When we encounter a``            ``// different characters``            ``else``            ``{``    ` `                ``// Increment the result``                ``result += count * (count + ``1``) / ``2``;``    ` `                ``// To repeat the whole``                ``// process set left equals``                ``// right and count variable to 1``                ``left = right;``                ``count = ``1``;``            ``}``    ` `            ``right++;``        ``}``    ` `        ``// Store the final``        ``// value of result``        ``result += count * (count + ``1``) / ``2``;``    ` `        ``System.out.println(result);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main (String[] args)``    ``{``        ``String s = ``"bbbcbb"``;` `        ``findNumbers(s);``    ``}``}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the above approach` `# Function to return the number of``# substrings of same characters``def` `findNumbers(s):``    ` `    ``# Size of the string``    ``n ``=` `len``(s)` `    ``# Initialize count to 1``    ``count ``=` `1``    ``result ``=` `0` `    ``# Initialize left to 0 and right to 1``    ``# to traverse the string``    ``left ``=` `0``    ``right ``=` `1` `    ``while` `(right < n):` `        ``# Checking if consecutive``        ``# characters are same and``        ``# increment the count``        ``if` `(s[left] ``=``=` `s[right]):``            ``count ``+``=` `1` `        ``# When we encounter a``        ``# different characters``        ``else``:` `            ``# Increment the result``            ``result ``+``=` `count ``*` `(count ``+` `1``) ``/``/` `2` `            ``# To repeat the whole``            ``# process set left equals``            ``# right and count variable to 1``            ``left ``=` `right``            ``count ``=` `1` `        ``right ``+``=` `1` `    ``# Store the final value of result``    ``result ``+``=` `count ``*` `(count ``+` `1``) ``/``/` `2` `    ``print``(result)` `# Driver code``s ``=` `"bbbcbb"` `findNumbers(s)` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ``// Function to return the``    ``// number of substrings of``    ``// same characters``    ``static` `void` `findNumbers(String s)``    ``{``        ``// Size of the string``        ``int` `n = s.Length;` `        ``// Initialize count to 1``        ``int` `count = 1;``        ``int` `result = 0;` `        ``// Initialize left to 0 and``        ``// right to 1 to traverse the``        ``// string``        ``int` `left = 0;``        ``int` `right = 1;` `        ``while` `(right < n)``        ``{``            ``// Checking if consecutive``            ``// characters are same and``            ``// increment the count``            ``if` `(s[left] == s[right])``                ``count++;` `            ``// When we encounter a``            ``// different characters``            ``else``            ``{``                ``// Increment the result``                ``result += count * (count + 1) / 2;` `                ``// To repeat the whole``                ``// process set left equals``                ``// right and count variable to 1``                ``left = right;``                ``count = 1;``            ``}``            ``right++;``        ``}` `        ``// Store the final``        ``// value of result``        ``result += count * (count + 1) / 2;` `        ``Console.WriteLine(result);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String s = ``"bbbcbb"``;` `        ``findNumbers(s);``    ``}``}` `// This code is contributed by``// sanjeev2552`

## Javascript

 ``

Output

```10
```

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

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