# Count of sub-strings that contain character X at least once

Given a string str and a character X. The task is to find the total number of sub-strings that contain the character X at least once.

Examples:

Input: str = “abcd”, X = ‘b’
Output: 6
“ab”, “abc”, “abcd”, “b”, “bc” and “bcd” are the required sub-strings.

Input: str = “geeksforgeeks”, X = ‘e’
Output: 66

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Total number of sub-strings are n * (n + 1) / 2, among them only those sub-strings need to be counted which contain character X. Character X is present in those sub-strings from position of X to the length of the string. For each character before X this sub-string must be counted.

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 sub-strings ` `int` `countSubStr(string str, ``int` `n, ``char` `x) ` `{ ` `    ``int` `res = 0, count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``if` `(str[i] == x) { ` ` `  `            ``// Number of sub-strings from position ` `            ``// of current x to the end of str ` `            ``res += ((count + 1) * (n - i)); ` ` `  `            ``// To store the number of characters ` `            ``// before x ` `            ``count = 0; ` `        ``} ` `        ``else` `            ``count++; ` `    ``} ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"abcabc"``; ` `    ``int` `n = str.length(); ` `    ``char` `x = ``'c'``; ` ` `  `    ``cout << countSubStr(str, n, x); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG  ` `{ ` ` `  `// Function to return the count of ` `// required sub-strings ` `static` `int` `countSubStr(String str, ``int` `n, ``char` `x) ` `{ ` `    ``int` `res = ``0``, count = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``if` `(str.charAt(i) == x) ` `        ``{ ` ` `  `            ``// Number of sub-strings from position ` `            ``// of current x to the end of str ` `            ``res += ((count + ``1``) * (n - i)); ` ` `  `            ``// To store the number of characters ` `            ``// before x ` `            ``count = ``0``; ` `        ``} ` `        ``else` `            ``count++; ` `    ``} ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String str = ``"abcabc"``; ` `    ``int` `n = str.length(); ` `    ``char` `x = ``'c'``; ` ` `  `    ``System.out.println(countSubStr(str, n, x)); ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python implementation of the approach  ` `  `  `# Function to return the count of  ` `# required sub-strings  ` `def` `countSubStr(``str``, n, x):  ` `    ``res ``=` `0``; count ``=` `0``;  ` `    ``for` `i ``in` `range``(n):  ` `        ``if` `(``str``[i] ``=``=` `x):  ` `  `  `            ``# Number of sub-strings from position  ` `            ``# of current x to the end of str  ` `            ``res ``+``=` `((count ``+` `1``) ``*` `(n ``-` `i));  ` `  `  `            ``# To store the number of characters  ` `            ``# before x  ` `            ``count ``=` `0``;  ` `        ``else``: ` `            ``count``+``=``1``;  ` `  `  `    ``return` `res;  ` ` `  `# Driver code  ` `str` `=` `"abcabc"``;  ` `n ``=` `len``(``str``);  ` `x ``=` `'c'``;  ` `  `  `print``(countSubStr(``str``, n, x)); ` ` `  `# This code contributed by PrinciRaj1992 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the count of ` `// required sub-strings ` `static` `int` `countSubStr(String str, ``int` `n, ``char` `x) ` `{ ` `    ``int` `res = 0, count = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``if` `(str[i] == x) ` `        ``{ ` ` `  `            ``// Number of sub-strings from position ` `            ``// of current x to the end of str ` `            ``res += ((count + 1) * (n - i)); ` ` `  `            ``// To store the number of characters ` `            ``// before x ` `            ``count = 0; ` `        ``} ` `        ``else` `            ``count++; ` `    ``} ` ` `  `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``String str = ``"abcabc"``; ` `    ``int` `n = str.Length; ` `    ``char` `x = ``'c'``; ` ` `  `    ``Console.WriteLine(countSubStr(str, n, x)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`

## PHP

 ` `

Output:

```15
```

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