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

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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 = “zambiatek”, X = ‘e’
Output: 66

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 <bits/stdc++.h>
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

<?php
// PHP implementation of the approach 
 
// Function to return the count of 
// required sub-strings 
function countSubStr($str, $n, $x) 
{ 
    $res = 0; $count = 0; 
    for ($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 
$str = "abcabc"; 
$n = strlen($str); 
$x = 'c'; 
 
echo countSubStr($str, $n, $x); 
 
// This code is contributed by Ryuga
?>

Javascript

<script>
    // Javascript implementation of the approach
     
    // Function to return the count of
    // required sub-strings
    function countSubStr(str, n, x)
    {
        let res = 0, count = 0;
        for (let 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;
    }
     
    let str = "abcabc";
    let n = str.length;
    let x = 'c';
   
    document.write(countSubStr(str, n, x));
 
// This code is contributed by divyeshrabadiya07.
</script>

Output:

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

Auxiliary Space: O(1), no extra space required so it is a constant.

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