Count all substrings having character K

Given a string str and a character K, the task is to find the count of all the substrings of str that contain the character K.
Examples:

Input: str = “geeks”, K = ‘g’ 
Output: 5 
“g”, “ge”, “gee”, “geek” and “geeks” are the valid substrings.
Input: str = “geeksforgeeks”, K = ‘k’ 
Output: 56 

Naive approach A simple approach will be to find all the substrings having character K of the given string and return the count;
Efficient approach: For every index i in the string, find the first index j such that i ? j and str[j] = K. Now, the substrings str[i…j], str[i…j + 1], str[i…j + 2], …, str[i…n – 1] will all contain the character K at least once. The approach seems to be O(n²) at first but the index j will not be calculated again for every index i, j will be a valid index for all the values of i less than j.
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 index of the
// next occurrence of character ch in str
// starting from the given index
int nextOccurrence(string str, int n,
                   int start, char ch)
{
    for (int i = start; i < n; i++) {

        // Return the index of the first
        // occurrence of ch
        if (str[i] == ch)
            return i;
    }

    // No occurrence found
    return -1;
}

// Function to return the count of all
// the substrings of str which contain
// the character ch at least one
int countSubStr(string str, int n, char ch)
{

    // To store the count of valid substrings
    int cnt = 0;

    // Index of the first occurrence of ch in str
    int j = nextOccurrence(str, n, 0, ch);
    for (int i = 0; i < n; i++) {
        while (j != -1 && j < i) {
            j = nextOccurrence(str, n, j + 1, ch);
        }

        // No occurrence of ch after index i in str
        if (j == -1)
            break;

        // Substrings starting at index i
        // and ending at indices j, j+1, ..., n-1
        // are all valid substring
        cnt += (n - j);
    }

    return cnt;
}

// Driver code
int main()
{

    string str = "geeksforgeeks";
    int n = str.length();
    char ch = 'k';

    cout << countSubStr(str, n, ch);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

    // Function to return the index of the
    // next occurrence of character ch in str
    // starting from the given index
    static int nextOccurrence(String str, int n,
                              int start, char ch)
    {
        for (int i = start; i < n; i++)
        {

            // Return the index of the first
            // occurrence of ch
            if (str.charAt(i) == ch)
                return i;
        }

        // No occurrence found
        return -1;
    }

    // Function to return the count of all
    // the substrings of str which contain
    // the character ch at least one
    static int countSubStr(String str,
                           int n, char ch)
    {

        // To store the count of valid substrings
        int cnt = 0;

        // Index of the first occurrence of ch in str
        int j = nextOccurrence(str, n, 0, ch);
        for (int i = 0; i < n; i++)
        {
            while (j != -1 && j < i)
            {
                j = nextOccurrence(str, n, j + 1, ch);
            }

            // No occurrence of ch after index i in str
            if (j == -1)
                break;

            // Substrings starting at index i
            // and ending at indices j, j+1, ..., n-1
            // are all valid substring
            cnt += (n - j);
        }
        return cnt;
    }

    // Driver code
    static public void main ( String []arg)
    {
        String str = "geeksforgeeks";
        int n = str.length();
        char ch = 'k';

        System.out.println(countSubStr(str, n, ch));
    }
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach

# Function to return the index of the
# next occurrence of character ch in strr
# starting from the given index
def nextOccurrence(strr, n, start, ch):
    for i in range(start, n):

        # Return the index of the first
        # occurrence of ch
        if (strr[i] == ch):
            return i

    # No occurrence found
    return -1

# Function to return the count of all
# the substrings of strr which contain
# the character ch at least one
def countSubStr(strr, n, ch):

    # To store the count of valid substrings
    cnt = 0

    # Index of the first occurrence of ch in strr
    j = nextOccurrence(strr, n, 0, ch)

    for i in range(n):
        while (j != -1 and j < i):
            j = nextOccurrence(strr, n, j + 1, ch)

        # No occurrence of ch after index i in strr
        if (j == -1):
            break

        # Substrings starting at index i
        # and ending at indices j, j+1, ..., n-1
        # are all valid substring
        cnt += (n - j)

    return cnt

# Driver code
strr = "geeksforgeeks"
n = len(strr)
ch = 'k'

print(countSubStr(strr, n, ch))

# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the index of the
    // next occurrence of character ch in str
    // starting from the given index
    static int nextOccurrence(String str, int n,
                               int start, char ch)
    {
        for (int i = start; i < n; i++)
        {

            // Return the index of the first
            // occurrence of ch
            if (str[i] == ch)
                return i;
        }

        // No occurrence found
        return -1;
    }

    // Function to return the count of all
    // the substrings of str which contain
    // the character ch at least one
    static int countSubStr(String str,
                           int n, char ch)
    {

        // To store the count of valid substrings
        int cnt = 0;

        // Index of the first occurrence of ch in str
        int j = nextOccurrence(str, n, 0, ch);
        for (int i = 0; i < n; i++)
        {
            while (j != -1 && j < i)
            {
                j = nextOccurrence(str, n, j + 1, ch);
            }

            // No occurrence of ch after index i in str
            if (j == -1)
                break;

            // Substrings starting at index i
            // and ending at indices j, j+1, ..., n-1
            // are all valid substring
            cnt += (n - j);
        }
        return cnt;
    }

    // Driver code
    static public void Main ()
    {
        String str = "geeksforgeeks";
        int n = str.Length;
        char ch = 'k';

        Console.WriteLine(countSubStr(str, n, ch));
    }
}

// This code is contributed by AnkitRai01

Javascript

<script>
      // JavaScript implementation of the approach
      // Function to return the index of the
      // next occurrence of character ch in str
      // starting from the given index
      function nextOccurrence(str, n, start, ch) {
        for (var i = start; i < n; i++) {
          // Return the index of the first
          // occurrence of ch
          if (str[i] === ch) return i;
        }

        // No occurrence found
        return -1;
      }

      // Function to return the count of all
      // the substrings of str which contain
      // the character ch at least one
      function countSubStr(str, n, ch) {
        // To store the count of valid substrings
        var cnt = 0;

        // Index of the first occurrence of ch in str
        var j = nextOccurrence(str, n, 0, ch);
        for (var i = 0; i < n; i++) {
          while (j !== -1 && j < i) {
            j = nextOccurrence(str, n, j + 1, ch);
          }

          // No occurrence of ch after index i in str
          if (j === -1) break;

          // Substrings starting at index i
          // and ending at indices j, j+1, ..., n-1
          // are all valid substring
          cnt += n - j;
        }
        return cnt;
      }

      // Driver code
      var str = "geeksforgeeks";
      var n = str.length;
      var ch = "k";

      document.write(countSubStr(str, n, ch));
    </script>

Output:

56