# Count of substrings which contains a given character K times

Given a string consisting of numerical alphabets, a character C and an integer K, the task is to find the number of possible substrings which contains the character C exactly Ktimes.

Examples:

```Input : str = "212", c = '2', k = 1
Output : 4
Possible substrings are {"2", "21", "12", "2"}
that contains 2 exactly 1 time.

Input  : str = "55555", c = '5', k = 4
Output : 2
Possible substrings are {"5555", "5555"}
that contains 5 exactly 4 times
```

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

Naive Approach: A simple solution is to generate all the substrings of string and count the number of substrings in which given character occurs exactly k times.

The time complexity of this solution is O(N2) where N is the length of the string.

Efficient Approach: An efficient solution is to use sliding window technique. Find the substring that contains character C exactly K times starting with character C. Count the number of characters on either side of the substring. Multiply the counts to get the number of possible substrings with given substring

Below is the implementation of the above approach:

## C++

 `// C++ program to count the number of substrings ` `// which contains the character C exactly K times ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to count the number of substrings ` `// which contains the character C exactly K times ` `int` `countSubString(string s, ``char` `c, ``int` `k) ` `{ ` `    ``// left and right counters for characters on ` `    ``// both sides of substring window ` `    ``int` `leftCount = 0, rightCount = 0; ` ` `  `    ``// left and right pointer on both sides ` `    ``// of substring window ` `    ``int` `left = 0, right = 0; ` ` `  `    ``// intialize the frequency ` `    ``int` `freq = 0; ` ` `  `    ``// result and length of string ` `    ``int` `result = 0, len = s.length(); ` ` `  `    ``// initialize the left pointer ` `    ``while` `(s[left] != c && left < len) { ` `        ``left++; ` `        ``leftCount++; ` `    ``} ` ` `  `    ``// initialize the right pointer ` `    ``right = left + 1; ` `    ``while` `(freq != (k - 1) && (right - 1) < len) { ` `        ``if` `(s[right] == c) ` `            ``freq++; ` `        ``right++; ` `    ``} ` ` `  `    ``// traverse all the window substrings ` `    ``while` `(left < len && (right - 1) < len) { ` ` `  `        ``// counting the characters on leftSide ` `        ``// of substring window ` `        ``while` `(s[left] != c && left < len) { ` `            ``left++; ` `            ``leftCount++; ` `        ``} ` ` `  `        ``// counting the characters on rightSide of ` `        ``// substring window ` `        ``while` `(right < len && s[right] != c) { ` `            ``if` `(s[right] == c) ` `                ``freq++; ` `            ``right++; ` `            ``rightCount++; ` `        ``} ` ` `  `        ``// Add the possible substrings on both ` `        ``// sides to result ` `        ``result = result + (leftCount + 1) * (rightCount + 1); ` ` `  `        ``// Setting the frequency for next ` `        ``// substring window ` `        ``freq = k - 1; ` ` `  `        ``// reset the left, right counters ` `        ``leftCount = 0; ` `        ``rightCount = 0; ` ` `  `        ``left++; ` `        ``right++; ` `    ``} ` ` `  `    ``return` `result; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"3123231"``; ` `    ``char` `c = ``'3'``; ` `    ``int` `k = 2; ` ` `  `    ``cout << countSubString(s, c, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to count the number of substrings  ` `// which contains the character C exactly K times  ` `class` `GFG  ` `{ ` ` `  `    ``// Function to count the number of substrings  ` `    ``// which contains the character C exactly K times  ` `    ``static` `int` `countSubString(String s, ``char` `c, ``int` `k)  ` `    ``{ ` `        ``// left and right counters for characters on  ` `        ``// both sides of substring window  ` `        ``int` `leftCount = ``0``, rightCount = ``0``; ` ` `  `        ``// left and right pointer on both sides  ` `        ``// of substring window  ` `        ``int` `left = ``0``, right = ``0``; ` ` `  `        ``// intialize the frequency  ` `        ``int` `freq = ``0``; ` ` `  `        ``// result and length of string  ` `        ``int` `result = ``0``, len = s.length(); ` ` `  `        ``// initialize the left pointer  ` `        ``while` `(s.charAt(left) != c && left < len)  ` `        ``{ ` `            ``left++; ` `            ``leftCount++; ` `        ``} ` ` `  `        ``// initialize the right pointer  ` `        ``right = left + ``1``; ` `        ``while` `(freq != (k - ``1``) && (right - ``1``) < len)  ` `        ``{ ` `            ``if` `(s.charAt(right) == c) ` `            ``{ ` `                ``freq++; ` `            ``} ` `            ``right++; ` `        ``} ` ` `  `        ``// traverse all the window substrings  ` `        ``while` `(left < len && (right - ``1``) < len) ` `        ``{ ` ` `  `            ``// counting the characters on leftSide  ` `            ``// of substring window  ` `            ``while` `(s.charAt(left) != c && left < len)  ` `            ``{ ` `                ``left++; ` `                ``leftCount++; ` `            ``} ` ` `  `            ``// counting the characters on rightSide of  ` `            ``// substring window  ` `            ``while` `(right < len && s.charAt(right) != c) ` `            ``{ ` `                ``if` `(s.charAt(right) == c) ` `                ``{ ` `                    ``freq++; ` `                ``} ` `                ``right++; ` `                ``rightCount++; ` `            ``} ` ` `  `            ``// Add the possible substrings on both  ` `            ``// sides to result  ` `            ``result = result + (leftCount + ``1``) * (rightCount + ``1``); ` ` `  `            ``// Setting the frequency for next  ` `            ``// substring window  ` `            ``freq = k - ``1``; ` ` `  `            ``// reset the left, right counters  ` `            ``leftCount = ``0``; ` `            ``rightCount = ``0``; ` ` `  `            ``left++; ` `            ``right++; ` `        ``} ` ` `  `        ``return` `result; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[])  ` `    ``{ ` `        ``String s = ``"3123231"``; ` `        ``char` `c = ``'3'``; ` `        ``int` `k = ``2``; ` ` `  `        ``System.out.println(countSubString(s, c, k)); ` `    ``} ` `} ` ` `  `/* This code is contributed by PrinciRaj1992 */`

## Python3

 `# Python3 program to count the number of substrings ` `# which contains the character C exactly K times ` ` `  `# Function to count the number of substrings ` `# which contains the character C exactly K times ` `def` `countSubString(s, c, k): ` ` `  `    ``# left and right counters for characters  ` `    ``# on both sides of subwindow ` `    ``leftCount ``=` `0` `    ``rightCount ``=` `0` ` `  `    ``# left and right poer on both sides ` `    ``# of subwindow ` `    ``left ``=` `0` `    ``right ``=` `0` ` `  `    ``# ialize the frequency ` `    ``freq ``=` `0` ` `  `    ``# result and Length of string ` `    ``result ``=` `0` `    ``Len` `=` `len``(s) ` ` `  `    ``# initialize the left poer ` `    ``while` `(s[left] !``=` `c ``and` `left < ``Len``): ` `        ``left ``+``=` `1` `        ``leftCount ``+``=` `1` ` `  `    ``# initialize the right poer ` `    ``right ``=` `left ``+` `1` `    ``while` `(freq !``=` `(k ``-` `1``) ``and` `(right ``-` `1``) < ``Len``): ` `        ``if` `(s[right] ``=``=` `c): ` `            ``freq ``+``=` `1` `        ``right ``+``=` `1` ` `  `    ``# traverse all the window substrings ` `    ``while` `(left < ``Len` `and` `(right ``-` `1``) < ``Len``): ` ` `  `        ``# counting the characters on leftSide ` `        ``# of subwindow ` `        ``while` `(s[left] !``=` `c ``and` `left < ``Len``): ` `            ``left ``+``=` `1` `            ``leftCount ``+``=` `1` `             `  `        ``# counting the characters on rightSide of ` `        ``# subwindow ` `        ``while` `(right < ``Len` `and` `s[right] !``=` `c): ` `            ``if` `(s[right] ``=``=` `c): ` `                ``freq ``+``=` `1` `            ``right ``+``=` `1` `            ``rightCount ``+``=` `1` ` `  `        ``# Add the possible substrings on both ` `        ``# sides to result ` `        ``result ``=` `(result ``+` `(leftCount ``+` `1``) ``*`  `                           ``(rightCount ``+` `1``)) ` ` `  `        ``# Setting the frequency for next ` `        ``# subwindow ` `        ``freq ``=` `k ``-` `1` ` `  `        ``# reset the left, right counters ` `        ``leftCount ``=` `0` `        ``rightCount ``=` `0` ` `  `        ``left ``+``=` `1` `        ``right ``+``=` `1` ` `  `    ``return` `result ` ` `  `# Driver code ` `s ``=` `"3123231"` `c ``=` `'3'` `k ``=` `2` ` `  `print``(countSubString(s, c, k)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# program to count the number of substrings  ` `// which contains the character C exactly K times  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `         `  `    ``// Function to count the number of substrings  ` `    ``// which contains the character C exactly K times  ` `    ``static` `int` `countSubString(``string` `s, ``char` `c, ``int` `k)  ` `    ``{  ` `        ``// left and right counters for characters on  ` `        ``// both sides of substring window  ` `        ``int` `leftCount = 0, rightCount = 0;  ` `     `  `        ``// left and right pointer on both sides  ` `        ``// of substring window  ` `        ``int` `left = 0, right = 0;  ` `     `  `        ``// intialize the frequency  ` `        ``int` `freq = 0;  ` `     `  `        ``// result and length of string  ` `        ``int` `result = 0, len = s.Length;  ` `     `  `        ``// initialize the left pointer  ` `        ``while` `(s[left] != c && left < len)  ` `        ``{  ` `            ``left++;  ` `            ``leftCount++;  ` `        ``}  ` `     `  `        ``// initialize the right pointer  ` `        ``right = left + 1;  ` `        ``while` `(freq != (k - 1) && (right - 1) < len)  ` `        ``{  ` `            ``if` `(s[right] == c)  ` `                ``freq++;  ` `            ``right++;  ` `        ``}  ` `     `  `        ``// traverse all the window substrings  ` `        ``while` `(left < len && (right - 1) < len) ` `        ``{  ` `     `  `            ``// counting the characters on leftSide  ` `            ``// of substring window  ` `            ``while` `(s[left] != c && left < len)  ` `            ``{  ` `                ``left++;  ` `                ``leftCount++;  ` `            ``}  ` `     `  `            ``// counting the characters on rightSide of  ` `            ``// substring window  ` `            ``while` `(right < len && s[right] != c)  ` `            ``{  ` `                ``if` `(s[right] == c)  ` `                    ``freq++;  ` `                ``right++;  ` `                ``rightCount++;  ` `            ``}  ` `     `  `            ``// Add the possible substrings on both  ` `            ``// sides to result  ` `            ``result = result + (leftCount + 1) * (rightCount + 1);  ` `     `  `            ``// Setting the frequency for next  ` `            ``// substring window  ` `            ``freq = k - 1;  ` `     `  `            ``// reset the left, right counters  ` `            ``leftCount = 0;  ` `            ``rightCount = 0;  ` `     `  `            ``left++;  ` `            ``right++;  ` `        ``}  ` `     `  `        ``return` `result;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``static` `public` `void` `Main ()  ` `    ``{  ` `        ``string` `s = ``"3123231"``;  ` `        ``char` `c = ``'3'``;  ` `        ``int` `k = 2;  ` `     `  `        ``Console.WriteLine(countSubString(s, c, k));  ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```8
```

Time Complexity: O(N)

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