# Count of Substrings that can be formed without using the given list of Characters

Given a string str and a list of characters L, the task is to count total numbers of substrings of the string str without using characters given in the list L.

Examples:

Input: str = “abcd”, L[] = {‘a’, ‘b’, ‘t’, ‘q’}
Output: 3
Explanation:
On ignoring characters ‘a’ and ‘b’ from the given string, substring “cd” is left.
Therefore, the total number of substrings formed with “cd” are:
(2 * (2 + 1)) / 2 = 3

Input: str = “abcpxyz”, L[] = {‘a’, ‘p’, ‘q’}
Output: 9
Explanation:
On ignoring characters ‘a’ and ‘p’ from the given string, substrings “bc” and “xyz” are left.
Therefore, total number of substrings formed with the substrings are:
(2*(2+1))/2 + (3*(3+1))/2 = 3 + 6 = 9

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

Approach: The total number of substrings for a given string of length N is given by the formula

`(N * (N + 1)) / 2`

The idea is to use the above formula and follow the below steps to compute the answer:

1. Traverse the string str character by character.
2. Count the number of characters till which a character from list L is not found. Let the count be N
3. Once a character from L is encountered, compute (N * (N + 1) / 2) and add it to the answer and reset count N to zero.

Below is the implementation of the above approach:

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find the Number of sub-strings ` `// without using given character ` `int` `countSubstring(string& S, ``char` `L[], ``int``& n) ` `{ ` `    ``int` `freq[26] = { 0 }, ans = 0; ` ` `  `    ``// Mark the given characters in ` `    ``// the freq array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``freq[(``int``)(L[i] - ``'a'``)] = 1; ` `    ``} ` ` `  `    ``// Count variable to store the count ` `    ``// of the characters until a character ` `    ``// from given L is encountered ` `    ``int` `count = 0; ` ` `  `    ``for` `(``auto` `x : S) { ` ` `  `        ``// If a character from L is encountered, ` `        ``// then the answer variable is incremented by ` `        ``// the value obtained by using ` `        ``// the mentioned formula and count is set to 0 ` `        ``if` `(freq[(``int``)(x - ``'a'``)]) { ` `            ``ans += (count * count + count) / 2; ` `            ``count = 0; ` `        ``} ` `        ``else` `            ``count++; ` `    ``} ` ` `  `    ``// For last remaining characters ` `    ``ans += (count * count + count) / 2; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``string S = ``"abcpxyz"``; ` `    ``char` `L[] = { ``'a'``, ``'p'``, ``'q'` `}; ` `    ``int` `n = ``sizeof``(L) / ``sizeof``(L[0]); ` ` `  `    ``cout << countSubstring(S, L, n); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to find the Number of sub-Strings ` `// without using given character ` `static` `int` `countSubString(``char` `[]S, ``char` `L[], ``int` `n) ` `{ ` `    ``int` `[]freq = ``new` `int``[``26``]; ` `    ``int` `ans = ``0``; ` ` `  `    ``// Mark the given characters in ` `    ``// the freq array ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``freq[(``int``)(L[i] - ``'a'``)] = ``1``; ` `    ``} ` ` `  `    ``// Count variable to store the count ` `    ``// of the characters until a character ` `    ``// from given L is encountered ` `    ``int` `count = ``0``; ` ` `  `    ``for` `(``int` `x : S) ` `    ``{ ` ` `  `        ``// If a character from L is encountered, ` `        ``// then the answer variable is incremented by ` `        ``// the value obtained by using ` `        ``// the mentioned formula and count is set to 0 ` `        ``if` `(freq[(``int``)(x - ``'a'``)] > ``0``)  ` `        ``{ ` `            ``ans += (count * count + count) / ``2``; ` `            ``count = ``0``; ` `        ``} ` `        ``else` `            ``count++; ` `    ``} ` ` `  `    ``// For last remaining characters ` `    ``ans += (count * count + count) / ``2``; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String S = ``"abcpxyz"``; ` `    ``char` `L[] = { ``'a'``, ``'p'``, ``'q'` `}; ` `    ``int` `n = L.length; ` ` `  `    ``System.out.print(countSubString(S.toCharArray(), L, n)); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

 `# Python3 implementation of the above approach ` ` `  `# Function to find the Number of sub-strings ` `# without using given character ` `def` `countSubstring(S, L,n): ` `    ``freq ``=` `[``0` `for` `i ``in` `range``(``26``)] ` `     `  `    ``# the freq array ` `    ``for` `i ``in` `range``(n): ` `        ``freq[(``ord``(L[i]) ``-` `ord``(``'a'``))] ``=` `1` ` `  `    ``# Count variable to store the count ` `    ``# of the characters until a character ` `    ``# from given L is encountered ` `    ``count,ans ``=` `0``,``0` ` `  `    ``for` `x ``in` `S: ` ` `  `        ``# If a character from L is encountered, ` `        ``# then the answer variable is incremented by ` `        ``# the value obtained by using ` `        ``# the mentioned formula and count is set to 0 ` `        ``if` `(freq[``ord``(x) ``-` `ord``(``'a'``)]): ` `            ``ans ``+``=` `(count ``*` `count ``+` `count) ``/``/` `2` `            ``count ``=` `0` `        ``else``: ` `            ``count ``+``=` `1` ` `  `    ``# For last remaining characters ` `    ``ans ``+``=` `(count ``*` `count ``+` `count) ``/``/` `2` ` `  `    ``return` `ans ` ` `  `# Driver code ` ` `  `S ``=` `"abcpxyz"` `L ``=` `[``'a'``, ``'p'``, ``'q'``] ` `n ``=` `len``(L) ` ` `  `print``(countSubstring(S, L, n)) ` ` `  `# This code is contributed by mohit kumar 29 `

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to find the Number of sub-Strings ` `    ``// without using given character ` `    ``static` `int` `countSubString(``char` `[]S, ``char` `[]L, ``int` `n) ` `    ``{ ` `        ``int` `[]freq = ``new` `int``[26]; ` `        ``int` `ans = 0; ` `     `  `        ``// Mark the given characters in ` `        ``// the freq array ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``freq[(``int``)(L[i] - ``'a'``)] = 1; ` `        ``} ` `     `  `        ``// Count variable to store the count ` `        ``// of the characters until a character ` `        ``// from given L is encountered ` `        ``int` `count = 0; ` `     `  `        ``foreach` `(``int` `x ``in` `S) ` `        ``{ ` `     `  `            ``// If a character from L is encountered, ` `            ``// then the answer variable is incremented by ` `            ``// the value obtained by using ` `            ``// the mentioned formula and count is set to 0 ` `            ``if` `(freq[(``int``)(x - ``'a'``)] > 0)  ` `            ``{ ` `                ``ans += (count * count + count) / 2; ` `                ``count = 0; ` `            ``} ` `            ``else` `                ``count++; ` `        ``} ` `     `  `        ``// For last remaining characters ` `        ``ans += (count * count + count) / 2; ` `     `  `        ``return` `ans; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``string` `S = ``"abcpxyz"``; ` `        ``char` `[]L = { ``'a'``, ``'p'``, ``'q'` `}; ` `        ``int` `n = L.Length; ` `     `  `        ``Console.WriteLine(countSubString(S.ToCharArray(), L, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Yash_R `

Output:
```9
```

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Improved By : mohit kumar 29, Rajput-Ji, Yash_R

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