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

**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:

- Traverse the string
**str**character by character. - Count the number of characters till which a character from list L is not found. Let the count be
**N** - 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++

`// C++ implementation of the above approach ` `#include <bits/stdc++.h> ` `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; ` `} ` |

*chevron_right*

*filter_none*

## Java

`// 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 ` |

*chevron_right*

*filter_none*

## Python3

`# 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 ` |

*chevron_right*

*filter_none*

## C#

`// 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 ` |

*chevron_right*

*filter_none*

**Output:**

9

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Count of substrings formed using a given set of characters only
- Strings formed from given characters without any consecutive repeating characters
- K length words that can be formed from given characters without repetition
- Check if a Palindromic String can be formed by concatenating Substrings of two given Strings
- Check if a given string can be formed using characters of adjacent cells of a Matrix
- Count of times second string can be formed from the characters of first string
- Number of ways in which the substring in range [L, R] can be formed using characters out of the range
- Check if a string can be split into substrings starting with N followed by N characters
- Count of numbers upto N digits formed using digits 0 to K-1 without any adjacent 0s
- Count of strings that can be formed using a, b and c under given constraints
- Count of substrings of a given Binary string with all characters same
- Check whether second string can be formed from characters of first string
- Count of strings that can be formed from another string using each character at-most once
- Find the count of numbers that can be formed using digits 3, 4 only and having length at max N.
- Minimum number of substrings the given string can be splitted into that satisfy the given conditions
- Maximum count of Equilateral Triangles that can be formed within given Equilateral Triangle
- Find the number of strings formed using distinct characters of a given string
- Get K-th letter of the decoded string formed by repeating substrings
- Count number of substrings with exactly k distinct characters
- Count substrings with same first and last characters

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.