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# Count of sub-strings that do not consist of the given character

• Difficulty Level : Medium
• Last Updated : 18 May, 2021

Given a string str and a character c. The task is to find the number of sub-strings that do not consist of the character c
Examples:

Input: str = “baa”, c = ‘b’
Output:
The sub-strings are “a”, “a” and “aa”
Input: str = “ababaa”, C = ‘b’
Output:

Approach: Initially take a counter that counts the number of characters continuously with no character c. Iterate in the string and increase the counter till str[i] != c. Once str[i] == c, the number of sub-strings from the contiguous length cnt will be (cnt * (cnt + 1)) / 2. After the complete traversal of the string also add (cnt *(cnt + 1)) / 2 to the result for the group of characters appearing after the last occurrence of c.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the number``// of sub-strings that do not contain``// the given character c``int` `countSubstrings(string s, ``char` `c)``{` `    ``// Length of the string``    ``int` `n = s.length();` `    ``int` `cnt = 0;``    ``int` `sum = 0;` `    ``// Traverse in the string``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``// If current character is different``        ``// from the given character``        ``if` `(s[i] != c)``            ``cnt++;``        ``else` `{` `            ``// Update the number of sub-strings``            ``sum += (cnt * (cnt + 1)) / 2;` `            ``// Reset count to 0``            ``cnt = 0;``        ``}``    ``}` `    ``// For the characters appearing``    ``// after the last occurrence of c``    ``sum += (cnt * (cnt + 1)) / 2;``    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``string s = ``"baa"``;``    ``char` `c = ``'b'``;``    ``cout << countSubstrings(s, c);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{``    ` `// Function to return the number``// of sub-strings that do not contain``// the given character c``static` `int` `countSubstrings(String s, ``char` `c)``{` `    ``// Length of the string``    ``int` `n = s.length();` `    ``int` `cnt = ``0``;``    ``int` `sum = ``0``;` `    ``// Traverse in the string``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{` `        ``// If current character is different``        ``// from the given character``        ``if` `(s.charAt(i) != c)``            ``cnt++;``        ``else``        ``{` `            ``// Update the number of sub-strings``            ``sum += (cnt * (cnt + ``1``)) / ``2``;` `            ``// Reset count to 0``            ``cnt = ``0``;``        ``}``    ``}` `    ``// For the characters appearing``    ``// after the last occurrence of c``    ``sum += (cnt * (cnt + ``1``)) / ``2``;``    ``return` `sum;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String s = ``"baa"``;``    ``char` `c = ``'b'``;``    ``System.out.println(countSubstrings(s, c));``}``}` `// This code is contributed by Code_Mech.`

## Python3

 `# Python3 implementation of the approach` `# Function to return the number``# of sub-strings that do not contain``# the given character c``def` `countSubstrings(s, c):` `    ``# Length of the string``    ``n ``=` `len``(s)` `    ``cnt ``=` `0``    ``Sum` `=` `0` `    ``# Traverse in the string``    ``for` `i ``in` `range``(n):` `        ``# If current character is different``        ``# from the given character``        ``if` `(s[i] !``=` `c):``            ``cnt ``+``=` `1``        ``else``:` `            ``# Update the number of sub-strings``            ``Sum` `+``=` `(cnt ``*` `(cnt ``+` `1``)) ``/``/` `2` `            ``# Reset count to 0``            ``cnt ``=` `0``        ` `    ``# For the characters appearing``    ``# after the last occurrence of c``    ``Sum` `+``=` `(cnt ``*` `(cnt ``+` `1``)) ``/``/` `2``    ``return` `Sum` `# Driver code``s ``=` `"baa"``c ``=` `'b'``print``(countSubstrings(s, c))` `# This code is contributed``# by mohit kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// Function to return the number``// of sub-strings that do not contain``// the given character c``static` `int` `countSubstrings(``string` `s, ``char` `c)``{` `    ``// Length of the string``    ``int` `n = s.Length;` `    ``int` `cnt = 0;``    ``int` `sum = 0;` `    ``// Traverse in the string``    ``for` `(``int` `i = 0; i < n; i++)``    ``{` `        ``// If current character is different``        ``// from the given character``        ``if` `(s[i] != c)``            ``cnt++;``        ``else``        ``{` `            ``// Update the number of sub-strings``            ``sum += (cnt * (cnt + 1)) / 2;` `            ``// Reset count to 0``            ``cnt = 0;``        ``}``    ``}` `    ``// For the characters appearing``    ``// after the last occurrence of c``    ``sum += (cnt * (cnt + 1)) / 2;``    ``return` `sum;``}` `// Driver code``public` `static` `void` `Main()``{``    ``string` `s = ``"baa"``;``    ``char` `c = ``'b'``;``    ``Console.Write(countSubstrings(s, c));``}``}` `// This code is contributed by Akanksha Rai`

## PHP

 ``

## Javascript

 ``
Output:
`3`

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