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# Count and Print the alphabets having ASCII value in the range [l, r]

Given a string str, the task is to count the number of alphabets having ASCII in the range [l, r].

Examples:

```Input: str = "geeksforgeeks"
l = 102, r = 111
Output: Count = 6, Characters = g, f, k, o
Characters - g, f, k, o have ascii values in the range [102, 111].

Input: str = "GeEkS"
l = 80, r = 111
Output: Count = 3, Characters = e, k, S ```

Approach: Start traversing the string and check if the current character has ASCII value less than equal to r and greater than equal to l. If yes then increment the count and print that element.

Below is the implementation of the an above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to count the number of``// characters whose ascii value is in range [l, r]``int` `CountCharacters(string str, ``int` `l, ``int` `r)``{``    ``// Initializing the count to 0``    ``int` `cnt = 0;` `    ``int` `len = str.length();``    ``for` `(``int` `i = 0; i < len; i++) {` `        ``// Increment the count``        ``// if the value is less``        ``if` `(l <= str[i] and str[i] <= r) {``            ``cnt++;``            ``cout << str[i] << ``" "``;``        ``}``    ``}` `    ``// return the count``    ``return` `cnt;``}` `// Driver code``int` `main()``{``    ``string str = ``"geeksforgeeks"``;``    ``int` `l = 102, r = 111;` `    ``cout << ``"Characters with ASCII values"``            ``" in the range [l, r] are \n"``;``    ``cout << ``"\nand their count is "` `<< CountCharacters(str, l, r);` `    ``return` `0;``}`

## Java

 `// Java implementation of the above approach``import` `java.util.*;``import` `java.lang.*;` `class` `GFG``{``// Function to count the number``// of characters whose ascii``// value is in range [l, r]``static` `int` `CountCharacters(String str,``                        ``int` `l, ``int` `r)``{``    ``// Initializing the count to 0``    ``int` `cnt = ``0``;` `    ``int` `len = str.length();``    ``for` `(``int` `i = ``0``; i < len; i++)``    ``{` `        ``// Increment the count``        ``// if the value is less``        ``if` `(l <= str.charAt(i) && str.charAt(i) <= r)``        ``{``            ``cnt++;``            ``System.out.print(str.charAt(i) + ``" "``);``        ``}``    ``}` `    ``// return the count``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``String str = ``"geeksforgeeks"``;``    ``int` `l = ``102``, r = ``111``;` `    ``System.out.print(``"Characters with ASCII values"` `+``                ``" in the range [l, r] are \n"``);``    ``System.out.print(``"\nand their count is "` `+``                ``CountCharacters(str, l, r));``}``}` `// This code is contributed``// by Akanksha Rai(Abby_akku)`

## Python3

 `# Python3 implementation of the above approach` `# Function to count the number of characters``# whose ascii value is in range [l, r]``def` `CountCharacters(str1, l, r):``    ` `    ``# Initializing the count to 0``    ``cnt ``=` `0``;` `    ``len1 ``=` `len``(str1)``    ``for` `i ``in` `str1:``        ` `        ``# Increment the count``        ``# if the value is less``        ``if` `(l <``=` `ord``(i) ``and` `ord``(i) <``=` `r):``            ``cnt ``=` `cnt ``+` `1``            ``print``(i, end ``=` `" "``)``        ` `    ``# return the count``    ``return` `cnt` `# Driver code``if` `__name__``=``=``'__main__'``:` `    ``str1 ``=` `"geeksforgeeks"``    ``l ``=` `102``    ``r ``=` `111``    ` `    ``print``(``"Characters with ASCII values "` `+``                ``"in the range [l, r] are"``)``    ``print``(``"\nand their count is "``,``            ``CountCharacters(str1, l, r))` `# This code is contributed by``# Kirti_Mangal`

## C#

 `// C# implementation of the above approach``using` `System;` `class` `GFG``{``// Function to count the number``// of characters whose ascii``// value is in range [l, r]``static` `int` `CountCharacters(``string` `str,``                           ``int` `l, ``int` `r)``{``    ``// Initializing the count to 0``    ``int` `cnt = 0;` `    ``int` `len = str.Length;``    ``for` `(``int` `i = 0; i < len; i++)``    ``{` `        ``// Increment the count``        ``// if the value is less``        ``if` `(l <= str[i] && str[i] <= r)``        ``{``            ``cnt++;``            ``Console.Write(str[i] + ``" "``);``        ``}``    ``}` `    ``// return the count``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `Main()``{``    ``string` `str = ``"geeksforgeeks"``;``    ``int` `l = 102, r = 111;` `    ``Console.Write(``"Characters with ASCII values"` `+``                   ``" in the range [l, r] are \n"``);``    ``Console.Write(``"\nand their count is "` `+``                  ``CountCharacters(str, l, r));``}``}` `// This code is contributed``// by Akanksha Rai(Abby_akku)`

## PHP

 ``

## Javascript

 ``

Output

```Characters with ASCII values in the range [l, r] are
g k f o g k
and their count is 6```

Complexity Analysis:

• Time Complexity: O(N), as we are using a loop to traverse N times so it will cost us O(N) time
• Auxiliary Space: O(1), as we are not using any extra space.

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