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# Find the count of substrings in alphabetic order

Given a string of length consisting of lowercase alphabets. The task is to find the number of such substrings whose characters occur in alphabetical order. Minimum allowed length of substring is 2.

Examples

```Input : str = "refjhlmnbv"
Output : 2
Substrings are: "ef", "mn"

Input : str = "qwertyuiopasdfghjklzxcvbnm"
Output : 3```

For a substring to be in alphabetical order its character is in the same sequence as they occur in English alphabets. Also, the ASCII value of consecutive characters in such substring differs by exactly 1. For finding a total number of substrings that are in alphabetical order traverse the given string and compare two neighboring characters, if they are in alphabetic order increment the result and then find the next character in the string which is not in alphabetic order to its former character.

Algorithm :

Iterate over string length:

• if str[i]+1 == str[i+1] -> increase the result by 1 and iterate the string till next character which is out of alphabetic order
• else continue

Below is the implementation of the above approach:

## C++

 `// CPP to find the number of substrings``// in alphabetical order` `#include ``using` `namespace` `std;` `// Function to find number of substrings``int` `findSubstringCount(string str)``{``    ``int` `result = 0;``    ``int` `n = str.size();` `    ``// Iterate over string length``    ``for` `(``int` `i = 0; i < n - 1; i++) {``        ``// if any two chars are in alphabetic order``        ``if` `(str[i] + 1 == str[i + 1]) {``            ``result++;``            ``// find next char not in order``            ``while` `(str[i] + 1 == str[i + 1]) {``                ``i++;``            ``}``        ``}``    ``}` `    ``// return the result``    ``return` `result;``}` `// Driver function``int` `main()``{``    ``string str = ``"alphabet"``;` `    ``cout << findSubstringCount(str) << endl;` `    ``return` `0;``}`

## Java

 `// Java to find the number of substrings``// in alphabetical order``import` `java.util.*;``class` `Solution``{``  ` `// Function to find number of substrings``static` `int` `findSubstringCount(String str)``{``    ``int` `result = ``0``;``    ``int` `n = str.length();``  ` `    ``// Iterate over string length``    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) {``        ``// if any two chars are in alphabetic order``        ``if` `(str.charAt(i) + ``1` `== str.charAt(i+``1``)) {``            ``result++;``            ``// find next char not in order``            ``while` `(str.charAt(i) + ``1` `== str.charAt(i+``1``)) {``                ``i++;``            ``}``        ``}``    ``}``  ` `    ``// return the result``    ``return` `result;``}``  ` `// Driver function``public` `static` `void` `main(String args[])``{``    ``String str = ``"alphabet"``;``  ` `    ``System.out.println(findSubstringCount(str));``  ` `}` `}``//contributed by Arnab Kundu`

## Python3

 `# Python3 to find the number of substrings``# in alphabetical order` `# Function to find number of substrings``def` `findSubstringCount(``str``):` `    ``result ``=` `0``    ``n ``=` `len` `(``str``)` `    ``# Iterate over string length``    ``for` `i ``in` `range` `(n ``-` `1``) :``        ` `        ``# if any two chars are in alphabetic order``        ``if` `(``ord``(``str``[i]) ``+` `1` `=``=` `ord``(``str``[i ``+` `1``])) :``            ``result ``+``=` `1``            ` `            ``# find next char not in order``            ``while` `(``ord``(``str``[i]) ``+` `1` `=``=` `ord``(``str``[i ``+` `1``])) :``                ``i ``+``=` `1` `    ``# return the result``    ``return` `result` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``str` `=` `"alphabet"` `    ``print``(findSubstringCount(``str``))` `# This code is contributed by ChitraNayal`

## C#

 `using` `System;` `// C# to find the number of substrings ``// in alphabetical order ``public` `class` `Solution``{` `// Function to find number of substrings ``public` `static` `int` `findSubstringCount(``string` `str)``{``    ``int` `result = 0;``    ``int` `n = str.Length;` `    ``// Iterate over string length ``    ``for` `(``int` `i = 0; i < n - 1; i++)``    ``{``        ``// if any two chars are in alphabetic order ``        ``if` `((``char``)(str[i] + 1) == str[i + 1])``        ``{``            ``result++;``            ``// find next char not in order ``            ``while` `((``char``)(str[i] + 1) == str[i + 1])``            ``{``                ``i++;``            ``}``        ``}``    ``}` `    ``// return the result ``    ``return` `result;``}` `// Driver function ``public` `static` `void` `Main(``string``[] args)``{``    ``string` `str = ``"alphabet"``;` `    ``Console.WriteLine(findSubstringCount(str));` `}` `}` `// This code is contributed by Shrikant13`

## Javascript

 ``

Output

`1`

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1)

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