# Count substrings that starts with character X and ends with character Y

Given a string str of n lowercase characters, the task is to count the number of substrings of str starting with character X and ending with character Y.

Examples:

```Input: str = "abbcaceghcak"
x = 'a', y = 'c'
Output: 5
abbc, abbcac, ac, abbcaceghc, aceghc

Input: str = "geeksforgeeks"
x = 'g', y = 'e'
Output: 6```

Approach:

• Initialize two counters i.e. tot_count to count the total number of substrings and count_x to count the number of strings that start with X.
• Start traversing the string.
• If the current character is X then increment the count of count_x.
• If the current character is Y, it means a string ends at Y so increment the count of tot_count i.e.
`tot_count = tot_count + count_x`
• It means that if there exists a Y then it will make a substring with all the X occurs before Y in the string. So, add the count of X to the total count.

Below is the implementation of above approach:

## C++

 `// C++ implementation to count substrings` `// starting with character X and ending` `// with character Y` `#include ` `using` `namespace` `std;`   `// function to count substrings starting with` `// character X and ending with character Y` `int` `countSubstr(string str, ``int` `n,` `                ``char` `x, ``char` `y)` `{` `    ``// to store total count of` `    ``// required substrings` `    ``int` `tot_count = 0;`   `    ``// to store count of character 'x'` `    ``// up to the point the string 'str'` `    ``// has been traversed so far` `    ``int` `count_x = 0;`   `    ``// traverse 'str' from left to right` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// if true, increment 'count_x'` `        ``if` `(str[i] == x)` `            ``count_x++;`   `        ``// if true accumulate 'count_x'` `        ``// to 'tot_count'` `        ``if` `(str[i] == y)` `            ``tot_count += count_x;` `    ``}`   `    ``// required count` `    ``return` `tot_count;` `}`   `// Driver code` `int` `main()` `{` `    ``string str = ``"abbcaceghcak"``;` `    ``int` `n = str.size();` `    ``char` `x = ``'a'``, y = ``'c'``;`   `    ``cout << ``"Count = "` `         ``<< countSubstr(str, n, x, y);`   `    ``return` `0;` `}`

## Java

 `// Java implementation to count ` `// substrings starting with ` `// character X and ending` `// with character Y` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;`   `class` `GFG` `{` `// function to count substrings ` `// starting with character X and` `// ending with character Y` `static` `int` `countSubstr(String str, ``int` `n,` `                       ``char` `x, ``char` `y)` `{` `    ``// to store total count of` `    ``// required substrings` `    ``int` `tot_count = ``0``;`   `    ``// to store count of character ` `    ``// 'x' up to the point the ` `    ``// string 'str' has been` `    ``// traversed so far` `    ``int` `count_x = ``0``;`   `    ``// traverse 'str' from` `    ``// left to right` `    ``for` `(``int` `i = ``0``; i < n; i++)` `    ``{`   `        ``// if true, increment 'count_x'` `        ``if` `(str.charAt(i) == x)` `            ``count_x++;`   `        ``// if true accumulate 'count_x'` `        ``// to 'tot_count'` `        ``if` `(str.charAt(i) == y)` `            ``tot_count += count_x;` `    ``}`   `    ``// required count` `    ``return` `tot_count;` `}`   `// Driver code` `public` `static` `void` `main(String args[])` `{` `    ``String str = ``"abbcaceghcak"``;` `    ``int` `n = str.length();` `    ``char` `x = ``'a'``, y = ``'c'``;`   `    ``System.out.print (``"Count = "` `+` `       ``countSubstr(str, n, x, y));` `}` `}`   `// This code is contributed` `// by Subhadeep`

## Python3

 `# Python 3 implementation to count substrings ` `# starting with character X and ending ` `# with character Y `   `# function to count substrings starting with ` `# character X and ending with character Y ` `def` `countSubstr(``str``, n, x, y):`   `    ``# to store total count of ` `    ``# required substrings ` `    ``tot_count ``=` `0`   `    ``# to store count of character 'x' ` `    ``# up to the point the string 'str' ` `    ``# has been traversed so far ` `    ``count_x ``=` `0`   `    ``# traverse 'str' from left to right` `    ``for` `i ``in` `range``(n):`   `        ``# if true, increment 'count_x' ` `        ``if` `str``[i] ``=``=` `x:` `            ``count_x ``+``=` `1`   `        ``# if true accumulate 'count_x' ` `        ``# to 'tot_count' ` `        ``if` `str``[i] ``=``=` `y:` `            ``tot_count ``+``=` `count_x` `    `  `    ``# required count` `    ``return` `tot_count`   `# Driver Code` `str` `=` `'abbcaceghcak'` `n ``=` `len``(``str``)` `x, y ``=` `'a'``, ``'c'` `print``(``'Count ='``, countSubstr(``str``, n, x, y))`   `# This code is contributed SamyuktaSHegde`

## C#

 `// C# implementation to count substrings ` `// starting with character X and ending` `// with character Y` `using` `System;`   `class` `GFG` `{` `// function to count substrings starting ` `// with character X and ending with character Y` `static` `int` `countSubstr(``string` `str, ``int` `n,` `                       ``char` `x, ``char` `y)` `{` `    ``// to store total count of` `    ``// required substrings` `    ``int` `tot_count = 0;`   `    ``// to store count of character 'x' up ` `    ``// to the point the string 'str' has ` `    ``// been traversed so far` `    ``int` `count_x = 0;`   `    ``// traverse 'str' from left to right` `    ``for` `(``int` `i = 0; i < n; i++)` `    ``{`   `        ``// if true, increment 'count_x'` `        ``if` `(str[i] == x)` `            ``count_x++;`   `        ``// if true accumulate 'count_x'` `        ``// to 'tot_count'` `        ``if` `(str[i] == y)` `            ``tot_count += count_x;` `    ``}`   `    ``// required count` `    ``return` `tot_count;` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``string` `str = ``"abbcaceghcak"``;` `    ``int` `n = str.Length;` `    ``char` `x = ``'a'``, y = ``'c'``;`   `    ``Console.Write(``"Count = "` `+` `    ``countSubstr(str, n, x, y));` `}` `}`   `// This code is contributed` `// by Akanksha Rai`

## PHP

 `

## Javascript

 ``

Output

`Count = 5`

Complexity Analysis:

• Time Complexity: O(n), to iterate over the array where n is the size of the given string
• Auxiliary Space: O(1),as no extra space is required

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next