# Find distinct characters in distinct substrings of a string

Given a string str, the task is to find the count of distinct characters in all the distinct sub-strings of the given string.

Examples:

Input: str = “ABCA”
Output: 18

Distinct sub-strings Distinct characters
A 1
AB 2
ABC 3
ABCA 3
B 1
BC 2
BCA 3
C 1
CA 2

Hence, 1 + 2 + 3 + 3 + 1 + 2 + 3 + 1 + 2 = 18

Input: str = “AAAB”
Output: 10

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Take all possible sub-strings of the given string and use a set to check whether the current sub-string has been processed before. Now, for every distinct sub-string, count the distinct characters in it (again set can be used to do so). The sum of this count for all the distinct sub-strings is the final answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of distinct ` `// characters in all the distinct ` `// sub-strings of the given string ` `int` `countTotalDistinct(string str) ` `{ ` `    ``int` `cnt = 0; ` ` `  `    ``// To store all the sub-strings ` `    ``set items; ` ` `  `    ``for` `(``int` `i = 0; i < str.length(); ++i) { ` ` `  `        ``// To store the current sub-string ` `        ``string temp = ``""``; ` ` `  `        ``// To store the characters of the ` `        ``// current sub-string ` `        ``set<``char``> ans; ` `        ``for` `(``int` `j = i; j < str.length(); ++j) { ` `            ``temp = temp + str[j]; ` `            ``ans.insert(str[j]); ` ` `  `            ``// If current sub-string hasn't ` `            ``// been stored before ` `            ``if` `(items.find(temp) == items.end()) { ` ` `  `                ``// Insert it into the set ` `                ``items.insert(temp); ` ` `  `                ``// Update the count of ` `                ``// distinct characters ` `                ``cnt += ans.size(); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `cnt; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"ABCA"``; ` ` `  `    ``cout << countTotalDistinct(str); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.HashSet; ` ` `  `class` `geeks ` `{ ` ` `  `    ``// Function to return the count of distinct ` `    ``// characters in all the distinct ` `    ``// sub-strings of the given string ` `    ``public` `static` `int` `countTotalDistinct(String str)  ` `    ``{ ` `        ``int` `cnt = ``0``; ` ` `  `        ``// To store all the sub-strings ` `        ``HashSet items = ``new` `HashSet<>(); ` ` `  `        ``for` `(``int` `i = ``0``; i < str.length(); ++i)  ` `        ``{ ` ` `  `            ``// To store the current sub-string ` `            ``String temp = ``""``; ` ` `  `            ``// To store the characters of the ` `            ``// current sub-string ` `            ``HashSet ans = ``new` `HashSet<>(); ` `            ``for` `(``int` `j = i; j < str.length(); ++j)  ` `            ``{ ` `                ``temp = temp + str.charAt(j); ` `                ``ans.add(str.charAt(j)); ` ` `  `                ``// If current sub-string hasn't ` `                ``// been stored before ` `                ``if` `(!items.contains(temp))  ` `                ``{ ` ` `  `                    ``// Insert it into the set ` `                    ``items.add(temp); ` ` `  `                    ``// Update the count of ` `                    ``// distinct characters ` `                    ``cnt += ans.size(); ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``return` `cnt; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``String str = ``"ABCA"``; ` `        ``System.out.println(countTotalDistinct(str)); ` `    ``} ` `} ` ` `  `// This code is contributed by ` `// sanjeev2552 `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the count of distinct  ` `# characters in all the distinct  ` `# sub-strings of the given string  ` `def` `countTotalDistinct(string) :  ` ` `  `    ``cnt ``=` `0``;  ` ` `  `    ``# To store all the sub-strings  ` `    ``items ``=` `set``();  ` ` `  `    ``for` `i ``in` `range``(``len``(string)) : ` ` `  `        ``# To store the current sub-string  ` `        ``temp ``=` `"";  ` ` `  `        ``# To store the characters of the  ` `        ``# current sub-string  ` `        ``ans ``=` `set``();  ` `        ``for` `j ``in` `range``(i, ``len``(string)) : ` `            ``temp ``=` `temp ``+` `string[j];  ` `            ``ans.add(string[j]);  ` ` `  `            ``# If current sub-string hasn't  ` `            ``# been stored before  ` `            ``if` `temp ``not` `in` `items : ` ` `  `                ``# Insert it into the set  ` `                ``items.add(temp);  ` ` `  `                ``# Update the count of  ` `                ``# distinct characters  ` `                ``cnt ``+``=` `len``(ans);  ` ` `  `    ``return` `cnt;  ` ` `  ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``string ``=` `"ABCA"``;  ` ` `  `    ``print``(countTotalDistinct(string));  ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `geeks ` `{ ` ` `  `    ``// Function to return the count of distinct ` `    ``// characters in all the distinct ` `    ``// sub-strings of the given string ` `    ``public` `static` `int` `countTotalDistinct(String str)  ` `    ``{ ` `        ``int` `cnt = 0; ` ` `  `        ``// To store all the sub-strings ` `        ``HashSet items = ``new` `HashSet(); ` ` `  `        ``for` `(``int` `i = 0; i < str.Length; ++i)  ` `        ``{ ` ` `  `            ``// To store the current sub-string ` `            ``String temp = ``""``; ` ` `  `            ``// To store the characters of the ` `            ``// current sub-string ` `            ``HashSet<``char``> ans = ``new` `HashSet<``char``>(); ` `            ``for` `(``int` `j = i; j < str.Length; ++j)  ` `            ``{ ` `                ``temp = temp + str[j]; ` `                ``ans.Add(str[j]); ` ` `  `                ``// If current sub-string hasn't ` `                ``// been stored before ` `                ``if` `(!items.Contains(temp))  ` `                ``{ ` ` `  `                    ``// Insert it into the set ` `                    ``items.Add(temp); ` ` `  `                    ``// Update the count of ` `                    ``// distinct characters ` `                    ``cnt += ans.Count; ` `                ``} ` `            ``} ` `        ``} ` ` `  `        ``return` `cnt; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{ ` `        ``String str = ``"ABCA"``; ` `        ``Console.WriteLine(countTotalDistinct(str)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```18
```

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