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
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 <bits/stdc++.h> 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<string> 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; } |
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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<String> 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<Character> 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 |
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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 |
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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<String> items = new HashSet<String>(); 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 |
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Output:
18
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