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Count of all unique substrings with non-repeating characters
• Difficulty Level : Medium
• Last Updated : 31 May, 2021

Given a string str consisting of lowercase characters, the task is to find the total numbers of unique substrings with non-repeating characters.
Examples:

Input: str = “abba”
Output:
Explanation:
There are 4 unique substrings. They are: “a”, “ab”, “b”, “ba”.
Input: str = “acbacbacaa”
Output: 10

Approach: The idea is to iterate over all the substrings. For every substring, check whether each particular character has previously occurred or not. If so, then increase the count of required substrings. In the end return this count as count of all unique substrings with non-repeating characters.
Below is the implementation of the above approach:

## CPP

 `// C++ program to find the count of``// all unique sub-strings with``// non-repeating characters` `#include ``using` `namespace` `std;` `// Function to count all unique``// distinct character substrings``int` `distinctSubstring(string& P, ``int` `N)``{``    ``// Hashmap to store all substrings``    ``unordered_set S;` `    ``// Iterate over all the substrings``    ``for` `(``int` `i = 0; i < N; ++i) {` `        ``// Boolean array to maintain all``        ``// characters encountered so far``        ``vector<``bool``> freq(26, ``false``);` `        ``// Variable to maintain the``        ``// substring till current position``        ``string s;` `        ``for` `(``int` `j = i; j < N; ++j) {` `            ``// Get the position of the``            ``// character in the string``            ``int` `pos = P[j] - ``'a'``;` `            ``// Check if the character is``            ``// encountred``            ``if` `(freq[pos] == ``true``)``                ``break``;` `            ``freq[pos] = ``true``;` `            ``// Add the current character``            ``// to the substring``            ``s += P[j];` `            ``// Insert substring in Hashmap``            ``S.insert(s);``        ``}``    ``}` `    ``return` `S.size();``}` `// Driver code``int` `main()``{``    ``string S = ``"abba"``;``    ``int` `N = S.length();` `    ``cout << distinctSubstring(S, N);` `    ``return` `0;``}`

## Java

 `// Java program to find the count of``// all unique sub-Strings with``// non-repeating characters``import` `java.util.*;` `class` `GFG{`` ` `// Function to count all unique``// distinct character subStrings``static` `int` `distinctSubString(String P, ``int` `N)``{``    ``// Hashmap to store all subStrings``    ``HashSet S = ``new` `HashSet();`` ` `    ``// Iterate over all the subStrings``    ``for` `(``int` `i = ``0``; i < N; ++i) {`` ` `        ``// Boolean array to maintain all``        ``// characters encountered so far``        ``boolean` `[]freq = ``new` `boolean``[``26``];`` ` `        ``// Variable to maintain the``        ``// subString till current position``        ``String s = ``""``;`` ` `        ``for` `(``int` `j = i; j < N; ++j) {`` ` `            ``// Get the position of the``            ``// character in the String``            ``int` `pos = P.charAt(j) - ``'a'``;`` ` `            ``// Check if the character is``            ``// encountred``            ``if` `(freq[pos] == ``true``)``                ``break``;`` ` `            ``freq[pos] = ``true``;`` ` `            ``// Add the current character``            ``// to the subString``            ``s += P.charAt(j);`` ` `            ``// Insert subString in Hashmap``            ``S.add(s);``        ``}``    ``}`` ` `    ``return` `S.size();``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``String S = ``"abba"``;``    ``int` `N = S.length();`` ` `    ``System.out.print(distinctSubString(S, N));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to find the count of``# all unique sub-strings with``# non-repeating characters` `# Function to count all unique``# distinct character substrings``def` `distinctSubstring(P, N):``    ` `    ``# Hashmap to store all substrings``    ``S ``=` `dict``()` `    ``# Iterate over all the substrings``    ``for` `i ``in` `range``(N):` `        ``# Boolean array to maintain all``        ``# characters encountered so far``        ``freq ``=` `[``False``]``*``26` `        ``# Variable to maintain the``        ``# subtill current position``        ``s ``=` `""` `        ``for` `j ``in` `range``(i,N):` `            ``# Get the position of the``            ``# character in the string``            ``pos ``=` `ord``(P[j]) ``-` `ord``(``'a'``)` `            ``# Check if the character is``            ``# encountred``            ``if` `(freq[pos] ``=``=` `True``):``                ``break` `            ``freq[pos] ``=` `True` `            ``# Add the current character``            ``# to the substring``            ``s ``+``=` `P[j]` `            ``# Insert subin Hashmap``            ``S[s] ``=` `1` `    ``return` `len``(S)` `# Driver code``S ``=` `"abba"``N ``=` `len``(S)` `print``(distinctSubstring(S, N))` `# This code is contributed by mohit kumar 29   `

## C#

 `// C# program to find the count of``// all unique sub-Strings with``// non-repeating characters``using` `System;``using` `System.Collections.Generic;` `class` `GFG{``  ` `// Function to count all unique``// distinct character subStrings``static` `int` `distinctSubString(String P, ``int` `N)``{``    ``// Hashmap to store all subStrings``    ``HashSet S = ``new` `HashSet();``  ` `    ``// Iterate over all the subStrings``    ``for` `(``int` `i = 0; i < N; ++i) {``  ` `        ``// Boolean array to maintain all``        ``// characters encountered so far``        ``bool` `[]freq = ``new` `bool``;``  ` `        ``// Variable to maintain the``        ``// subString till current position``        ``String s = ``""``;``  ` `        ``for` `(``int` `j = i; j < N; ++j) {``  ` `            ``// Get the position of the``            ``// character in the String``            ``int` `pos = P[j] - ``'a'``;``  ` `            ``// Check if the character is``            ``// encountred``            ``if` `(freq[pos] == ``true``)``                ``break``;``  ` `            ``freq[pos] = ``true``;``  ` `            ``// Add the current character``            ``// to the subString``            ``s += P[j];``  ` `            ``// Insert subString in Hashmap``            ``S.Add(s);``        ``}``    ``} ``    ``return` `S.Count;``}``  ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``String S = ``"abba"``;``    ``int` `N = S.Length;``  ` `    ``Console.Write(distinctSubString(S, N));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:

`4`

Time Complexity: O(N2) where N is the length of the string.

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