Count of all unique substrings with non-repeating characters
Last Updated :
24 Mar, 2023
Given a string str consisting of lowercase characters, the task is to find the total number of unique substrings with non-repeating characters.
Examples:
Input: str = "abba"
Output: 4
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 <bits/stdc++.h>
using namespace std;
// Function to count all unique
// distinct character substrings
int distinctSubstring(string& P, int N)
{
// Hashmap to store all substrings
unordered_set<string> 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 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<String> S = new HashSet<String>();
// 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 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
# substring till 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 substring in 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# 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<String> S = new HashSet<String>();
// 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[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[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
<script>
// JavaScript program to find the count of
// all unique sub-strings with
// non-repeating characters
// Function to count all unique
// distinct character substrings
function distinctSubstring(P, N)
{
// Hashmap to store all substrings
var S = new Set();
// Iterate over all the substrings
for (var i = 0; i < N; ++i) {
// Boolean array to maintain all
// characters encountered so far
var freq = Array(26).fill(false);
// Variable to maintain the
// substring till current position
var s = "";
for (var j = i; j < N; ++j) {
// Get the position of the
// character in the string
var pos = P[j].charCodeAt(0) - 'a'.charCodeAt(0);
// 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.size;
}
// Driver code
var S = "abba";
var N = S.length;
document.write( distinctSubstring(S, N));
</script>
Time Complexity: O(N2) where N is the length of the string.
Auxiliary Space: O(N2), to store all substrings in hashmap.
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