Count number of Distinct Substring in a String (original) (raw)

Last Updated : 06 Mar, 2025

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Given a string, count all distinct substrings of the given string.

**Examples:

**Input : abcd
**Output : abcd abc ab a bcd bc b cd c d
All Elements are Distinct

**Input : aaa
**Output : aaa aa a aa a a
All elements are not Distinct

**Prerequisite : Print subarrays of a given array

The idea is to use hash table (HashSet in Java) to store all generated substrings. Finally we return size of the HashSet.

**Implementation:

C++ `

// C++ program to count all distinct substrings in a string #include<bits/stdc++.h> using namespace std;

int distinctSubstring(string str) { // Put all distinct substring in a HashSet set result ;

// List All Substrings
for (int i = 0; i <= str.length(); i++)
{
    for (int j = 1; j <= str.length()-i; j++)
    {

        // Add each substring in Set
        result.insert(str.substr(i, j));
    }
}

// Return size of the HashSet
return result.size();

}

// Driver Code int main() { string str = "aaaa"; cout << (distinctSubstring(str)); }

// This code is contributed by Rajput-Ji

Java

// Java program to count all distinct substrings in a string import java.util.HashSet; import java.util.Iterator; import java.util.Set;

public class DistinctSubstring {

public static int distinctSubstring(String str)
{
    // Put all distinct substring in a HashSet
    Set<String> result = new HashSet<String>();

    // List All Substrings
    for (int i = 0; i <= str.length(); i++) {
        for (int j = i + 1; j <= str.length(); j++) {

            // Add each substring in Set
            result.add(str.substring(i, j));
        }
    }

    // Return size of the HashSet
    return result.size();
}

// Driver Code
public static void main(String[] args)
{
    String str = "aaaa";
    System.out.println(distinctSubstring(str));
}

}

Python

Python3 program to count all distinct substrings in a string

def distinctSubstring(str): # Put all distinct substring in a HashSet result = set()

# List All Substrings
for i in range(len(str)+1):
    for j in range( i + 1, len(str)+1):

        # Add each substring in Set
        result.add(str[i:j]);
    # Return size of the HashSet
return len(result);

Driver Code

if name == 'main': str = "aaaa"; print(distinctSubstring(str));

This code has been contributed by 29AjayKumar

C#

// C# program to count all distinct // substrings in a string using System; using System.Collections.Generic;

class DistinctSubstring { public static int distinctSubstring(String str) { // Put all distinct substring in a HashSet HashSet result = new HashSet();

    // List All Substrings
    for (int i = 0; i <= str.Length; i++) 
    {
        for (int j = i + 1; j <= str.Length; j++) 
        {

            // Add each substring in Set
            result.Add(str.Substring(i, j - i));
        }
    }

    // Return size of the HashSet
    return result.Count;
}

// Driver Code
public static void Main(String[] args)
{
    String str = "aaaa";
    Console.WriteLine(distinctSubstring(str));
}

}

// This code is contributed by 29AjayKumar

JavaScript

function distinctSubstring(str) { // Use a Set to store distinct substrings const result = new Set();

// List all substrings
for (let i = 0; i < str.length; i++) {
    for (let j = 1; j <= str.length - i; j++) {
        // Add each substring to the Set
        result.add(str.substring(i, i + j));
    }
}

// Return the size of the Set
return result.size;

}

// Driver Code const str = "aaaa"; console.log(distinctSubstring(str));

`

**Complexity Analysis:

• **Time Complexity: O(n 3 logn)
• **Auxiliary Space: O(n), since n extra space has been taken.

**How to print the distinct substrings?

C++ `

// C++ program to count all distinct // substrings in a string #include <bits/stdc++.h> using namespace std;

set distinctSubstring(string str) {

// Put all distinct substrings
// in the Hashset
set<string> result;

// List all substrings
for(int i = 0; i <= str.length(); i++)
{
    for(int j = i + 1; j <= str.length(); j++)
    {

        // Add each substring in Set
        result.insert(str.substr(i, j));
    }
}

// Return the hashset
return result;

}

// Driver code int main() { string str = "aaaa"; set subs = distinctSubstring(str);

cout << "Distinct Substrings are: \n";
for(auto i : subs)
    cout << i << endl;

}

// This code is contributed by Ronak Mangal

Java

// Java program to count all distinct substrings in a string import java.util.HashSet; import java.util.Iterator; import java.util.Set;

public class DistinctSubstring {

public static Set<String> distinctSubstring(String str)
{

    // Put all distinct substring in a HashSet
    Set<String> result = new HashSet<String>();

    // List All Substrings
    for (int i = 0; i <= str.length(); i++) {
        for (int j = i + 1; j <= str.length(); j++) {

            // Add each substring in Set
            result.add(str.substring(i, j));
        }
    }

    // Return the HashSet
    return result;
}

// Driver Code
public static void main(String[] args)
{
    String str = "aaaa";
    Set<String> subs = distinctSubstring(str);

    System.out.println("Distinct Substrings are: ");
    for (String s : subs) {
        System.out.println(s);
    }
}

}

Python

Python3 program to count all distinct

substrings in a string

def distinctSubstring(str):

# Put all distinct substring in a HashSet
result = set();

# List All Substrings
for i in range(len(str)):
    for j in range(i + 1, len(str) + 1):

        # Add each substring in Set
        result.add(str[i:j]);

    # Return the HashSet
return result;

Driver Code

if name == 'main':

str = "aaaa";
subs = distinctSubstring(str);

print("Distinct Substrings are: ");
for s in subs:
    print(s);

This code is contributed by 29AjayKumar

C#

// C# program to count all distinct // substrings in a string using System; using System.Collections.Generic;

class GFG { public static HashSet distinctSubstring(String str) {

    // Put all distinct substring in a HashSet
    HashSet<String> result = new HashSet<String>();

    // List All Substrings
    for (int i = 0; i <= str.Length; i++) 
    {
        for (int j = i + 1; j <= str.Length; j++) 
        {

            // Add each substring in Set
            result.Add(str.Substring(i, j - i));
        }
    }

    // Return the HashSet
    return result;
}

// Driver Code
public static void Main(String[] args)
{
    String str = "aaaa";
    HashSet<String> subs = distinctSubstring(str);

    Console.WriteLine("Distinct Substrings are: ");
    foreach (String s in subs) 
    {
        Console.WriteLine(s);
    }
}

}

// This code is contributed by 29AjayKumar

JavaScript

function distinctSubstring(str) { // Use a Set to store distinct substrings const result = new Set();

// List all substrings
for (let i = 0; i < str.length; i++) {
    for (let j = i + 1; j <= str.length; j++) {
        // Add each substring to the Set
        result.add(str.substring(i, j));
    }
}

// Return the Set containing distinct substrings
return result;

}

// Driver Code const str = "aaaa"; const subs = distinctSubstring(str);

console.log("Distinct Substrings are: "); subs.forEach(substring => { console.log(substring); });

`

Output

Distinct Substrings are: a aa aaa aaaa

**Complexity Analysis:

• **Time Complexity: O(n 3 logn)
• **Auxiliary Space: O(n)

**Optimization: We can further optimize the above code. The substr() function works in linear time. We can use append current character to previous substring to get the current substring.

**Implementation:

C++ `

// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;

// Function to return the count of // valid sub-strings void printSubstrings(string s) {

// To store distinct output substrings
unordered_set<string> us;

// Traverse through the given string and
// one by one generate substrings beginning
// from s[i].
for (int i = 0; i < s.size(); ++i) {

    // One by one generate substrings ending
    // with s[j]
    string ss = "";
    for (int j = i; j < s.size(); ++j) {

        ss = ss + s[j];
        us.insert(ss);
    }
}

// Print all substrings one by one
for (auto s : us)
    cout << s << " ";

}

// Driver code int main() { string str = "aaabc"; printSubstrings(str); return 0; }

Java

// Java implementation of the approach import java.util.*;

class GFG {

// Function to return the count of // valid sub-Strings static void printSubStrings(String s) {

// To store distinct output subStrings
HashSet<String> us = new HashSet<String>();

// Traverse through the given String and
// one by one generate subStrings beginning
// from s[i].
for (int i = 0; i < s.length(); ++i)
{

    // One by one generate subStrings ending
    // with s[j]
    String ss = "";
    for (int j = i; j < s.length(); ++j) 
    {
        ss = ss + s.charAt(j);
        us.add(ss);
    }
}

// Print all subStrings one by one
for (String str : us)
    System.out.print(str + " ");

}

// Driver code public static void main(String[] args) { String str = "aaabc"; printSubStrings(str); } }

// This code is contributed by Rajput-Ji

Python

Python3 implementation of the approach

Function to return the count of

valid sub-Strings

def printSubStrings(s):

# To store distinct output subStrings
us = set();

# Traverse through the given String and
# one by one generate subStrings beginning
# from s[i].
for i in range(len(s)):

    # One by one generate subStrings ending
    # with s[j]
    ss = "";
    for j in range(i, len(s)):
        ss = ss + s[j];
        us.add(ss);
    
# Print all subStrings one by one
for str in us:
    print(str, end=" ");

Driver code

if name == 'main': str = "aaabc"; printSubStrings(str);

This code is contributed by 29AjayKumar

C#

// C# implementation of the approach using System; using System.Collections.Generic;

class GFG {

// Function to return the count of // valid sub-Strings static void printSubStrings(String s) {

// To store distinct output subStrings
HashSet<String> us = new HashSet<String>();

// Traverse through the given String and
// one by one generate subStrings 
// beginning from s[i].
for (int i = 0; i < s.Length; ++i)
{

    // One by one generate subStrings
    // ending with s[j]
    String ss = "";
    for (int j = i; j < s.Length; ++j) 
    {
        ss = ss + s[j];
        us.Add(ss);
    }
}

// Print all subStrings one by one
foreach (String str in us)
    Console.Write(str + " ");

}

// Driver code public static void Main(String[] args) { String str = "aaabc"; printSubStrings(str); } }

// This code is contributed by Rajput-Ji

JavaScript

function printSubstrings(s) { // Set to store distinct substrings const us = new Set();

// Traverse through the given string and generate substrings
for (let i = 0; i < s.length; i++) {
    let ss = "";
    for (let j = i; j < s.length; j++) {
        ss += s[j];  // Append characters to form the substring
        us.add(ss);  // Add the substring to the set
    }
}

// Print all distinct substrings
us.forEach(substring => {
    console.log(substring);
});

}

// Driver code const str = "aaabc"; printSubstrings(str);

`

Output

bc b abc ab aabc aa aaa c a aaab aab aaabc

**Complexity Analysis:

• **Time Complexity: O(n 2 )
• **Auxiliary Space: O(n)

**Space Optimization using **Trie Data Structure (when we just need count of distinct substrings)

The above approach makes use of hashing which may lead to memory limit exceeded (MLE) in case of very large strings. The approximate space complexity of them is around O(n^3) as there can be n(n+1)/2 substrings which is around O(n^2) and each substring can be at least of 1 length or n length, i.e O(n/2) average case. This makes the total space complexity to be O(n^3).

We can improve this using Trie. The idea is to insert characters that are not already present in the Trie. And when such addition happens we know that this string is occurring for the first time and thus we print it. And if some characters of the string is already present we just move on to the next node without reading them which helps us on saving space.

The time complexity for this approach is O(n^2) similar to previous approach but the space reduces to O(n)*26.

**Implementation:

C++ `

#include <bits/stdc++.h> using namespace std;

class TrieNode { public: bool isWord; TrieNode* child[26];

TrieNode()
{
    isWord = 0;
    for (int i = 0; i < 26; i++) {
        child[i] = 0;
    }
}

};

int countDistinctSubstring(string str) { TrieNode* head = new TrieNode();

// will hold the count of unique substrings
int count = 0;
// included count of substr " "

for (int i = 0; i < str.length(); i++) {
    TrieNode* temp = head;

    for (int j = i; j < str.length(); j++) {
        // when char not present add it to the trie
        if (temp->child[str[j] - 'a'] == NULL) {
            temp->child[str[j] - 'a'] = new TrieNode();
            temp->isWord = 1;
            count++;
        }
        // move on to the next char
        temp = temp->child[str[j] - 'a'];
    }
}

return count;

}

int main() { int count = countDistinctSubstring("aaabc");

cout << "Count of Distinct Substrings: " << count
     << endl;

return 0;

}

Java

/*package whatever //do not write package name here */

import java.io.*;

class GFG { static class TrieNode { TrieNode children[]; boolean isEnd;

    TrieNode()
    {
        this.children = new TrieNode[26];
        this.isEnd = false;
    }
}

static TrieNode root = new TrieNode();

static void insert(String str)
{
    TrieNode cur = root;
    for (char ch : str.toCharArray()) {
        int idx = ch - 'a';

        if (cur.children[idx] == null)
            cur.children[idx] = new TrieNode();

        cur = cur.children[idx];
    }
    cur.isEnd = true;
}

public static int distinctSubstringCount(String str)
{
    // will hold the count of unique substrings
    int cnt = 0;
    for (int i = 0; i <= str.length(); i++) {
        // start from root of trie each time as new
        // starting point
        TrieNode temp = root;
        for (int j = i; j < str.length(); j++) {
            char ch = str.charAt(j);
            // when char not present add it to the trie
            if (temp.children[ch - 'a'] == null) {
                temp.children[ch - 'a']
                    = new TrieNode();
                temp.isEnd = true;
                cnt++;
            }
            // move on to the next char
            temp = temp.children[ch - 'a'];
        }
    }
    return cnt;
}

public static void main(String[] args)
{
    int cnt = distinctSubstringCount("aaabc");
    System.out.println("Count of distinct substrings: "
                       + cnt);
}

}

Python

Python code for the above approach:

class TrieNode: def init(self): self.isWord = False self.child = [None] * 26

def countDistinctSubstring(string): head = TrieNode()

# will hold the count of unique substrings
count = 0
# included count of substr " "

for i in range(len(string)):
    temp = head

    for j in range(i, len(string)):
        # when char not present add it to the trie
        if not temp.child[ord(string[j]) - ord('a')]:
            temp.child[ord(string[j]) - ord('a')] = TrieNode()
            temp.isWord = True
            count += 1
        # move on to the next char
        temp = temp.child[ord(string[j]) - ord('a')]

return count

count = countDistinctSubstring("aaabc") print("Count of Distinct Substrings:", count)

This code is contributed by lokesh.

C#

using System; using System.Collections.Generic;

class GFG {

class TrieNode { public bool isWord; public TrieNode[] child = new TrieNode[26]; public TrieNode() { this.isWord=false; for (int i = 0; i < 26; i++) { child[i] = null; } } }

static int countDistinctSubstring(string str) { TrieNode head = new TrieNode();

// will hold the count of unique substrings
int count = 0;
// included count of substr " "

for (int i = 0; i < str.Length; i++) {
  TrieNode temp = head;

  for (int j = i; j < str.Length; j++)
  {
    
    // when char not present add it to the trie
    if (temp.child[str[j] - 'a'] == null) {
      temp.child[str[j] - 'a'] = new TrieNode();
      temp.isWord = true;
      count++;
    }
    // move on to the next char
    temp = temp.child[str[j] - 'a'];
  }
}

return count;

}

public static void Main() { int count = countDistinctSubstring("aaabc");

Console.Write("Count of Distinct Substrings: " + count);

} }

// This code is contributed by poojaagarwal2.

JavaScript

class TrieNode { constructor() { this.isWord = false; this.child = new Array(26).fill(null); } }

function countDistinctSubstring(str) { let head = new TrieNode(); let count = 0; for (let i = 0; i < str.length; i++) { let temp = head; for (let j = i; j < str.length; j++) { if (temp.child[str.charCodeAt(j) - 'a'.charCodeAt(0)] === null) { temp.child[str.charCodeAt(j) - 'a'.charCodeAt(0)] = new TrieNode(); temp.isWord = true; count++; } temp = temp.child[str.charCodeAt(j) - 'a'.charCodeAt(0)]; } } return count; }

console.log("Count of Distinct Substrings: " + countDistinctSubstring("aaabc"));

`

Output

Count of Distinct Substrings: 12

**Complexity Analysis:

• **Time Complexity: O(n 2 )
• **Auxiliary Space: O(n 2 )