Count of substrings having all distinct characters

Given a string str consisting of lowercase alphabets, the task is to find the number of possible substrings (not necessarily distinct) that consists of distinct characters only.
Examples: 
 

Input: Str = “gffg” 
Output:
Explanation: 
All possible substrings from the given string are, 
( “g“, “gf“, “gff”, “gffg”, “f“, “ff”, “ffg”, “f“, “fg“, “g” ) 
Among them, the highlighted ones consists of distinct characters only.

Input: str = “gfg” 
Output:
Explanation: 
All possible substrings from the given string are, 
( “g“, “gf“, “gfg”, “f“, “fg“, “g” ) 
Among them, the highlighted consists of distinct characters only. 
 

Naive Approach: 
The simplest approach is to generate all possible substrings from the given string and check for each substring whether it contains all distinct characters or not. If the length of string is N, then there will be N*(N+1)/2 possible substrings. 
Time complexity: O(N3) 
Auxiliary Space: O(1)

Efficient Approach: 
The problem can be solved in linear time using Two Pointer Technique, with the help of counting frequencies of characters of the string.
Detailed steps for this approach are as follows: 

Below is the implementation of the above approach: 



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// C++ Program to implement
// the above appraoach
#include <bits/stdc++.h>
using namespace std;
  
// Function to count total
// number of valid substrings
long long int countSub(string str)
{
    int n = (int)str.size();
  
    // Stores the count of
    // substrings
    long long int ans = 0;
  
    // Stores the frequency
    // of characters
    int cnt[26];
  
    memset(cnt, 0, sizeof(cnt));
  
    // Initialised both pointers
    // to beginning of the string
    int i = 0, j = 0;
  
    while (i < n) {
  
        // If all characters in
        // substring from index i
        // to j are distinct
        if (j < n && (cnt[str[j] - 'a']
                      == 0)) {
  
            // Increment count of j-th
            // character
            cnt[str[j] - 'a']++;
  
            // Add all substring ending
            // at j and starting at any
            // index between i and j
            // to the answer
            ans += (j - i + 1);
  
            // Increment 2nd pointer
            j++;
        }
  
        // If some characters are repeated
        // or j pointer has reached to end
        else {
  
            // Decrement count of j-th
            // character
            cnt[str[i] - 'a']--;
  
            // Increment first pointer
            i++;
        }
    }
  
    // Return the final
    // count of substrings
    return ans;
}
  
// Driver Code
int main()
{
    string str = "gffg";
  
    cout << countSub(str);
  
    return 0;
}
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// Java program to implement
// the above appraoach
import java.util.*;
  
class GFG{
  
// Function to count total
// number of valid subStrings
static int countSub(String str)
{
    int n = (int)str.length();
  
    // Stores the count of
    // subStrings
    int ans = 0;
  
    // Stores the frequency
    // of characters
    int []cnt = new int[26];
  
    // Initialised both pointers
    // to beginning of the String
    int i = 0, j = 0;
  
    while (i < n)
    {
  
        // If all characters in
        // subString from index i
        // to j are distinct
        if (j < n && 
           (cnt[str.charAt(j) - 'a'] == 0))
        {
  
            // Increment count of j-th
            // character
            cnt[str.charAt(j) - 'a']++;
  
            // Add all subString ending
            // at j and starting at any
            // index between i and j
            // to the answer
            ans += (j - i + 1);
  
            // Increment 2nd pointer
            j++;
        }
  
        // If some characters are repeated
        // or j pointer has reached to end
        else 
        {
  
            // Decrement count of j-th
            // character
            cnt[str.charAt(i) - 'a']--;
  
            // Increment first pointer
            i++;
        }
    }
  
    // Return the final
    // count of subStrings
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    String str = "gffg";
  
    System.out.print(countSub(str));
}
}
  
// This code is contributed by amal kumar choubey
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// C# program to implement
// the above appraoach
using System;
  
class GFG{
  
// Function to count total
// number of valid subStrings
static int countSub(String str)
{
    int n = (int)str.Length;
  
    // Stores the count of
    // subStrings
    int ans = 0;
  
    // Stores the frequency
    // of characters
    int []cnt = new int[26];
  
    // Initialised both pointers
    // to beginning of the String
    int i = 0, j = 0;
  
    while (i < n)
    {
  
        // If all characters in
        // subString from index i
        // to j are distinct
        if (j < n && 
           (cnt[str[j] - 'a'] == 0))
        {
  
            // Increment count of j-th
            // character
            cnt[str[j] - 'a']++;
  
            // Add all subString ending
            // at j and starting at any
            // index between i and j
            // to the answer
            ans += (j - i + 1);
  
            // Increment 2nd pointer
            j++;
        }
  
        // If some characters are repeated
        // or j pointer has reached to end
        else
        {
  
            // Decrement count of j-th
            // character
            cnt[str[i] - 'a']--;
  
            // Increment first pointer
            i++;
        }
    }
  
    // Return the final
    // count of subStrings
    return ans;
}
  
// Driver Code
public static void Main(String[] args)
{
    String str = "gffg";
  
    Console.Write(countSub(str));
}
}
  
// This code is contributed by amal kumar choubey
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Output: 
6

Time complexity: O(N) 
Auxiliary Space: O(1)
 

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