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: 6 
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: 5 
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: 

• Consider two pointers i and j, initially both pointing to the first character of the string i.e. i = j = 0.
• Initialize an array Cnt[ ] to store the count of characters in substring from index i to j both inclusive.
• Now, keep on incrementing j pointer until some a repeated character is encountered. While incrementing j, add the count of all the substrings ending at j^th index and starting at any index between i and j to the answer. All these substrings will contain distinct characters as no character is repeated in them.
• If some repeated character is encountered in substring between index i to j, to exclude this repeated character, keep on incrementing the i pointer until repeated character is removed and keep updating Cnt[ ] array accordingly.
• Continue this process until j reaches the end of string. Once the string is traversed completely, print the answer.

Below is the implementation of the above approach: 

// C++ Program to implement
// the above approach
#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;
}

// Java program to implement
// the above approach
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

# Python3 program to implement 
# the above approach 

# Function to count total 
# number of valid substrings 
def countSub(Str): 

    n = len(Str)

    # Stores the count of 
    # substrings 
    ans = 0 

    # Stores the frequency 
    # of characters 
    cnt = 26 * [0]

    # Initialised both pointers 
    # to beginning of the string 
    i, j = 0, 0 

    while (i < n):

        # If all characters in 
        # substring from index i 
        # to j are distinct 
        if (j < n and 
           (cnt[ord(Str[j]) - ord('a')] == 0)): 

            # Increment count of j-th 
            # character 
            cnt[ord(Str[j]) - ord('a')] += 1 

            # 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 += 1 

        # If some characters are repeated 
        # or j pointer has reached to end 
        else: 

            # Decrement count of j-th 
            # character 
            cnt[ord(Str[i]) - ord('a')] -= 1 

            # Increment first pointer 
            i += 1 

    # Return the final 
    # count of substrings 
    return ans

# Driver code
Str = "gffg"

print(countSub(Str))

# This code is contributed by divyeshrabadiya07

// C# program to implement
// the above approach
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

<script>

// JavaScript Program to implement
// the above approach

// Function to count total
// number of valid substrings
function countSub(str)
{
    var n = str.length;

    // Stores the count of
    // substrings
    var ans = 0;

    // Stores the frequency
    // of characters
    var cnt = Array(26).fill(0);

    // Initialised both pointers
    // to beginning of the string
    var 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].charCodeAt(0) - 'a'.charCodeAt(0)]
                      == 0)) {

            // Increment count of j-th
            // character
            cnt[str[j].charCodeAt(0) - 'a'.charCodeAt(0)]++;

            // 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].charCodeAt(0) - 'a'.charCodeAt(0)]--;

            // Increment first pointer
            i++;
        }
    }

    // Return the final
    // count of substrings
    return ans;
}

// Driver Code
var str = "gffg";
document.write( countSub(str));

</script> 

6

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