Count of substrings having all distinct characters - GeeksforGeeks

14 Jan, 2022
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 jth 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++
Java
Python3
C#
Javascript
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std; // Function to count total// number of valid substringslong 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 Codeint main(){    string str = "gffg";     cout << countSub(str);     return 0;}
// Java program to implement// the above approachimport java.util.*; class GFG{ // Function to count total// number of valid subStringsstatic 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 Codepublic 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 substringsdef 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 codeStr = "gffg" print(countSub(Str)) # This code is contributed by divyeshrabadiya07
// C# program to implement// the above approachusing System; class GFG{ // Function to count total// number of valid subStringsstatic 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 Codepublic 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 substringsfunction 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 Codevar str = "gffg";document.write( countSub(str)); </script>
6
 
Time complexity: O(N) Auxiliary Space: O(1) 
Amal Kumar Choubey
divyeshrabadiya07
noob2000
sumitgumber28
frequency-counting
substring
two-pointer-algorithm
Searching
Strings
two-pointer-algorithm
Searching
Strings
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