Count of Substrings that can be formed without using the given list of Characters

Given a string str and a list of characters L, the task is to count total numbers of substrings of the string str without using characters given in the list L.

Examples:

Input: str = “abcd”, L[] = {‘a’, ‘b’, ‘t’, ‘q’}
Output: 3
Explanation:
On ignoring characters ‘a’ and ‘b’ from the given string, substring “cd” is left.
Therefore, the total number of substrings formed with “cd” are:
(2 * (2 + 1)) / 2 = 3

Input: str = “abcpxyz”, L[] = {‘a’, ‘p’, ‘q’}
Output: 9
Explanation:
On ignoring characters ‘a’ and ‘p’ from the given string, substrings “bc” and “xyz” are left.
Therefore, total number of substrings formed with the substrings are:
(2*(2+1))/2 + (3*(3+1))/2 = 3 + 6 = 9

Approach: The total number of substrings for a given string of length N is given by the formula



(N * (N + 1)) / 2

The idea is to use the above formula and follow the below steps to compute the answer:

  1. Traverse the string str character by character.
  2. Count the number of characters till which a character from list L is not found. Let the count be N
  3. Once a character from L is encountered, compute (N * (N + 1) / 2) and add it to the answer and reset count N to zero.

Below is the implementation of the above approach:

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// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the Number of sub-strings
// without using given character
int countSubstring(string& S, char L[], int& n)
{
    int freq[26] = { 0 }, ans = 0;
  
    // Mark the given characters in
    // the freq array
    for (int i = 0; i < n; i++) {
        freq[(int)(L[i] - 'a')] = 1;
    }
  
    // Count variable to store the count
    // of the characters until a character
    // from given L is encountered
    int count = 0;
  
    for (auto x : S) {
  
        // If a character from L is encountered,
        // then the answer variable is incremented by
        // the value obtained by using
        // the mentioned formula and count is set to 0
        if (freq[(int)(x - 'a')]) {
            ans += (count * count + count) / 2;
            count = 0;
        }
        else
            count++;
    }
  
    // For last remaining characters
    ans += (count * count + count) / 2;
  
    return ans;
}
  
// Driver code
int main()
{
  
    string S = "abcpxyz";
    char L[] = { 'a', 'p', 'q' };
    int n = sizeof(L) / sizeof(L[0]);
  
    cout << countSubstring(S, L, n);
  
    return 0;
}
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// Java implementation of the above approach
import java.util.*;
  
class GFG
{
  
// Function to find the Number of sub-Strings
// without using given character
static int countSubString(char []S, char L[], int n)
{
    int []freq = new int[26];
    int ans = 0;
  
    // Mark the given characters in
    // the freq array
    for (int i = 0; i < n; i++) 
    {
        freq[(int)(L[i] - 'a')] = 1;
    }
  
    // Count variable to store the count
    // of the characters until a character
    // from given L is encountered
    int count = 0;
  
    for (int x : S)
    {
  
        // If a character from L is encountered,
        // then the answer variable is incremented by
        // the value obtained by using
        // the mentioned formula and count is set to 0
        if (freq[(int)(x - 'a')] > 0
        {
            ans += (count * count + count) / 2;
            count = 0;
        }
        else
            count++;
    }
  
    // For last remaining characters
    ans += (count * count + count) / 2;
  
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    String S = "abcpxyz";
    char L[] = { 'a', 'p', 'q' };
    int n = L.length;
  
    System.out.print(countSubString(S.toCharArray(), L, n));
}
}
  
// This code is contributed by Rajput-Ji
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# Python3 implementation of the above approach
  
# Function to find the Number of sub-strings
# without using given character
def countSubstring(S, L,n):
    freq = [0 for i in range(26)]
      
    # the freq array
    for i in range(n):
        freq[(ord(L[i]) - ord('a'))] = 1
  
    # Count variable to store the count
    # of the characters until a character
    # from given L is encountered
    count,ans = 0,0
  
    for x in S:
  
        # If a character from L is encountered,
        # then the answer variable is incremented by
        # the value obtained by using
        # the mentioned formula and count is set to 0
        if (freq[ord(x) - ord('a')]):
            ans += (count * count + count) // 2
            count = 0
        else:
            count += 1
  
    # For last remaining characters
    ans += (count * count + count) // 2
  
    return ans
  
# Driver code
  
S = "abcpxyz"
L = ['a', 'p', 'q']
n = len(L)
  
print(countSubstring(S, L, n))
  
# This code is contributed by mohit kumar 29
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// C# implementation of the above approach
using System;
  
class GFG
{
  
    // Function to find the Number of sub-Strings
    // without using given character
    static int countSubString(char []S, char []L, int n)
    {
        int []freq = new int[26];
        int ans = 0;
      
        // Mark the given characters in
        // the freq array
        for (int i = 0; i < n; i++) 
        {
            freq[(int)(L[i] - 'a')] = 1;
        }
      
        // Count variable to store the count
        // of the characters until a character
        // from given L is encountered
        int count = 0;
      
        foreach (int x in S)
        {
      
            // If a character from L is encountered,
            // then the answer variable is incremented by
            // the value obtained by using
            // the mentioned formula and count is set to 0
            if (freq[(int)(x - 'a')] > 0) 
            {
                ans += (count * count + count) / 2;
                count = 0;
            }
            else
                count++;
        }
      
        // For last remaining characters
        ans += (count * count + count) / 2;
      
        return ans;
    }
      
    // Driver code
    public static void Main()
    {
        string S = "abcpxyz";
        char []L = { 'a', 'p', 'q' };
        int n = L.Length;
      
        Console.WriteLine(countSubString(S.ToCharArray(), L, n));
    }
}
  
// This code is contributed by Yash_R
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Output:
9

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Improved By : mohit kumar 29, Rajput-Ji, Yash_R

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