Given a integer k and a string str, the task is to count the number of distinct sub-strings such that each sub-string does not contain some specific characters more than k times. The specific characters are given as another string.
Examples:
Input: str = “ababab”, anotherStr = “bcd”, k = 1
Output: 5
All valid sub-strings are “a”, “b”, “ab”, “ba” and “aba”Input: str = “acbacbacaa”, anotherStr = “ycb”, k = 2
Output: 8
Approach:
- Store characters of anotherStr in a boolean array of size 256 for quick lookip
- Traverse through all substrings of given string. For every substring, keep the count of illegal characters in anotherStr.
- If the count of these characters exceeds the value of k then break out of the inner loop.
- Else, store this sub-string in an hash table to keep distinct substrings.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;const int MAX_CHAR = 256;// Function to return the count of valid sub-stringsint countSubStrings(string s, string anotherStr, int k){ // Store all characters of anotherStr in a // direct index table for quick lookup. bool illegal[MAX_CHAR] = { false }; for (int i = 0; i < anotherStr.size(); i++) illegal[anotherStr[i]] = true; // 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 = ""; int count = 0; for (int j = i; j < s.size(); ++j) { // If character is illegal if (illegal[s[j]]) ++count; ss = ss + s[j]; // If current substring is valid if (count <= k) { us.insert(ss); } // If current substring is invalid, // adding more characters would not // help. else break; } } // Return the count of distinct sub-strings return us.size();}// Driver codeint main(){ string str = "acbacbacaa"; string anotherStr = "abcdefghijklmnopqrstuvwxyz"; int k = 2; cout << countSubStrings(str, anotherStr, k); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG { static int MAX_CHAR = 256; // Function to return the count of valid sub-strings static int countSubStrings(String s, String anotherStr, int k) { // Store all characters of anotherStr in a // direct index table for quick lookup. boolean illegal[] = new boolean[MAX_CHAR]; for (int i = 0; i < anotherStr.length(); i++) { illegal[anotherStr.charAt(i)] = true; } // 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 = ""; int count = 0; for (int j = i; j < s.length(); ++j) { // If character is illegal if (illegal[s.charAt(j)]) { ++count; } ss = ss + s.charAt(j); // If current substring is valid if (count <= k) { us.add(ss); } // If current substring is invalid, // adding more characters would not // help. else { break; } } } // Return the count of distinct sub-strings return us.size(); } // Driver code public static void main(String[] args) { String str = "acbacbacaa"; String anotherStr = "abcdefghijklmnopqrstuvwxyz"; int k = 2; System.out.println(countSubStrings(str, anotherStr, k)); }}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach MAX_CHAR = 256# Function to return the count# of valid sub-strings def countSubStrings(s, anotherStr, k) : # Store all characters of anotherStr in # a direct index table for quick lookup. illegal = [False] * MAX_CHAR for i in range(len(anotherStr)) : illegal[ord(anotherStr[i])] = True # 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 = "" count = 0 for j in range(i, len(s)) : # If character is illegal if (illegal[ord(s[j])]) : count += 1 ss = ss + s[j] # If current substring is valid if (count <= k) : us.add(ss) # If current substring is invalid, # adding more characters would not # help. else : break # Return the count of distinct # sub-strings return len(us)# Driver code if __name__ == "__main__" : string = "acbacbacaa" anotherStr = "abcdefghijklmnopqrstuvwxyz" k = 2 print(countSubStrings(string, anotherStr, k))# This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG{ static int MAX_CHAR = 256; // Function to return the count // of valid sub-strings static int countSubStrings(String s, String anotherStr, int k) { // Store all characters of anotherStr in a // direct index table for quick lookup. bool []illegal = new bool[MAX_CHAR]; for (int i = 0; i < anotherStr.Length; i++) { illegal[anotherStr[i]] = true; } // 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 = ""; int count = 0; for (int j = i; j < s.Length; ++j) { // If character is illegal if (illegal[s[j]]) { ++count; } ss = ss + s[j]; // If current substring is valid if (count <= k) { us.Add(ss); } // If current substring is invalid, // adding more characters would not // help. else { break; } } } // Return the count of distinct sub-strings return us.Count; } // Driver code public static void Main() { String str = "acbacbacaa"; String anotherStr = "abcdefghijklmnopqrstuvwxyz"; int k = 2; Console.WriteLine(countSubStrings(str, anotherStr, k)); }}//This code is contributed by 29AjayKumar |
Javascript
<script>// js implementation of the approachlet MAX_CHAR = 256;// Function to return the count of valid sub-stringsfunction countSubStrings( s, anotherStr, k){ // Store all characters of anotherStr in a // direct index table for quick lookup. let illegal = []; for(let i = 0;i<256;i++) illegal.push(false); for (let i = 0; i < anotherStr.length; i++) illegal[anotherStr[i]] = true; // To store distinct output substrings let us = new Set(); // Traverse through the given string and // one by one generate substrings beginning // from s[i]. for (let i = 0; i < s.length; ++i) { // One by one generate substrings ending // with s[j] let ss = ""; let count = 0; for (let j = i; j < s.length; ++j) { // If character is illegal if (illegal[s[j]]) ++count; ss = ss + s[j]; // If current substring is valid if (count <= k) { us.add(ss); } // If current substring is invalid, // adding more characters would not // help. else break; } } // Return the count of distinct sub-strings return us.size;}// Driver codelet str = "acbacbacaa";let anotherStr = "abcdefghijklmnopqrstuvwxyz";let k = 2;document.write(countSubStrings(str, anotherStr, k));</script> |
Output:
8
Time Complexity: O(max(n2, m), where n and m are the length of string “str” and “anotherstr” respectively,
Auxiliary Space: O(max(n2, MAX_CHAR))
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