Count of substrings of length K with exactly K distinct characters
Given a string str of lowercase alphabets and an integer K, the task is to count all substrings of length K which have exactly K distinct characters.
Example:
Input: str = “abcc”, K = 2
Output: 2
Explanation:
Possible substrings of length K = 2 are
ab : 2 distinct characters
bc : 2 distinct characters
cc : 1 distinct character
Only two valid substrings exist {“ab”, “bc”}.Input: str = “aabab”, K = 3
Output: 0
Explanation:
Possible substrings of length K = 3 are
aab : 2 distinct characters
aba : 2 distinct characters
bab : 2 distinct characters
No substrings of length 3 exists with exactly 3 distinct characters
Naive approach:
The idea is to generate all substrings of length K and for each substring count number of distinct characters. If the length of a string is N, then there can be N – K + 1 substrings of length K. Generating these substrings will require O(N) complexity, and for checking each substring requires O(K) complexity, hence making the overall complexity as O(N*K).
Efficient approach:
The idea is to use Window Sliding Technique. Maintain a window of size K and keep a count of all the characters in the window using a HashMap. Traverse through the string reducing the count of the first character of the previous window and adding the frequency of the last character of the current window in the HashMap. If the count of distinct characters in a window of length K is equal to K, increment the answer by 1.
Below is the implementation of the above approach:
C++
// C++ program to find the // count of k length substrings // with k distinct characters // using sliding window #include <bits/stdc++.h> using namespace std; // Function to return the // required count of substrings int countSubstrings(string str, int K) { int N = str.size(); // Store the count int answer = 0; // Store the count of // distinct characters // in every window unordered_map<char, int> map; // Store the frequency of // the first K length substring for (int i = 0; i < K; i++) { // Increase frequency of // i-th character map[str[i]]++; } // If K distinct characters // exist if (map.size() == K) answer++; // Traverse the rest of the // substring for (int i = K; i < N; i++) { // Increase the frequency // of the last character // of the current substring map[str[i]]++; // Decrease the frequency // of the first character // of the previous substring map[str[i - K]]--; // If the character is not present // in the current substring if (map[str[i - K]] == 0) { map.erase(str[i - K]); } // If the count of distinct // characters is 0 if (map.size() == K) { answer++; } } // Return the count return answer; } // Driver code int main() { // string str string str = "aabcdabbcdc"; // integer K int K = 3; // Print the count of K length // substrings with k distinct characters cout << countSubstrings(str, K) << endl; return 0; } |
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Java
// Java program to find the count // of k length substrings with k // distinct characters using // sliding window import java.util.*; class GFG{ // Function to return the // required count of substrings public static int countSubstrings(String str, int K) { int N = str.length(); // Store the count int answer = 0; // Store the count of // distinct characters // in every window Map<Character, Integer> map = new HashMap<Character, Integer>(); // Store the frequency of // the first K length substring for(int i = 0; i < K; i++) { // Increase frequency of // i-th character if (map.get(str.charAt(i)) == null) { map.put(str.charAt(i), 1); } else { map.put(str.charAt(i), map.get(str.charAt(i)) + 1); } } // If K distinct characters // exist if (map.size() == K) answer++; // Traverse the rest of the // substring for(int i = K; i < N; i++) { // Increase the frequency // of the last character // of the current substring if (map.get(str.charAt(i)) == null) { map.put(str.charAt(i), 1); } else { map.put(str.charAt(i), map.get(str.charAt(i)) + 1); } // Decrease the frequency // of the first character // of the previous substring map.put(str.charAt(i - K), map.get(str.charAt(i - K)) - 1); // If the character is not present // in the current substring if (map.get(str.charAt(i - K)) == 0) { map.remove(str.charAt(i - K)); } // If the count of distinct // characters is 0 if (map.size() == K) { answer++; } } // Return the count return answer; } // Driver code public static void main(String[] args) { // string str String str = "aabcdabbcdc"; // integer K int K = 3; // Print the count of K length // substrings with k distinct characters System.out.println(countSubstrings(str, K)); } } // This code is contributed by grand_master |
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Python3
# Python3 program to find the # count of k length substrings # with k distinct characters # using sliding window # Function to return the # required count of substrings def countSubstrings(str, K): N = len(str) # Store the count answer = 0 # Store the count of # distinct characters # in every window map = {} # Store the frequency of # the first K length substring for i in range(K): # Increase frequency of # i-th character map[str[i]] = map.get(str[i], 0) + 1 # If K distinct characters # exist if (len(map) == K): answer += 1 # Traverse the rest of the # substring for i in range(K, N): # Increase the frequency # of the last character # of the current substring map[str[i]] = map.get(str[i], 0) + 1 # Decrease the frequency # of the first character # of the previous substring map[str[i - K]] -= 1 # If the character is not present # in the current substring if (map[str[i - K]] == 0): del map[str[i - K]] # If the count of distinct # characters is 0 if (len(map) == K): answer += 1 # Return the count return answer # Driver code if __name__ == '__main__': str = "aabcdabbcdc" # Integer K K = 3 # Print the count of K length # substrings with k distinct characters print(countSubstrings(str, K)) # This code is contributed by mohit kumar 29 |
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C#
// C# program to find the count // of k length substrings with k // distinct characters using // sliding window using System; using System.Collections.Generic; class GFG{ // Function to return the // required count of substrings public static int countSubstrings(string str, int K) { int N = str.Length; // Store the count int answer = 0; // Store the count of // distinct characters // in every window Dictionary<char, int> map = new Dictionary<char, int>(); // Store the frequency of // the first K length substring for(int i = 0; i < K; i++) { // Increase frequency of // i-th character if(!map.ContainsKey(str[i])) { map[str[i]] = 1; } else { map[str[i]]++; } } // If K distinct characters // exist if (map.Count == K) answer++; // Traverse the rest of the // substring for(int i = K; i < N; i++) { // Increase the frequency // of the last character // of the current substring if(!map.ContainsKey(str[i])) { map[str[i]] = 1; } else { map[str[i]]++; } // Decrease the frequency // of the first character // of the previous substring map[str[i - K]]--; // If the character is not present // in the current substring if (map[str[i - K]] == 0) { map.Remove(str[i - K]); } // If the count of distinct // characters is 0 if (map.Count == K) { answer++; } } // Return the count return answer; } // Driver code public static void Main(string[] args) { // string str string str = "aabcdabbcdc"; // integer K int K = 3; // Print the count of K length // substrings with k distinct characters Console.Write(countSubstrings(str, K)); } } // This code is contributed by rutvik_56 |
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Output:
5
Time Complexity: O(N)
