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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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;  
}  

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  

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  

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 

Output:

Time Complexity: O(N)

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

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