Count of substrings of length K with exactly K distinct characters

Given string str of the lowercase alphabet 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 exist with exactly 3 distinct characters. 

Naive approach:
The idea is to generate all substrings of length K and, for each substring count, a number of distinct characters. If the length of a string is N, then there can be N – K + 1 substring of length K. Generating these substrings will require O(N) complexity, and checking each substring requires O(K) complexity, hence making the overall complexity like 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 reduces the count of the first character of the previous window and adds 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

Javascript

<script>

// Javascript program to find the
// count of k length substrings
// with k distinct characters
// using sliding window

// Function to return the
// required count of substrings
function countSubstrings(str, K)
{
    var N = str.length;
    // Store the count
    var answer = 0;

    // Store the count of
    // distinct characters
    // in every window
    var map = new Map();

    // Store the frequency of
    // the first K length substring
    for (var i = 0; i < K; i++) {

        // Increase frequency of
        // i-th character
        if(map.has(str[i]))
            map.set(str[i], map.get(str[i])+1)
        else
            map.set(str[i], 1)
    }

    // If K distinct characters
    // exist
    if (map.size == K)
        answer++;

    // Traverse the rest of the
    // substring
    for (var i = K; i < N; i++) {

        // Increase the frequency
        // of the last character
        // of the current substring
        if(map.has(str[i]))
            map.set(str[i], map.get(str[i])+1)
        else
            map.set(str[i], 1)
        // Decrease the frequency
        // of the first character
        // of the previous substring
        if(map.has(str[i-K]))
            map.set(str[i-K], map.get(str[i-K])-1)


        // If the character is not present
        // in the current substring
        if (map.has(str[i - K]) && map.get(str[i-K])==0) {
            map.delete(str[i - K]);
        }

        // If the count of distinct
        // characters is 0
        if (map.size == K) {
            answer++;
        }
    }

    // Return the count
    return answer;
}

// Driver code
// string str
var str = "aabcdabbcdc";
// integer K
var K = 3;
// Print the count of K length
// substrings with k distinct characters
document.write( countSubstrings(str, K) );

</script>

Output:

5

Time Complexity: O(N)
Auxiliary Space: O(1)