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).
// 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 s, int K)
{
int n = s.size();
int count = 0;
// Generate all the subarray of size K
for (int i = 0; i < n - K + 1; i++) {
string s1 = s.substr(i, K);
unordered_map<char, int> unmap;
for (auto x : s1)
unmap[x]++;
// Check for any duplicate
if (unmap.size() == s1.size())
count++;
}
return count;
}
// 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 program to find the
// count of k length substrings
// with k distinct characters
// using sliding window
import java.util.HashMap;
public class Main {
static int count = 0;
// Function to return the
// required count of substrings
static int countSubstrings(String s, int K) {
int n = s.length();
// Generate all the subarray of size K
for (int i = 0; i < n - K + 1; i++) {
String s1 = s.substring(i, i + K);
HashMap<Character, Integer> unmap = new HashMap<>();
for (char c : s1.toCharArray()) {
unmap.put(c, unmap.getOrDefault(c, 0) + 1);
}
// Check for any duplicate
if (unmap.size() == K) {
count++;
}
}
return count;
}
// 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 Utkarsh
# Python3 program to find the
# count of k length substrings
# with k distinct characters
# using sliding window
def countSubstrings(s: str, K: int) -> int:
n = len(s)
count = 0
# Generate all the subarray of size K
for i in range(n - K + 1):
s1 = s[i:i+K]
unmap = {}
for c in s1:
if c in unmap:
unmap[c] += 1
else:
unmap[c] = 1
# Check for any duplicate
if len(unmap) == len(s1):
count += 1
return count
# Driver code
if __name__ == "__main__":
# string str
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 ik_9
// C# program to find the
// count of k length substrings
// with k distinct characters
// using sliding window
using System;
using System.Collections.Generic;
class MainClass {
static int count = 0;
// Function to return the
// required count of substrings
static int CountSubstrings(string s, int K) {
int n = s.Length;
// Generate all the subarray of size K
for (int i = 0; i < n - K + 1; i++) {
string s1 = s.Substring(i, K);
Dictionary<char, int> unmap = new Dictionary<char, int>();
foreach (char c in s1.ToCharArray()) {
if (unmap.ContainsKey(c)) {
unmap[c]++;
} else {
unmap.Add(c, 1);
}
}
// Check for any duplicate
if (unmap.Count == K) {
count++;
}
}
return count;
}
// 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.WriteLine(CountSubstrings(str, K));
}
}
// This code is contributed by Prajwal Kandekar
// 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(s, K)
{
let n = s.length;
let count = 0;
// Generate all the subarray of size K
for (let i = 0; i < n - K + 1; i++) {
let s1 = s.substring(i, i+K);
let unmap = new Map();
for (let c of s1)
{
if(unmap.has(c))
unmap.set(c,unmap.get(c)+1);
else
unmap.set(c, 1);
}
// unmap[c]++;
// Check for any duplicate
if (unmap.size == s1.length)
count++;
}
return count;
}
// Driver code
// string str
let str = "aabcdabbcdc";
// integer K
let K = 3;
// Print the count of K length
// substrings with k distinct characters
document.write(countSubstrings(str, K));
Output
5
Time Complexity: O(N*K)
Auxiliary Space: O(K)
Count of substrings of length K with exactly K distinct characters using Sliding Window Technique:
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++ 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 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 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# 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
<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(K)
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