Count of substrings of given string with frequency of each character at most K

Given a string str, the task is to calculate the number of substrings of the given string such that the frequency of each element of the string is almost K.

Examples:

Input: str = “abab”, K = 1
Output: 7
Explanation: The substrings such that the frequency of each character is atmost 1 are “a”, “b”, “a”, “b”, “ab”, “ba” and “ab”.

Input: str[] = “xxyyzzxx”, K = 2
Output: 33

Approach: The given problem can be solved using the two-pointer technique. Iterate through each character of the string in the range [0, N) and maintain the frequency of each character in an unordered map. Create a variable ptr, which stores the index of the starting point of the current window. Initially, the value of ptr is 0. For index i, if the frequency of str[i] is less than or equal to K, add (i – ptr + 1) into the substring count, otherwise, increment the value of ptr until str[i] > K. 

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the count of
// substrings such that frequency
// of each character is atmost K
int cntSubstr(string str, int K)
{
    // Stores the size of string
    int N = str.size();

    // Stores the final count
    int ans = 0;

    // Stores the starting index
    int ptr = 0;

    // Stores the frequency of
    // characters of string
    unordered_map<char, int> m;

    // Loop to iterate through string
    for (int i = 0; i < N; i++) {

        // Increment the frequency of
        // the current character
        m[str[i]]++;

        // While the frequency of char is
        // greater than K, increment ptr
        while (m[str[i]] > K && ptr <= i) {
            m[str[ptr]]--;
            ptr++;
        }

        // Update count
        ans += (i - ptr + 1);
    }

    // Return Answer
    return ans;
}

// Driver Code
int main()
{
    string str = "abab";
    int K = 1;

    cout << cntSubstr(str, K);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the count of
// substrings such that frequency
// of each character is atmost K
static int cntSubstr(String str, int K)
{

    // Stores the size of string
    int N = str.length();

    // Stores the final count
    int ans = 0;

    // Stores the starting index
    int ptr = 0;

    // Stores the frequency of
    // characters of string
    HashMap<Character,
            Integer> m = new HashMap<Character,
                                     Integer>();

    // Loop to iterate through string
    for(int i = 0; i < N; i++)
    {

        // Increment the frequency of
        // the current character
        int count = 0;
        if (m.containsKey(str.charAt(i)))
        {
            count = m.get(str.charAt(i));
        }
        m.put(str.charAt(i), count + 1);

        // While the frequency of char is
        // greater than K, increment ptr
        while (m.get(str.charAt(i)) > K && ptr <= i)
        {
            m.put(str.charAt(ptr),
                  m.get(str.charAt(ptr)) - 1);
            ptr++;
        }

        // Update count
        ans += (i - ptr + 1);
    }

    // Return Answer
    return ans;
}

// Driver Code
public static void main(String[] args)
{
    String str = "abab";
    int K = 1;

    System.out.println(cntSubstr(str, K));
}
}

// This code is contributed by ukasp

Python3

# Python Program to implement
# the above approach

# Function to find the count of
# substrings such that frequency
# of each character is atmost K
def cntSubstr(str, K):

    # Stores the size of string
    N = len(str)

    # Stores the final count
    ans = 0

    # Stores the starting index
    ptr = 0

    # Stores the frequency of
    # characters of string
    m = {}

    # Loop to iterate through string
    for i in range(N) :

        # Increment the frequency of
        # the current character
        if (str[i] in m):
            m[str[i]] += 1
        else:
            m[str[i]] = 1

        # While the frequency of char is
        # greater than K, increment ptr
        while (m[str[i]] > K and ptr <= i):
            m[str[ptr]] -= 1
            ptr += 1

        # Update count
        ans += (i - ptr + 1)

    # Return Answer
    return ans

# Driver Code
str = "abab"
K = 1
print(cntSubstr(str, K))

# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

// Function to find the count of
// substrings such that frequency
// of each character is atmost K
static int cntSubstr(string str, int K)
{

    // Stores the size of string
    int N = str.Length;

    // Stores the final count
    int ans = 0;

    // Stores the starting index
    int ptr = 0;

    // Stores the frequency of
    // characters of string
    Dictionary<char, int> m =
          new Dictionary<char, int>();

    // Loop to iterate through string
    for (int i = 0; i < N; i++) {

        // Increment the frequency of
        // the current character
        int count = 0;
        if (m.ContainsKey(str[i])) {

            count = m[str[i]];
        }
        m[str[i]] = count + 1;

        // While the frequency of char is
        // greater than K, increment ptr
        while (m[str[i]] > K && ptr <= i) {
            m[str[ptr]]--;
            ptr++;
        }

        // Update count
        ans += (i - ptr + 1);
    }

    // Return Answer
    return ans;
}

// Driver Code
public static void Main()
{
    string str = "abab";
    int K = 1;

    Console.Write(cntSubstr(str, K));

}
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>

       // JavaScript Program to implement
       // the above approach

       // Function to find the count of
       // substrings such that frequency
       // of each character is atmost K
       function cntSubstr(str, K) {
           // Stores the size of string
           let N = str.length;

           // Stores the final count
           let ans = 0;

           // Stores the starting index
           let ptr = 0;

           // Stores the frequency of
           // characters of string
           let m = new Map();

           // Loop to iterate through string
           for (let i = 0; i < N; i++) {

               // Increment the frequency of
               // the current character
               if (m.has(str.charAt(i)))
                   m.set(str[i], m.get(str[i]) + 1);
               else
                   m.set(str[i], 1);

               // While the frequency of char is
               // greater than K, increment ptr
               while (m.get(str[i]) > K && ptr <= i) {
                   m.set(str[ptr], m.get(str[ptr]) - 1);
                   ptr++;
               }

               // Update count
               ans += (i - ptr + 1);
           }

           // Return Answer
           return ans;
       }

       // Driver Code
       let str = "abab";
       let K = 1;

       document.write(cntSubstr(str, K));

   // This code is contributed by Potta Lokesh
   </script>

Output

7

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