Count of substrings of given string with frequency of each character at most K
Last Updated :
14 Dec, 2021
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 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
# 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# 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.
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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