给定字符串的子串计数，每个字符的频率最多为 K

给定一个字符串str ，任务是计算给定字符串的子字符串数，使得字符串中每个元素的频率几乎为K 。

例子：

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

方法：给定的问题可以使用两指针技术来解决。遍历 [0, N) 范围内字符串的每个字符，并保持每个字符在无序映射中出现的频率。创建一个变量ptr ，它存储当前窗口起点的索引。最初， ptr 的值为0 。对于索引i ，如果str[i]的频率小于或等于K ，则将(i – ptr + 1)添加到子字符串计数中，否则，递增 ptr 的值直到str[i] > K 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
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 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 m = new HashMap();
 
    // 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 m =
          new Dictionary();
           
    // 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

 
       // 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
   

输出

7

时间复杂度： O(N)
辅助空间： O(N)