字谜子串总数

给定字符串低字母字符，计算该字符串的总子串，它们彼此是字谜。

例子：

Input  : str = “xyyx”
Output : 4
Total substrings of this string which
are anagram to each other are 4 which 
can be enumerated as,
{“x”, “x”}, {"y", "y"}, {“xy”, “yx”}, 
{“xyy”, “yyx”}

Input  : str = "geeg"
Output : 4

这个想法是创建一个地图。我们使用字符频率作为键，使用相应的计数作为值。我们可以通过遍历所有子字符串并计算每个子字符串中字符的频率来解决这个问题。我们可以在遍历子字符串时更新字符的频率，即不会有额外的循环来计算字符的频率。
在下面的代码中，键“向量类型”和值“整数类型”的映射用于存储子字符串字符的“长度为 26 的频率数组”的出现。一旦存储了每个频率数组的出现“o”，总字谜将是所有不同频率数组的 o*(o-1)/2 的总和，因为如果特定子字符串在字符串中具有“o”字谜，则总数 o*(o -1)/2 字谜对可以形成。

下面是上述想法的实现。

C++

// C++ program to count total anagram
// substring of a string
#include 
using namespace std;
  
// Total number of lowercase characters
#define MAX_CHAR 26
  
// Utility method to return integer index
// of character 'c'
int toNum(char c)
{
    return (c - 'a');
}
  
// Returns count of total number of anagram
// substrings of string str
int countOfAnagramSubstring(string str)
{
    int N = str.length();
  
    // To store counts of substrings with given
    // set of frequencies.
    map, int> mp;
  
    // loop for starting index of substring
    for (int i=0; i freq(MAX_CHAR, 0);
  
        // loop for length of substring
        for (int j=i; jsecond;
        result += ((freq) * (freq-1))/2;
    }
    return result;
}
  
//  Driver code to test above methods
int main()
{
    string str = "xyyx";
    cout << countOfAnagramSubstring(str) << endl;
    return 0;
}

Java

import java.util.Arrays;
import java.util.HashMap;
  
public class anagramPairCount {
    public static void main(String[] args) {
        subString("kkkk");
    }
  
    static void subString(String s){
        HashMap map= new HashMap<>();
  
        for(int i = 0; i < s.length(); i++){
            for(int j = i; j < s.length(); j++){
                char[] valC = s.substring(i, j+1).toCharArray();
                Arrays.sort(valC);
                String val = new String(valC);
                if (map.containsKey(val)) 
                    map.put(val, map.get(val)+1);
                else 
                    map.put(val, 1);
            }
        }
        int anagramPairCount = 0;
        for(String key: map.keySet()){
            int n = map.get(key);
            anagramPairCount += (n * (n-1))/2;
        }
        System.out.println(anagramPairCount);
    }
}

Python3

# Python3 program to count total anagram
# substring of a string
def countOfAnagramSubstring(s):
      
    # Returns total number of anagram
    # substrings in s
    n = len(s)
    mp = dict()
      
    # loop for length of substring
    for i in range(n):
        sb = ''
        for j in range(i, n):
            sb = ''.join(sorted(sb + s[j]))
            mp[sb] = mp.get(sb, 0)
              
            # increase count corresponding
            # to this dict array
            mp[sb] += 1
  
    anas = 0
      
    # loop over all different dictionary 
    # items and aggregate substring count
    for k, v in mp.items():
        anas += (v*(v-1))//2
    return anas
  
# Driver Code
s = "xyyx"
print(countOfAnagramSubstring(s))
  
# This code is contributed by fgaim

C#

using System;
using System.Collections.Generic;
  
  
public class anagramPairCount {
    public static void Main() {
        subString("kkkk");
    }
  
    static void subString(String s){
        Dictionary map= new Dictionary();
  
        for(int i = 0; i < s.Length; i++){
            for(int j = i; j < s.Length; j++){
                char[] valC = s.Substring(i, j+1-i).ToCharArray();
                Array.Sort(valC);
                string val = new string(valC);
                if (map.ContainsKey(val)) 
                    map[val]=map[val]+1;
                else
                    map.Add(val, 1);
            }
        }
        int anagramPairCount = 0;
        foreach(string key in map.Keys){
            int n = map[key];
            anagramPairCount += (n * (n-1))/2;
        }
        Console.Write(anagramPairCount);
    }
}
  
// This code is contributed by AbhiThakur

输出：