当前字符串的字符的频率在另一个字符串总结

给定两个长度分别为M和N 的字符串S1和S2 ，任务是计算字符串字符出现的频率之和 S1的字符串S2英寸

例子：

Input: S1 = “pPKf”, S2 = “KKKttsdppfP”
Output: 7
Explanation:
The character ‘p’ occurs twice in the string S2.
The character ‘P’ occurs once in the string S2.
The character ‘K’ occurs thrice in the string S2.
The character ‘f’ occurs once in the string S2.
Therefore, in total, characters of the string S1 occurs 7 times in the string S2.

Input: S1 = “geEksFOR”, S2 = “GeEksforgeEKS”
Output: 7

朴素的方法：最简单的方法是遍历字符串S1 的每个字符，计算其在字符串S2 中的频率。

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：上述方法可以通过使用 Hashing 进行优化。请按照以下步骤解决问题：

• 初始化整数计数以存储所需的总和。
• 将字符串S1 的所有字符插入一个 Set 中，比如bset 。
• 遍历字符串S2和检查的字符，如果当前字符存在于该组，BSET，或没有。如果发现为真，则将计数加一。
• 完成以上步骤后，打印count的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include
using namespace std;
 
// Function to find sum of frequencies
// of characters of S1 present in S2
void countTotalFrequencies(string S1, string S2)
{
 
  // Insert all characters of
  // string S1 in the set
  set bset;
  for(auto x:S1)
    bset.insert(x);
  int count = 0;
 
  // Traverse the string S2
  for (auto x: S2)
  {
 
    // Check if X is present
    // in bset or not
    if (bset.find(x) != bset.end())
 
      // Increment count by 1
      count += 1;
  }
 
  // Finally, print the count
  cout << count << endl;
 
}
 
// Driver Code
int main()
{
   
  // Given strings
  string S1 = "geEksFOR";
  string S2 = "GeEksforgeEKS";
 
  countTotalFrequencies(S1, S2);
 
}
 
// This code is contributed by ipg2016107.

Java

// Java program for the above approach
import java.util.HashSet;
 
class GFG{
 
// Function to find sum of frequencies
// of characters of S1 present in S2
static void countTotalFrequencies(String S1, String S2)
{
     
    // Insert all characters of
    // string S1 in the set
    HashSet bset = new HashSet();
    char[] S1arr = S1.toCharArray();
    char[] S2arr = S2.toCharArray();
     
    for(char x : S1arr)
        bset.add(x);
         
    int count = 0;
 
    // Traverse the string S2
    for(char x : S2arr)
    {
         
        // Check if X is present
        // in bset or not
        if (bset.contains(x))
 
            // Increment count by 1
            count += 1;
    }
 
    // Finally, print the count
    System.out.print(count);
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given strings
    String S1 = "geEksFOR";
    String S2 = "GeEksforgeEKS";
 
    countTotalFrequencies(S1, S2);
}
}
 
// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach
 
# Function to find sum of frequencies
# of characters of S1 present in S2
def countTotalFrequencies(S1, S2):
 
    # Insert all characters of
    # string S1 in the set
    bset = set(S1)
    count = 0
 
    # Traverse the string S2
    for x in S2:
 
        # Check if X is present
        # in bset or not
        if x in bset:
 
            # Increment count by 1
            count += 1
 
    # Finally, print the count
    print(count)
 
# Driver Code
 
# Given strings
S1 = "geEksFOR"
S2 = "GeEksforgeEKS"
 
countTotalFrequencies(S1, S2)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG {
     
// Function to find sum of frequencies
// of characters of S1 present in S2
static void countTotalFrequencies(string S1,
                                  string S2)
{
 
    // Insert all characters of
    // string S1 in the set
    HashSet bset = new HashSet();
 
    foreach(char x in S1)
        bset.Add(x);
         
    int count = 0;
 
    // Traverse the string S2
    foreach(char x in S2)
    {
 
        // Check if X is present
        // in bset or not
        if (bset.Contains(x))
 
            // Increment count by 1
            count += 1;
    }
     
    // Finally, print the count
    Console.Write(count);
}
 
// Driver code
static void Main()
{
     
    // Given strings
    string S1 = "geEksFOR";
    string S2 = "GeEksforgeEKS";
 
    countTotalFrequencies(S1, S2);
}
}
 
// This code is contributed by abhinavjain194

Javascript

 
// Javascript program for the above approach
 
// Function to find sum of frequencies
// of characters of S1 present in S2
function countTotalFrequencies(S1, S2)
{
     
    // Insert all characters of
    // string S1 in the set
    var bset = new Set();
    for(var i = 0; i < S1.length; i++)
    {
        bset.add(S1[i]);
    }
     
    var count = 0;
     
    // Traverse the string S2
    for(var i = 0; i < S2.length; i++)
    {
     
        // Check if X is present
        // in bset or not
        if (bset.has(S2[i]))
         
            // Increment count by 1
            count += 1;
    }
     
    // Finally, print the count
    document.write(count);
}
 
// Driver Code
 
// Given strings
var S1 = "geEksFOR";
var S2 = "GeEksforgeEKS";
 
countTotalFrequencies(S1, S2);
 
// This code is contributed by rutvik_56
 

输出：

7

时间复杂度： O(N)
辅助空间： O(N)

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