检查一个字符串中字符的频率是否是另一个字符串中相同字符的频率的因子或倍数

给定两个字符串，任务是检查一个字符（对于每个字符）在一个字符串中的频率是多倍还是另一个字符串的一个因子。如果是，则输出“YES”，否则输出“NO”。
例子：

Input: s1 = “aabccd”, s2 = “bbbaaaacc” 
Output: YES 
Frequency of ‘a’ in s1 and s2 are 2 and 4 respectively, and 2 is a factor of 4 
Frequency of ‘b’ in s1 and s2 are 1 and 3 respectively, and 1 is a factor of 3 
Frequency of ‘c’ in s1 and s2 are same hence it also satisfies. 
Frequency of ‘d’ in s1 and s2 are 1 and 0 respectively, but 0 is a multiple of every number, hence satisfied. 
Hence, the answer YES.
Input: s1 = “hhdwjwqq”, s2 = “qwjdddhhh” 
Output: NO

方法：

1. 将 s1 中字符的频率存储在第一个映射 STL 中。
2. 将 s2 中字符的频率存储在第二个映射 STL 中。
3. 设字符在第一张地图中出现的频率为 F1。我们还假设这个字符在第二张地图中出现的频率是 F2。
4. 检查 F1%F2 和 F2%F1（模运算）。如果其中一个为 0，则满足条件。
5. 检查所有字符。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// Function that checks if the frequency of character
// are a factor or multiple of each other
bool multipleOrFactor(string s1, string s2)
{
    // map store frequency of each character
    map m1, m2;
    for (int i = 0; i < s1.length(); i++)
        m1[s1[i]]++;
 
    for (int i = 0; i < s2.length(); i++)
        m2[s2[i]]++;
 
    map::iterator it;
 
    for (it = m1.begin(); it != m1.end(); it++) {
 
        // if any frequency is 0, then continue
        // as condition is satisfied
        if (m2.find((*it).first) == m2.end())
            continue;
 
        // if factor or multiple, then condition satified
        if (m2[(*it).first] % (*it).second == 0
            || (*it).second % m2[(*it).first] == 0)
            continue;
 
        // if condition not satisfied
        else
            return false;
    }
}
 
// Driver code
int main()
{
    string s1 = "geeksforgeeks";
    string s2 = "geeks";
 
    multipleOrFactor(s1, s2) ? cout << "YES"
                             : cout << "NO";
 
    return 0;
}

Java

// Java implementation of above approach
import java.util.HashMap;
 
class GFG
{
 
    // Function that checks if the frequency of character
    // are a factor or multiple of each other
    public static boolean multipleOrFactor(String s1, String s2)
    {
         
        // map store frequency of each character
        HashMap m1 = new HashMap<>();
        HashMap m2 = new HashMap<>();
 
        for (int i = 0; i < s1.length(); i++)
        {
            if (m1.containsKey(s1.charAt(i)))
            {
                int x = m1.get(s1.charAt(i));
                m1.put(s1.charAt(i), ++x);
            }
            else
                m1.put(s1.charAt(i), 1);
        }
 
        for (int i = 0; i < s2.length(); i++)
        {
            if (m2.containsKey(s2.charAt(i)))
            {
                int x = m2.get(s2.charAt(i));
                m2.put(s2.charAt(i), ++x);
            }
            else
                m2.put(s2.charAt(i), 1);
        }
 
        for (HashMap.Entry entry : m1.entrySet())
        {
             
            // if any frequency is 0, then continue
            // as condition is satisfied
            if (!m2.containsKey(entry.getKey()))
                continue;
 
            // if factor or multiple, then condition satified
            if (m2.get(entry.getKey()) != null &&
                (m2.get(entry.getKey()) % entry.getValue() == 0
                || entry.getValue() % m2.get(entry.getKey()) == 0))
                continue;
             
            // if condition not satisfied
            else
                return false;
        }
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s1 = "geeksforgeeks", s2 = "geeks";
        if (multipleOrFactor(s1, s2))
            System.out.println("Yes");
        else
            System.out.println("No");
 
    }
}
 
// This code is contributed by
// sanjeev2552

Python3

# Python3 implementation of above approach
from collections import defaultdict
 
# Function that checks if the frequency of
# character are a factor or multiple of each other
def multipleOrFactor(s1, s2):
  
    # map store frequency of each character
    m1 = defaultdict(lambda:0)
    m2 = defaultdict(lambda:0)
    for i in range(0, len(s1)):
        m1[s1[i]] += 1
 
    for i in range(0, len(s2)):
        m2[s2[i]] += 1
 
    for it in m1: 
 
        # if any frequency is 0, then continue
        # as condition is satisfied
        if it not in m2:
            continue
 
        # if factor or multiple, then condition satified
        if (m2[it] % m1[it] == 0 or
            m1[it] % m2[it] == 0):
            continue
 
        # if condition not satisfied
        else:
            return False
             
    return True
 
# Driver code
if __name__ == "__main__":
  
    s1 = "geeksforgeeks"
    s2 = "geeks"
 
    if multipleOrFactor(s1, s2): print("YES")
    else: print("NO")
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function that checks if the
    // frequency of character are
    // a factor or multiple of each other
    public static Boolean multipleOrFactor(String s1,
                                           String s2)
    {
         
        // map store frequency of each character
        Dictionary m1 = new Dictionary();
        Dictionary m2 = new Dictionary();
 
        for (int i = 0; i < s1.Length; i++)
        {
            if (m1.ContainsKey(s1[i]))
            {
                var x = m1[s1[i]];
                m1[s1[i]]= ++x;
            }
            else
                m1.Add(s1[i], 1);
        }
 
        for (int i = 0; i < s2.Length; i++)
        {
            if (m2.ContainsKey(s2[i]))
            {
                var x = m2[s2[i]];
                m2[s2[i]]= ++x;
            }
            else
                m2.Add(s2[i], 1);
        }
 
        foreach(KeyValuePair entry in m1)
        {
             
            // if any frequency is 0, then continue
            // as condition is satisfied
            if (!m2.ContainsKey(entry.Key))
                continue;
 
            // if factor or multiple, then condition satified
            if (m2[entry.Key] != 0 &&
               (m2[entry.Key] % entry.Value == 0 ||
                   entry.Value % m2[entry.Key] == 0))
                continue;
             
            // if condition not satisfied
            else
                return false;
        }
        return true;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        String s1 = "geeksforgeeks", s2 = "geeks";
        if (multipleOrFactor(s1, s2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

 
// JavaScript implementation of above approach
 
// Function that checks if the frequency of character
// are a factor or multiple of each other
function multipleOrFactor(s1, s2){
    // map store frequency of each character
    let m1 = new Map();
    let m2 = new Map();
    for (let i = 0; i < s1.length; i++){
        if(m1[s1[i]])
        m1[s1[i]]++;
        else
        m1[s1[i]] = 1
    }
    for (let i = 0; i < s2.length; i++){
       if(m2[s2[i]])
        m2[s2[i]]++;
        else
        m2[s2[i]] = 1
    }
     
 
    for (var it in m1) {
 
        // if any frequency is 0, then continue
        // as condition is satisfied
        if (!(m2[it]))
            continue;
 
        // if factor or multiple, then condition satified
        if (m2[it] % m1[it] == 0
            || m1[it] % m2[it] == 0)
            continue;
 
        // if condition not satisfied
        else
            return false;
    }
    return true;
}
 
// Driver code
let s1 = "geeksforgeeks";
let s2 = "geeks";
 
multipleOrFactor(s1, s2) ?document.write( "YES")
                             : document.write("NO");
                              

输出：

YES

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