Check if frequency of character in one string is a factor or multiple of frequency of same character in other string

Given two strings, the task is to check whether the frequencies of a character(for each character) in one string is a multiple or a factor in another string. If it is, then output “YES”, otherwise output “NO”.

Examples:

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

Approach:



  1. Store frequency of characters in s1 in first map STL.
  2. Store frequency of characters in s2 in second map STL.
  3. Let the frequency of a character in first map be F1. Let us also assume the frequency of this character in second map is F2.
  4. Check F1%F2 and F2%F1(modulo operation). If either of them is 0, then the condition is satisfied.
  5. Check it for all the characters.

Below is the implementation of the above approach:

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// C++ implementation of above approach
#include <bits/stdc++.h>
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<char, int> 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<char, int>::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;
}
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// 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<Character, Integer> m1 = new HashMap<>();
        HashMap<Character, Integer> 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<Character, Integer> 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
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# 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
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// 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<char, int> m1 = new Dictionary<char, int>();
        Dictionary<char, int> m2 = new Dictionary<char, int>();
  
        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<char, int> 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 
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Output:
YES

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