Open In App

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

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Share
Report issue
Report

Given two strings, the task is to check whether the frequencies of a character(for each character) in one string are 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 the first map be F1. Let us also assume the frequency of this character in the 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:

C++




// 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 satisfied
        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<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 satisfied
            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 satisfied
        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<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 satisfied
            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




<script>
 
// 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 satisfied
        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");
                              
</script>


Output

YES

Complexity Analysis:

  • Time Complexity: O((n+m)*log(n+m)), where n is the size of string s1 and m is the size of string s2
  • Auxiliary Space: O(n)


Last Updated : 06 Mar, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads