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

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  • Difficulty Level : Hard
  • Last Updated : 28 Sep, 2022
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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(1), because constant size hashmap m1 and m2 are used

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