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:

C++

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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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Python3

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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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Output:

YES


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