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 chracter 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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Output:

YES


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