# 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.

Input: s1 = “hhdwjwqq”, s2 = “qwjdddhhh”
Output: NO

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

 `// C++ implementation of above approach ` `#include ` `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; ` `} `

Output:

```YES
```

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