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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
• Difficulty Level : Hard
• Last Updated : 09 Aug, 2019

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 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++

 `// 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; ` `} `

## 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 m1 = ``new` `HashMap<>(); ` `        ``HashMap 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 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 satified ` `            ``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 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 `

## 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 satified ` `            ``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  `

Output:

```YES
```

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