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:
- Store frequency of characters in s1 in first map STL.
- Store frequency of characters in s2 in second map STL.
- 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.
- Check F1%F2 and F2%F1(modulo operation). If either of them is 0, then the condition is satisfied.
- Check it for all the characters.
Below is the implementation of the above approach:
// 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 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 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# 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 |
<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> |
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(n)