# Untitled

unknown
plain_text
a year ago
1.5 kB
24
Indexable
Never
```public class Solution {
public int solve(ArrayList<String> A, ArrayList<String> B) {
// Find the hashvalue for each string in a ArrayList A
int n=A.size();
HashMap<Long,Integer> hm1=new HashMap<Long,Integer>();
for(int i=0;i<n;i++)
{
int n1=A.get(i).length();
long hash=0;
for(int j=0;j<n1;j++)
{
hash+=((A.get(i).charAt(j)-'a'+1)*(((long)Math.pow(29,n1-j-1))));
}
hm1.put(hash,hm1.getOrDefault(hash,0)+1);
}

// Find the hashvalue for each string in Arraylist B

int m=B.size();
HashMap<Long,Integer> hm2=new HashMap<Long,Integer>();
for(int i=0;i<m;i++)
{
int m1=B.get(i).length();
long hash=0;
for(int j=0;j<m1;j++)
{
hash+=((B.get(i).charAt(j) - 'a'+1)*(((long)Math.pow(29,m1-j-1))));
}
hm2.put(hash,hm2.getOrDefault(hash,0)+1);
}

// Use advance forloop to take each element in the hashmap1 and check its freq is 1 then check the same is in hashmap2
int ans=0;
// System.out.println("hashmap1 : "+hm1+"hashmap 2 :"+hm2);
for(Map.Entry<Long,Integer> set : hm1.entrySet())
{
if(set.getValue() == 1 && hm2.containsKey(set.getKey()) && hm2.get(set.getKey()) == 1)
{
ans++;
}
}

return ans;
}
}
```