# Anagram

## program:-

Given two strings, a and b , that may or may not be of the same length, determine the minimum number of character deletions required to make a and b anagrams. Any characters can be deleted from either of the strings.
Input :

• test cases,t
• two strings a and b, for each test case

Output:
Desired O/p
Constraints :
string lengths<=10000
Note :
Anagram of a word is formed by rearranging the letters of the word.
For e.g. -> For the word RAM – MAR,ARM,AMR,RMA etc. are few anagrams.

SAMPLE INPUT

`1cdeabc`
SAMPLE OUTPUT

`4`
Time Limit:1.0 sec(s) for each input file.
Memory Limit:256 MB
Source Limit:1024 KB

## solotion:-

#include<stdio.h>
#include<string.h>
void main()
{
char a,b;
int l1,l2,s=0,t,i,j,k,l,sum,s1=0,s2=0;
scanf(“%d”,&t);
for(i=1;i<=t;i++)
{
scanf(“%s”,a);
scanf(“%s”,b);
l1=strlen(a);
l2=strlen(b);
for(j=0;j<=l11;j++)
{
for(k=0;k<=l11;k++)
{
if(a[j]==a[k])
{
s1=s1+1;
if(k!=j)
a[k]=1;
}
}
for(k=0;k<=l21;k++)
{
if(a[j]==b[k])
s2=s2+1;
}
if(s1>s2)
s=s+s2;
else
s=s+s1;
s1=0;
s2=0;
}
l=l1+l2;
sum=2*s;
printf(“%dn”,(lsum));
s=0;
}
}