In this HackerEarth Xsquare And Double Strings problem solution, Xsquare got bored playing with the arrays all the time. Therefore, he has decided to play with the strings. Xsquare called a string P a "double string" if string P is not empty and can be broken into two strings A and B such that A + B = P and A = B. for eg : strings like "baba" , "blabla" , "lolo" are all double strings whereas strings like "hacker" , "abc" , "earth" are not double strings at all.

Today, Xsquare has a special string S consisting of lower case English letters. He can remove as many characters ( possibly zero ) as he wants from his special string S. Xsquare wants to know , if its possible to convert his string S to a double string or not.

Help him in accomplishing this task.

## HackerEarth Xsquare And Double Strings problem solution.

```#include <bits/stdc++.h>
using namespace std ;
#define LL long long int
#define ft first
#define sd second
#define PII pair<int,int>
#define MAXN 100005
#define MAXM 10000001
#define mp make_pair
#define f_in(st) freopen(st,"r",stdin)
#define f_out(st) freopen(st,"w",stdout)
#define sc(x) scanf("%d",&x)
#define scll(x) scanf("%lld",&x)
#define pr(x) printf("%d\n",x)
#define pb push_back
#define MOD 1000000007
#define MAX 1000010
#define ull long long
#define prime 37
#define pb push_back
#define ASST(x,y,z) assert(x >= y && x <= z)

int t , n;
string s ;
int main(){
f_in("in02.txt") ;
f_out("out02.txt") ;
sc(t) ;
ASST(t,1,100) ;
while(t --){
cin >> s ;
n = s.length() ;
ASST(n,1,100) ;
int M[26]={0} , c = 0;
for(int i=0;i<n;i++) ASST(s[i],'a','z') ;
bool ok = false ;
for(int i=0;i<n;i++) M[s[i]-'a'] ++ ;
for(int i=0;i<26;i++) if(M[i] > 1) ok = true  ;
puts(ok ? "Yes" : "No") ;
}
return 0 ;
}```

### Second solution

```#include <bits/stdc++.h>

using namespace std;

int cnt[32];

int main()
{
int t;
string s;
cin >> t;
while ( t-- ) {
cin >> s;
memset(cnt, 0, sizeof(cnt));
for ( int i = 0; i < s.size(); i++ ) cnt[s[i]-'a']++;
for ( int i = 0; i < 26; i++ ) {
if ( cnt[i] >= 2 ) {
printf("Yes\n");
goto p1;
}
}
printf("No\n");
p1: { }
}
return 0;
}```