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={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;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;}`