ABABAABA

• Problem Description

  
  You are given a uniformly randomly generated string S, consisting of letters from the set {""A"", ""B""}. Your task is to find a string T that appears in S as a subsequence exactly twice.
  In other words, you need to find such a string T, that there exist exactly two sets of indexes i1, i2, ..., i|T| and j1, j2, ..., j|T| such that there exists some k, where ik jk and S{i1...i|T|} = S{j1...j|T|} = T.
  
  Input
  The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
  The first and only line of each test case contains a single string S.
  The string S was generated randomly. For a generating string S, we first choose an integer N denoting a length of S. After that every symbol of the string S is chosen randomly from the set {""A"", ""B""} and the both symbols have equal probability to be chosen. Note that N is not choosen randomly.
  
  Output
  For each test case, output a string that occurs exactly twice as a subsequence in S, or output -1 if there is no such string. If there are more than one possible subsequences occurring exactly two times, you can print any one of them.
  
  Constraints
  1 <=T<= 10
  
  Explanation
  Test case #1: 
  The string ""AAAA"" appears once as a subsequence in itself.
  The string ""AAA"" appears four times as a subsequence in ""AAAA""; possible positions: {2, 3, 4}, {1, 3, 4}, {1, 2, 4}, {1, 2, 3}.
  The strings ""AA"" and ""A"" also appear in ""AAAA"" as a subsequence strictly more than twice.
  
  So, there is no string of ""AAAA"", which appears exactly twice. Hence answer is -1.
  Test case #2: Two occurrences of ""B"" in ""BAB"" are {1} and {3} (1-based indexing)."

• CODING ARENA::

• #include <stdlib.h> 

  #include<stdio.h>

  #include<string.h>

  #include<math.h> 

  int main() 

  {int t;

   scanf("%d",&t);

   while(t--)

  {char ch[5010];

   scanf("%s",ch);

   int i,j,x=0,y=0;

   for(i=0;ch[i]!='\0';i++)

  {if(ch[i]=='A')

       x++;

   else

       y++;

   }

   if(x==2)

   {printf("A\n");continue;

   }

   if(y==2)

  {printf("B\n");continue;

  }

   if(x==1 || y==1)

   {printf("-1\n");continue;

   }

   for(j=0;j<i;j++)

   {int ans=0;

   while(ch[j]!='A' && j<i)

   {ans++;

       j++;

   }

    if(ans==2)

    {for(x=0;x<j-2;x++)

     {if(ch[x]=='A')

        printf("%c",ch[x]);

     }

     printf("B");

     for(x=j;x<i;x++)

     {if(ch[x]=='A')

        printf("%c",ch[x]);

         

     }

       printf("\n");

     j=-1;

    }

    if(j==-1)

        break;

   }

   if(j==-1)

       {continue;

       }

    for(j=0;j<i;j++)

   {int ans=0;

   while(ch[j]!='B' && j<i)

   {ans++;

       j++;

   }

    if(ans==2)

    {for(x=0;x<j-2;x++)

     {if(ch[x]=='B')

        printf("%c",ch[x]);

     }

     printf("A");

     for(x=j;x<i;x++)

     {if(ch[x]=='B')

        printf("%c",ch[x]);

         

     }

     printf("\n");

     j=-1;

    }

    if(j==-1)

        break;

   }

   if(j==-1)

       {continue;

       }

   printf("-1\n");

  } 

      return 0;

  }   

• Test Case 1

  
  Input (stdin)

  2
  
  AAAA
  
  BAB
  
  

  Expected Output

  -1
  
  B

• Test Case 2

  
  Input (stdin)

  3
  
  AAAAA
  
  BABBA
  
  AAA
  
  

  Expected Output

  -1
  
  A
  
  -1