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 <stdio.h>

  #include <string.h>

  int main()

  {int t;

   scanf("%d",&t);

   while(t--)

   {char c[5010];

    scanf("%s",c);

    int i,j,x=0,y=0;

    for(i=0;c[i]!='\0';i++)

    {if(c[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(c[j]!='A' && j<i)

     {ans++;

        j++;

     }

      if(ans==2)

      {for(x=0;x<j-2;x++)

        {if(c[x]=='A')

            printf("%c",c[x]);

        }

        printf("B");

        for(x=j;x<i;x++)

        {if(c[x]=='A')

          printf("%c",c[x]);

         

        }

       printf("\n");

       j=-1;

      }

     if(j==-1)

       break;

    }

    if(j==-1)

    {continue;

    }

    for(j=0;j<i;j++)

    {int ans=0;

     while(c[j]!='B' && j<i)

     {ans++;

      j++;

     }

     if(ans==2)

     {for(x=0;x<j-2;x++)

     {if(c[x]=='B')

       printf("%c",c[x]);

     }

     printf("A");

      for(x=j;x<i;x++)

      {if(c[x]=='B')

        printf("%c",c[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