ABABAABA

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)." 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

Program :

#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;

}