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;
    }