### Encoded String codechef January long challenge solution || Encoded String codechef January long challenge solution Editorial 2021

## Encoded String codechef January long challenge solution || Encoded String codechef January long challenge solution Editorial 2021-

### Problem statement-

An encoder encodes the first lowercase English letters using bits each. The first bit (from the left) of the code is if the letter lies among the first letters, else it is , signifying that it lies among the last letters. The second bit of the code is if the letter lies among the first letters of those letters found in the previous step, else it's , signifying that it lies among the last letters of those letters. Similarly, the third and the fourth bit each signify the half in which the letter lies.

For example, the letter would be encoded as :

- Among , appears in the second half. So the first bit of its encoding is .
- Now, among , appears in the first half. So the second bit of its encoding is .
- Now, among , appears in the first half. So the third bit of its encoding is .
- Now, among , appears in the second half. So the fourth and last bit of its encoding is .

So 's encoding is ,

Given a binary encoded string , of length at most , decode the string. That is, the first 4 bits are the encoding of the first letter of the secret message, the next 4 bits encode the second letter, and so on. It is guaranteed that the string's length is a multiple of 4.

### Input:

- The first line of the input contains an integer , denoting the number of test cases.
- The first line of each test case contains an integer , the length of the encoded string.
- The second line of each test case contains the encoded string .

### Output:

For each test case, print the decoded string, in a separate line.

### Constraints

- The length of the encoded string is a multiple of .

### Subtasks

- points : Original constraints.

### Sample Input:

```
3
4
0000
8
00001111
4
1001
```

### Sample Output:

```
a
ap
j
```

### Explanation:

- Sample Case :

The first bit is , so the letter lies among the first letters, i.e., among . The second bit is , so it lies among the first four of these, i.e., among .

The third bit is , so it again lies in the first half, i.e., it's either or . Finally, the fourth bit is also , so we know that the letter is .

- Sample Case :

Each four bits correspond to a character. Just like in sample case , is equivalent to . Similarly, is equivalent to . So, the decoded string is .

- Sample Case :

The first bit is , so the letter lies among the last letters, i.e., among . The second bit is , so it lies among the first four of these, i.e., among .

The third bit is , so it again lies in the first half, i.e., it's either or . Finally, the fourth bit is , so we know that the letter is .

### Solution-

if(s=="0000"){ | |

return 'a'; | |

} | |

if(s=="0001"){ | |

return 'b'; | |

} | |

if(s=="0010"){ | |

return 'c'; | |

} | |

if(s=="0011"){ | |

return 'd'; | |

} | |

if(s=="0100"){ | |

return 'e'; | |

} | |

if(s=="0101"){ | |

return 'f'; | |

} | |

if(s=="0110"){ | |

return 'g'; | |

} | |

if(s=="0111"){ | |

return 'h'; | |

} | |

if(s=="1000"){ | |

return 'i'; | |

} | |

if(s=="1001"){ | |

return 'j'; | |

} | |

if(s=="1010"){ | |

return 'k'; | |

} | |

if(s=="1011"){ | |

return 'l'; | |

} | |

if(s=="1100"){ | |

return 'm'; | |

} | |

if(s=="1101"){ | |

return 'n'; | |

} | |

if(s=="1110"){ | |

return 'o'; | |

} | |

if(s=="1111"){ | |

return 'p'; | |

} Now you have to iterate over string in set of 4 character. finally print ans. follow my code |

## Comments

## Post a Comment