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Given the following input:

• An integer n where n > 0.
• A string s where s is not empty and s~=[0-9A-Z]+ (alpha-numeric capitals only).

Using a standard, simplified QWERTY keyboard (as shown below):

1234567890 QWERTYUIOP ASDFGHJKL ZXCVBNM 

Perform the following operation:

• Find the original row that each character is in on the keyboard.
• Replace the letter with the correct shifted equivalent for n based on its original position + n.
  • E.G. s="AB" and n=2: A would become D and B would become M.
• If keyboard_row[position + n] > keyboard_row.length, wrap back to the start.
  • E.G. s="0P" and n=2: 0 would become 2 and P would become W.

#Examples:

f("0PLM",1) = 1QAZ f("ZXCVB",2) = CVBNM f("HELLO",3) = LYDDW f("0PLM",11) = 1QSV f("0PLM",2130) = 0PHX 

#Rules

• This is code-golf, lowest byte-count wins.

I think this will be interesting because numbers can be handled differently than letters.

Given the following input:

• An integer n where n > 0.
• A string s where s is not empty and s~=[0-9A-Z]+ (alpha-numeric capitals only).

Using a standard, simplified QWERTY keyboard (as shown below):

1234567890 QWERTYUIOP ASDFGHJKL ZXCVBNM 

Perform the following operation:

• Find the original row that each character is in on the keyboard.
• Replace the letter with the correct shifted equivalent for n based on its original position + n.
  • E.G. s="AB" and n=2: A would become D and B would become M.
• If keyboard_row[position + n] > keyboard_row.length, wrap back to the start.
  • E.G. s="0P" and n=2: 0 would become 2 and P would become W.

#Examples:

f("0PLM",1) = 1QAZ f("ZXCVB",2) = CVBNM f("HELLO",3) = LYDDW f("0PLM",11) = 1QSV f("0PLM",2130) = 0PHX 

#Rules

• This is code-golf, lowest byte-count wins.

This is slightly more difficult than it seems at first glance.

Given the following input:

• An integer n where n > 0.
• A string s where s is not empty and s~=[0-9A-Z]+ (alpha-numeric capitals only).

Using a standard, simplified QWERTY keyboard (as shown below):

1234567890 QWERTYUIOP ASDFGHJKL ZXCVBNM 

Perform the following operation:

• Find the original row that each character is in on the keyboard.
• Replace the letter with the correct shifted equivalent for n based on its original position + n.
  • E.G. s="AB" and n=2: A would become D and B would become M.
• If keyboard_row[position + n] > keyboard_row.length, wrap back to the start.
  • E.G. s="0P" and n=2: 0 would become 2 and P would become W.

#Examples:

f("0PLM",1)=1QAZ f("ZXCVB",2)=CVBNM f("HELLO",3)=LYDDW 

#Rules

• This is code-golf, lowest byte-count wins.

I think this will be interesting because numbers can be handled differently than letters.

Given the following input:

• An integer n where n > 0.
• A string s where s is not empty and s~=[0-9A-Z]+ (alpha-numeric capitals only).

Using a standard, simplified QWERTY keyboard (as shown below):

1234567890 QWERTYUIOP ASDFGHJKL ZXCVBNM 

Perform the following operation:

• Find the original row that each character is in on the keyboard.
• Replace the letter with the correct shifted equivalent for n based on its original position + n.
  • E.G. s="AB" and n=2: A would become D and B would become M.
• If keyboard_row[position + n] > keyboard_row.length, wrap back to the start.
  • E.G. s="0P" and n=2: 0 would become 2 and P would become W.

#Examples:

f("0PLM",1) = 1QAZ f("ZXCVB",2) = CVBNM f("HELLO",3) = LYDDW f("0PLM",11) = 1QSV f("0PLM",2130) = 0PHX 

#Rules

• This is code-golf, lowest byte-count wins.

I think this will be interesting because numbers can be handled differently than letters.

Given the following input:

• An integer n where n > 0.
• A string s where s is not empty and s~=[0-9A-Z]+ (non-empty alpha-numeric capitalized).

Using a standard, simplified QWERTY keyboard (as shown below):

1234567890 QWERTYUIOP ASDFGHJKL ZXCVBNM 

Perform the following operation:

• Find the original row that each character is in on the keyboard.
• Replace the letter with the correct shifted equivalent for n based on its original position + n.
  • E.G. s="AB" and n=2: A would become D and B would become M.
• If keyboard_row[position + n] > keyboard_row.length, wrap back to the start.
  • E.G. s="0P" and n=2: 0 would become 2 and P would become W.

#Examples:

f("0PLM",1)=1QAZ f("ZXCVB",2)=CVBNM f("HELLO",3)=LYDDW 

#Rules

• This is code-golf, lowest byte-count wins.

I think this will be interesting because numbers can be handled differently than letters.

Given the following input:

• An integer n where n > 0.
• A string s where s is not empty and s~=[0-9A-Z]+ (alpha-numeric capitals only).

Using a standard, simplified QWERTY keyboard (as shown below):

1234567890 QWERTYUIOP ASDFGHJKL ZXCVBNM 

Perform the following operation:

• Find the original row that each character is in on the keyboard.
• Replace the letter with the correct shifted equivalent for n based on its original position + n.
  • E.G. s="AB" and n=2: A would become D and B would become M.
• If keyboard_row[position + n] > keyboard_row.length, wrap back to the start.
  • E.G. s="0P" and n=2: 0 would become 2 and P would become W.

#Examples:

f("0PLM",1)=1QAZ f("ZXCVB",2)=CVBNM f("HELLO",3)=LYDDW 

#Rules

• This is code-golf, lowest byte-count wins.

I think this will be interesting because numbers can be handled differently than letters.