#challenge

💻 Letter Distance | #easy

Given two words, the letter distance is calculated by taking the absolute value of the difference in character codes and summing up the difference.

If one word is longer than another, add the difference in lengths towards the score.

To illustrate:

 "fly") = dist("h", "f") + dist("o", "l") + dist("u", "y") + dist(house.Length, fly.Length)

= |104 - 102| + |111 - 108| + |117 - 121| + |5 - 3|
= 2 + 3 + 4 + 2
= 11

Examples:

LetterDistance("sharp", "sharq") ➞ 1
LetterDistance("abcde", "Abcde") ➞ 32
LetterDistance("abcde", "bcdef") ➞ 5

🏆 Leave your solutions in the comments. The solution will be posted below in a couple of hours 👇

#interview