Don't Get Volunteered!

Problem

As a henchman on Commander Lambda’s space station, you’re expected to be resourceful, smart, and a quick thinker. It’s not easy building a doomsday device and ordering the bunnies around at the same time, after all! In order to make sure that everyone is sufficiently quick-witted, Commander Lambda has installed new flooring outside the henchman dormitories. It looks like a chessboard, and every morning and evening you have to solve a new movement puzzle in order to cross the floor. That would be fine if you got to be the rook or the queen, but instead, you have to be the knight. Worse, if you take too much time solving the puzzle, you get “volunteered” as a test subject for the LAMBCHOP doomsday device!

To help yourself get to and from your bunk every day, write a function called solution(src, dest) which takes in two parameters: the source square, on which you start, and the destination square, which is where you need to land to solve the puzzle. The function should return an integer representing the smallest number of moves it will take for you to travel from the source square to the destination square using a chess knight’s moves (that is, two squares in any direction immediately followed by one square perpendicular to that direction, or vice versa, in an “L” shape). Both the source and destination squares will be an integer between 0 and 63, inclusive, and are numbered like the example chessboard below:

01234567
89101112131415
1617181920212223
2425262728293031
3233343536373839
4041424344454647
4849505152535455
5657585960616263

Test Cases

Your code should pass the following test cases. Note that it may also be run against hidden test cases not shown here.

``````Input:
solution.solution(0, 1)
Output:
3

Input:
solution.solution(19, 36)
Output:
1
``````

My Solution

This wasn’t an easy problem to solve, and the solution I found it’s perhaps not the most optimal, but it works!

My solution consists on reviewing each possible movement combination one level of depth at a time.

One easy to do performance optimization to this code would be to not re-process cells we already checked before.

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