# Finding all points in a 2D plane that are k manhattan distance away.

Given an $$N * M$$ matrix representing a 2D plane and a start point $$(sx, sy)$$ and another constant $$k$$ find all points in the matrix such that it has a Manhattan distance equal to $$k$$.

This is how I solved it.

Let $$(tx, ty)$$ be the target point.

$$|sx - tx| + |sy - ty| = k$$

Since $$(sx, sy)$$ and $$k$$ are already known, we can group known terms

$$|-tx - ty| = k - |sx - sy|$$

$$|tx + ty| = k - |sx - sy|$$

$$|tx + ty| =constant$$

All the possible values $$ty$$ can take are

$$ty = constant - tx_i$$ where $$1

Since the range of $$tx$$ is $$1 < tx < N$$ and $$ty$$ is $$1 < ty < M$$

I'm not sure if this is correct. Could someone help me out with this one?

• No, it's not correct. The mistake is when you grouped the terms. Absolute value is not parenthesis – Andrei May 19 at 5:17