Extra Corners#540

Solution

If $2 k < n$ , the removed corner squares do not overlap in their interiors. At each corner, two boundary segments of length $k$ disappear and two new segments of the same length appear, so the total perimeter remains $4 n$ .

If $2 k = n$ , the removed squares touch at a single point, and two square pieces of side length $n - k$ n-k remain. We count the entire boundary of each piece. Their total perimeter is $8 (n - k) = 4 n$ 8(n-k)=4n, which agrees with the second example.

If $2 k > n$ 2k>n, the remaining paper consists of two separate squares of side length $n - k$ n-k. Their total perimeter is $2 \cdot 4 (n - k) = 8 (n - k) .$ 2\cdot4(n-k)=8(n-k).

Thus the answer is $4 n$ when $2 k \leq n$ 2k\le n, and $8 (n - k)$ 8(n-k) otherwise.

Complexity

$O (1)$ time and $O (1)$ extra memory.