(Analysis by Nick Wu)

In this problem, we have two rectangles drawn on a grid and want to draw a square that covers both of the original rectangles and is as small as possible.

Instead of trying to compute the number directly, let's reword the question slightly. Let's instead ask the similar question: is it possible to draw a square with side length $s$ that covers both rectangles?

For some values of $s$, the answer is obviously no. Take the example given in the problem: we know that the square must stretch from $x=1$ all the way over to $x=8$. This means that the side length of the square must be at least $8-1$, or $7$. Similarly, the square must stretch from $y=6$ to $y=9$, so the side length of the square must also be at least $9-6$, or $3$.

In the example given, the answer happened to be $7^2$. Is the answer always the square of the distance that the square is required to stretch?

It turns out the answer to that question is yes. If we look back at that example, we already know exactly how far the square has to stretch: it must stretch $7$ units in the $x$-direction and $3$ units in the $y$-direction. That means that if we make the square $7$ units long and place it directly inside the column from $x=1$ to $x=8$, we can slide it until it covers the desired horizontal region from $y=6$ and $y=9$.

Therefore, what we need to do is compute the distance the square must stretch in both the $x$-direction and the $y$-direction, take the maximum of those, and print out the square of that result.

Here is my Java code:

import java.io.*;
import java.util.*;
public class square {
	public static void main(String[] args) throws IOException {
		// initialize file I/O
		BufferedReader br = new BufferedReader(new FileReader("square.in"));
		PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("square.out")));

		// track the bottom, top, left, and rightmost points that need to be covered 
		int smallestX = 10;
		int largestX = 0;
		int smallestY = 10;
		int largestY = 0;

		// read in two lines, store corners of the pastures
		for(int i = 0; i < 2; i++) {
			StringTokenizer st = new StringTokenizer(br.readLine());
			int xLow = Integer.parseInt(st.nextToken());
			int yLow = Integer.parseInt(st.nextToken());
			int xHigh = Integer.parseInt(st.nextToken());
			int yHigh = Integer.parseInt(st.nextToken());
			if(xLow < smallestX) {
				smallestX = xLow;
			if(xHigh > largestX) {
				largestX = xHigh;
			if(yLow < smallestY) {
				smallestY = yLow;
			if(yHigh > largestY) {
				largestY = yHigh;
		// compute the desired side length of the square
		int desiredSideLength = largestX - smallestX;
		if(largestY - smallestY > largestX - smallestX) {
			desiredSideLength = largestY - smallestY; 
		// print the answer
		pw.println(desiredSideLength * desiredSideLength);