Rectangle Geometry

Most problems in this category include only two or three squares or rectangles, in which case you can simply draw out cases on paper. This should logically lead to a solution.

Example - Fence Painting

Slow Solution

Since all the intervals lie between the range, $[0, 100]$, we can mark each interval of length $1$ contained within each interval as painted using a loop. Then the answer will be the number of marked intervals.

Time Complexity: $\mathcal{O}(\text{max coordinate})$

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#include <bits/stdc++.h>
using namespace std;

const int MAX_POS = 100;

int main() {
	freopen("paint.in", "r", stdin);
	freopen("paint.out", "w", stdout);

	int a, b, c, d;

However, this solution would not work for higher constraints (ex. if the coordinates were up to $10^9$).

Fast Solution

Calculate the answer by adding the original lengths and subtracting the intersection length.

The official analysis splits computing the intersection length into several cases. However, we can do it in a simpler way. An interval $[x,x+1]$ is contained within both $[a,b]$ and $[c,d]$ if $a\le x$, $c\le x$, $x<b$, and $x<d$, or in other words if, $\max(a,c)\le x$ and $x<\min(b,d)$. So the length of the intersection is $\min(b,d)-\max(a,c)$ if this quantity is positive and zero otherwise!

Time Complexity: $\mathcal{O}(1)$

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#include <bits/stdc++.h>
using namespace std;

int main() {
	freopen("paint.in", "r", stdin);
	freopen("paint.out", "w", stdout);

	int a, b, c, d;
	cin >> a >> b >> c >> d;

Example - Blocked Billboard

Think of this as the 2D analog of the previous example.

Slow Solution

Time Complexity: $\mathcal{O}((\text{max coordinate})^2)$

Since all coordinates are in the range $[-1000,1000]$, we can simply go through each of the $2000^2$ possible visible squares and check which ones are visible using nested for loops.

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#include <bits/stdc++.h>
using namespace std;

const int MAX_POS = 2000;
bool visible[MAX_POS][MAX_POS];

int main() {
	freopen("billboard.in", "r", stdin);
	freopen("billboard.out", "w", stdout);
	for (int i = 0; i < 3; ++i) {

Of course, this wouldn't suffice if the coordinates were changed to be up to $10^9$.

Fast Solution

Time Complexity: $\mathcal{O}(1)$

Official Analysis

Note how creating a struct Rect to represent a rectangle makes the code easier to understand.

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#include <bits/stdc++.h>
using namespace std;

struct Rect {
	int x1, y1, x2, y2;
	int area() { return (y2 - y1) * (x2 - x1); }
};

int intersect(Rect p, Rect q) {
	int xOverlap = max(0, min(p.x2, q.x2) - max(p.x1, q.x1));

Common Formulas

Certain tasks show up often in rectangle geometry problems. For example, many problems involve finding the overlapping area of two or more rectangles based on their coordinate points, or determining whether two rectangles intersect. Here, we'll discuss these formulas.

Note that these formulas only apply to rectangles which have sides parallel to the coordinate axes.

Finding area

The formula for finding the area of an individual rectangle is $w \cdot l$.

$\texttt{length}$ is the length of the vertical sides, and $\texttt{width}$ is the length of the horizontal sides.

1. $\texttt{width} = \texttt{tr}_x - \texttt{bl}_x$
2. $\texttt{length} = \texttt{tr}_y - \texttt{bl}_y$
3. $\texttt{area} = \texttt{width} \cdot \texttt{length}$

Implementation

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long long area(int bl_x, int bl_y, int tr_x, int tr_y) {
	long long length = tr_y - bl_y;
	long long width = tr_x - bl_x;
	return length * width;
}

Checking if two rectangles intersect

Given two rectangles $a$ and $b$, there are only two cases where they do not intersect:

1. $\texttt{tr}_{a_y}$ $\leq$ $\texttt{bl}_{b_y}$ or $\texttt{bl}_{a_y}$ $\geq$ $\texttt{tr}_{b_y}$.
2. $\texttt{bl}_{a_x}$ $\geq$ $\texttt{tr}_{b_x}$ or $\texttt{tr}_{a_x}$ $\leq$ $\texttt{bl}_{b_x}$.

In all other cases, the rectangles intersect.

Implementation

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bool intersect(vector<int> s1, vector<int> s2) {
	int bl_a_x = s1[0], bl_a_y = s1[1], tr_a_x = s1[2], tr_a_y = s1[3];
	int bl_b_x = s2[0], bl_b_y = s2[1], tr_b_x = s2[2], tr_b_y = s2[3];

	// no overlap
	if (bl_a_x >= tr_b_x || tr_a_x <= bl_b_x || bl_a_y >= tr_b_y ||
	    tr_a_y <= bl_b_y) {
		return false;
	} else {
		return true;
	}
}

Finding area of intersection

We'll assume that the shape formed by the intersection of two rectangles is itself a rectangle.

First, we'll find this rectangle's length and width. $\texttt{width} = \min(\texttt{tr}_{a_x}, \texttt{tr}_{b_x}) - \max(\texttt{bl}_{a_x}, \texttt{bl}_{b_x})$. $\texttt{length} = \min(\texttt{tr}_{a_y}, \texttt{tr}_{b_y}) - \max(\texttt{bl}_{a_y}, \texttt{bl}_{b_y})$.

If either of these values are negative, the rectangles do not intersect. If they are zero, the rectangles intersect at a single point. Multiply the length and width to find the overlapping area.

Implementation

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int inter_area(vector<int> s1, vector<int> s2) {
	int bl_a_x = s1[0], bl_a_y = s1[1], tr_a_x = s1[2], tr_a_y = s1[3];
	int bl_b_x = s2[0], bl_b_y = s2[1], tr_b_x = s2[2], tr_b_y = s2[3];

	return ((min(tr_a_x, tr_b_x) - max(bl_a_x, bl_b_x)) *
	        (min(tr_a_y, tr_b_y) - max(bl_a_y, bl_b_y)));
}

Problems