PROBLEM LINK:
Author:Hasan Jaddouh
Tester:Misha Chorniy
Editorialist:Pushkar Mishra
DIFFICULTY:
Simple
PREREQUISITES:
Greedy
PROBLEM:
Given an array of $A$ of $N$ integers which represent stick lengths, we have to report the maximal area over all the rectangles that we can form with the given sticks without breaking any.
EXPLANATION:
The problem is a simple one based on a very basic property of all rectangles: the opposite sides of a rectangle are parallel, and hence, equal in length. The area of a rectangle is given by base $B$ multiplied by height $H$. Now, to maximise the area, we simply need to maximise the length of sticks we choose as our $B$ and $H$. This is pretty intuitive but we can also invoke the exchange argument method (page 3 of link) to give a more formal argument as proof of why this works (left to reader as a simple learning exercise).
Now we come to the implementation detail of the above mentioned algorithm. Note that $N$ and $A_{max}$ both range between 1 and $10^3$. So, one way is that we simply keep an array $Count$ of length $10^3$ wherein $Count[i]$ is the number of sticks of length $i$ that we have. Once we have populated this structure, we can simply scan from $10^3$ down to 1. If while scanning, we can find one field which has more than 3 sticks, then that $i$ can be our height and base both; at this point we terminate our scan. If we find a field that has more than 1 stick, we can make it either our height or base and continue our scan to find a value for the other one. If at the end of scan we don't find at least two $i$ for which the count is more than 1, then we output -1 since no rectangle can be formed.
The other implementation can be by using a map instead of an array. In that case, if the $N$ is much smaller than $A_{max}$, we can get a better performance. Both of the implementations easily pass for the given constraints.
Please see editorialist's/setter's program for implementation details.
COMPLEXITY:
$\mathcal{O}(N\log N)$ or $\mathcal{O}(A_{max})$ per test case.