Get Minimum Perimeter of Rectangle Whose Area is N in C++

Suppose we have a number n. We are developing a project to build a new data center. The plot of this data center will be a rectangle with an area of exactly n square meters. Each side of the data center must be an integer. We want to minimize the impact of the external environment on the data center. For this reason, we want to minimize the length of the perimeter of the data center (that is, the sum of the lengths of its four sides). We have to find the minimum perimeter of a rectangular data center with an area of exactly n square meters.

Problem Category

Various problems in programming can be solved through different techniques. To solve a problem, we have to devise an algorithm first, and to do that we have to study the particular problem in detail. A recursive approach can be used if there is a recurring appearance of the same problem over and over again; alternatively, we can use iterative structures also. Control statements such as if-else and switch cases can be used to control the flow of logic in the program. Efficient usage of variables and data structures provides an easier solution and a lightweight, low-memory-requiring program. We have to look at the existing programming techniques, such as Divide-and-conquer, Greedy Programming, Dynamic Programming, and find out if they can be used. This problem can be solved by some basic logic or a brute-force approach. Follow the following contents to understand the approach better.

So, if the input of our problem is like n = 36, then the output will be 24, because the required shape of the data center will be 6×6 square. And the perimeter is 6 + 6 + 6 + 6 = 24.

Steps

To solve this, we will follow these steps −

ans := inf
for initialize i := 1, when i * i <= n, update (increase i by 1), do:
   if n mod i is same as 0, then:
      tmp := n / i
      ans := minimum of ans and 2 * (tmp + i)
return ans

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(int n){
   int ans = 9999;
   for (int i = 1; i * i <= n; i++){
      if (n % i == 0){
         int tmp = n / i;
         ans = min(ans, 2 * (tmp + i));
      }
   }
   return ans;
}
int main(){
   int n = 36;
   cout << solve(n) << endl;
}

Input

36

Output

24