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# Minimum Perimeter of n blocks

We are given n blocks of size 1 x 1, we need to find the minimum perimeter of the grid made by these blocks.
Examples :

```Input : n = 4
Output : 8
Minimum possible perimeter with 4 blocks
is 8. See below explanation.

Input : n = 11
Output : 14
The square grid of above examples would be as``` ` `

Let us take an example to see a pattern. Let us say that we have 4 blocks, following are different possibilities

```  +--+--+--+--+
|  |  |  |  |  Perimeter = 10
+--+--+--+--+

+--+--+--+
|  |  |  |     Perimeter = 10
+--+--+--+
|  |
+--+

+--+--+--+
|  |  |  |     Perimeter = 10
+--+--+--+
|  |
+--+

+--+--+
|  |  |        Perimeter = 8
+--+--+
|  |  |
+--+--+```

If we do some examples using pen and paper, we can notice that the perimeter becomes minimum when the shape formed is closest to a square. The reason for this is, we want maximum sides of blocks to face inside the shape so that perimeter of the shape becomes minimum.
If the Number of blocks is a perfect square then the perimeter would simply be 4*sqrt(n).
But, if the Number of blocks is not a perfect square root then we calculate number of rows and columns closest to square root. After arranging the blocks in a rectangular we still have blocks left then we will simply add 2 to the perimeter because only 2 extra side would be left.
The implementation of the above idea is given below.

## C++

 `// CPP program to find minimum ``// perimeter using n blocks.``#include ``using` `namespace` `std;`` ` `int` `minPerimeter(``int` `n)``{``    ``int` `l = ``sqrt``(n);``    ``int` `sq = l * l;`` ` `    ``// if n is a perfect square``    ``if` `(sq == n) ``        ``return` `l * 4;``    ``else``    ``{``        ``// Number of rows ``        ``long` `long` `int` `row = n / l; `` ` `        ``// perimeter of the ``        ``// rectangular grid ``        ``long` `long` `int` `perimeter ``                      ``= 2 * (l + row); `` ` `        ``// if there are blocks left ``        ``if` `(n % l != 0) ``            ``perimeter += 2;``        ``return` `perimeter;``    ``}``}`` ` `// Driver code``int` `main()``{``    ``int` `n = 10;``    ``cout << minPerimeter(n);``    ``return` `0;``}`

## Java

 `// JAVA Code to find minimum ``// perimeter using n blocks``import` `java.util.*;`` ` `class` `GFG ``{``    ``public` `static` `long` `minPerimeter(``int` `n)``    ``{``        ``int` `l = (``int``) Math.sqrt(n);``        ``int` `sq = l * l;``     ` `        ``// if n is a perfect square``        ``if` `(sq == n) ``            ``return` `l * ``4``;``        ``else``        ``{``            ``// Number of rows ``            ``long` `row = n / l; ``     ` `            ``// perimeter of the ``            ``// rectangular grid ``            ``long` `perimeter ``                  ``= ``2` `* (l + row); ``     ` `            ``// if there are blocks left ``            ``if` `(n % l != ``0``) ``                ``perimeter += ``2``;``            ``return` `perimeter;``        ``}``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `main(String[] args) ``    ``{``        ``int` `n = ``10``;``        ``System.out.println(minPerimeter(n));``    ``}``}`` ` `// This code is contributed by Arnav Kr. Mandal`

## Python3

 `# Python3 program to find minimum ``# perimeter using n blocks.``import` `math`` ` `def` `minPerimeter(n):``    ``l ``=` `math.sqrt(n)``    ``sq ``=` `l ``*` `l``  ` `    ``# if n is a perfect square``    ``if` `(sq ``=``=` `n): ``        ``return` `l ``*` `4``    ``else` `:``        ``# Number of rows ``        ``row ``=` `n ``/` `l``  ` `        ``# perimeter of the ``        ``# rectangular grid ``        ``perimeter ``=` `2` `*` `(l ``+` `row)``                       ` `        ``# if there are blocks left ``        ``if` `(n ``%` `l !``=` `0``): ``            ``perimeter ``+``=` `2``        ``return` `perimeter`` ` `# Driver code``n ``=` `10``print``(``int``(minPerimeter(n)))`` ` `# This code is contributed by ``# Prasad Kshirsagar`

## C#

 `// C# Code to find minimum ``// perimeter using n blocks``using` `System;`` ` `class` `GFG ``{``    ``public` `static` `long` `minPerimeter(``int` `n)``    ``{``        ``int` `l = (``int``) Math.Sqrt(n);``        ``int` `sq = l * l;``     ` `        ``// if n is a perfect square``        ``if` `(sq == n) ``            ``return` `l * 4;``        ``else``        ``{``            ``// Number of rows ``            ``long` `row = n / l; ``         ` `            ``// perimeter of the ``            ``// rectangular grid ``            ``long` `perimeter``                  ``= 2 * (l + row); ``     ` `            ``// if there are blocks left ``            ``if` `(n % l != 0) ``                ``perimeter += 2;``            ``return` `perimeter;``        ``}``    ``}``     ` `    ``// Driver code``    ``public` `static` `void` `Main() ``    ``{``        ``int` `n = 10;``        ``Console.Write(minPerimeter(n));``    ``}``}`` ` `// This code is contributed by nitin mittal`

## PHP

 ``

## Javascript

 ``

Output :

`14`

Time complexity : O(logn)
Auxiliary Space : O(1)

References :
http://mathforum.org/library/drmath/view/61595.html
intermath.coe.uga.edu/tweb/gcsu-geo-spr06/aheath/aheath_rectperimeter.doc
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