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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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 <bits/stdc++.h> 
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

<?php 
// PHP program to find minimum  
// perimeter using n blocks. 
  
function minPerimeter($n) 
{ 
    $l = floor(sqrt($n)); 
    $sq = $l * $l; 
  
    // if n is a perfect square 
    if ($sq == $n)  
        return $l * 4; 
    else
    { 
        // Number of rows  
        $row = floor($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 
$n = 10; 
echo minPerimeter($n); 
  
// This code is contributed  
// by nitin mittal. 
?> 

Output :

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

This article is contributed by Sarthak Kohli. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Minimum Perimeter of n blocks

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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