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 <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 = 10print(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. ?> |
Javascript
<script> // JavaScript program for the // above approach function minPerimeter(n) { let l = Math.sqrt(n); let sq = l * l; // if n is a perfect square if (sq == n) return l * 4; else { // Number of rows let row = n / l; // perimeter of the // rectangular grid let perimeter = 2 * (l + row); // if there are blocks left if (n % l != 0) perimeter += 2; return perimeter; } } // Driver Code let n = 10; document.write(Math.floor(minPerimeter(n))) </script> |
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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