Number of unique rectangles formed using N unit squares

You are given N unit squares (squares with side length 1 unit), and you are asked to make rectangles using these squares. You have to count the number of rotationally unique rectangles than you can make. What does rotationally unique mean? Well, two rectangles are rotationally unique if one can’t be rotated to become equivalent to the other one. 

Example – The 4×2 rectangle can be rotated 90 degrees clockwise to make it the exact same as the 2×4 rectangle and so these are not rotationally unique. 

Examples :

Input : N = 4
Output : 5
We can make following five rectangles 
1 x 1, 1 x 2, 2 x 2, 1 x 3 and 1 x 4

Input : N = 5
Output : 6

Input : 6
Output : 8

So how do we solve this problem? 
Every rectangle is uniquely determined by its length and its height. 
A rectangle of length = l and height = h then l * h <= n is considered equivalent to a rectangle with length = h and height = l provided l is not equal to h. If we can have some sort of “ordering” in these pairs then we can avoid counting (l, h) and (h, l) as different rectangles. One way to define such an ordering is: 

Assume that length <= height and count for all such pairs such that length*height <= n. 
We have, length <= height 
or, length*length <= length*height 
or, length*length <= n 
or, length <= sqrt(n) 

C++

// C++ program to count rotationally equivalent
// rectangles with n unit squares
#include<bits/stdc++.h>
using namespace std;

int countRect(int n)
{
    int ans = 0;
    for (int length = 1; length <= sqrt(n); ++length)
    for (int height = length; height*length <= n; ++height)
            // height >= length is maintained
        ans++;
    return ans;
}

// Driver code
int main() {
    int n = 5;
    printf("%d", countRect(n));
    return 0;
}

Java

// Java program to count rotationally equivalent
// rectangles with n unit squares
class GFG {

    static int countRect(int n)
    {
        int ans = 0;

        for (int length = 1; length <= Math.sqrt(n);
                                           ++length)
            for (int height = length; height*length <= n;
                                                ++height)
                // height >= length is maintained
                ans++;

        return ans;
    }

    //driver code
    public static void main (String[] args)
    {
        int n = 5;

        System.out.print(countRect(n));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python3 program to count rotationally
# equivalent rectangles with n unit squares
import math

def countRect(n):

    ans = 0
    for length in range(1, int(math.sqrt(n)) + 1):
        height = length
        while(height * length <= n):

            # height >= length is maintained
            ans += 1
            height += 1
    return ans

# Driver code
n = 5
print(countRect(n))

# This code is contributed by Anant Agarwal.

C#

// C# program to count rotationally
// equivalent rectangles with n unit
// squares
using System;

class GFG {

    static int countRect(int n)
    {
        int ans = 0;
        for (int length = 1;
            length <= Math.Sqrt(n); ++length)

            for (int height = length;
                     height*length <= n; ++height)
                ans++;

        return ans;
    }

    //driver code
    public static void Main()
    {

        int n = 5;

        Console.Write(countRect(n));
    }
}

//This code is contributed by Anant Agarwal.

PHP

<?php
// PHP program to count
// rotationally equivalent
// rectangles with n unit squares
function countRect($n)
{
    $ans = 0;
    for ($length = 1;
         $length <= sqrt($n); $length++)
    for ($height = $length;
         $height * $length <= $n; $height++)
            // height >= length is maintained
        $ans++;
    return $ans;
}

// Driver code
$n = 5;
echo countRect($n);

// This code is contributed by @ajit
?>

Javascript

<script>

// Javascript program to count rotationally
// equivalent rectangles with n unit
// squares
function countRect(n)
{
    let ans = 0;
    for(let length = 1;
            length <= parseInt(Math.sqrt(n), 10);
          ++length)

        for(let height = length;
                height * length <= n;
              ++height)
            ans++;

    return ans;
}

// Driver code
let n = 5;

document.write(countRect(n));

// This code is contributed by divyesh072019

</script>

Output :

6 

Time Complexity: O(n√n) 
Auxiliary Space: O(1)

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