Maximum number of squares that can fit in a right angle isosceles triangle

You are given an isosceles (a triangle with at-least two equal sides) right angle triangle with base b, we need to find the maximum number of squares of side m, which can be fitted into given triangle.
Examples: 

Input : b = 6, m = 2
Output : 3

Input : b = 4, m = 1
Output : 6

Let’s consider a right angle triangle XYZ, where YZ is the base of triangle. Suppose length of the base is b. If we consider the position of first square with the vertex Y, we will have (b / m-1) squares in the base, and we will be left with another isosceles right angle triangle having base length (b – m).
Illustration : 

Let f(b, m) = Number of squares which can be fitted in triangle having base length b. 
then f(b, m) = (b / m – 1) + f(b – m, m)
We can calculate f(b) using above recursion, and with use of memoization. Later we can answer each query in O(1) time. We can do it for even and odd numbers separately with the base case if (b < 2 * m) f(b, m) = 0.
The given recursion can be solved as :
f(b, m) = b / m – 1 + f(b – m, m) = b / m – 1 + (b – m) / m – 1 + f(b – 2m, m) 
f(b, m) = b / m – 1 + b / m – 2 + f(b – 3m, m) +…+ f(b – (b / m)m, m) 
f(b) = b / m – 1 + b / m – 2 + b / m – 3 +…..+ 1 + 0 
With conditions, if (b < 2 * m) f(b, m) = 0 
f(b) = sum of first (b / m – 1) natural numbers 
= (b / m – 1) * (b / m) / 2 
This formula can be used to reduce the time complexity upto O(1).

C++

// CPP program for finding maximum squares
// that can fit in right angle isosceles
// triangle
#include<bits/stdc++.h>
using namespace std;

// function for finding max squares
int maxSquare(int b, int m)
{
    // return in O(1) with derived
    // formula
    return (b / m - 1) * (b / m) / 2;
}

// driver program
int main()
{
    int b = 10, m = 2;
    cout << maxSquare (b,m);
    return 0;
}

Java

// Java program for finding maximum squares
// that can fit in right angle isosceles
// triangle
public class GFG
{
    // function for finding max squares
    static int maxSquare(int b, int m)
    {
        // return in O(1) with derived
        // formula
        return (b / m - 1) * (b / m) / 2;
    }

    // driver program
    public static void main(String args[])
    {
        int b = 10, m = 2;
        System.out.println(maxSquare (b,m));
    }
}

// This code is contribute by Sumit Ghosh

Python3

# Python3 program for
# finding maximum squares
# that can fit in
# right angle isosceles
# triangle

# function for finding max squares
def maxSquare(b, m):

    # return in O(1) with derived
    # formula
    return (b / m - 1) * (b / m) / 2


# driver program
b = 10
m = 2
print(int(maxSquare (b,m)))

# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program for finding maximum squares
// that can fit in right angle isosceles
// triangle
using System;

public class GFG
{
    // function for finding max squares
    static int maxSquare(int b, int m)
    {
        // return in O(1) with derived
        // formula
        return (b / m - 1) * (b / m) / 2;
    }

    // driver program
    public static void Main()
    {
        int b = 10, m = 2;
        Console.WriteLine(maxSquare (b, m));
    }
}

// This code is contribute by vt_m

PHP

<?php
// PHP program for finding
// maximum squares that can
// fit in right angle isosceles
// triangle

// function for finding
// max squares
function maxSquare($b, $m)
{

    // return in O(1) with
    // derived formula
    return ($b / $m - 1) *
           ($b / $m) / 2;
}

    // Driver Code
    $b = 10; $m = 2;
    echo maxSquare($b,$m);
// This code is contribute by vt_m
?>

Javascript

<script>

// Javascript program for finding maximum squares
// that can fit in right angle isosceles
// triangle

// function for finding max squares
function maxSquare(b, m)
{

    // return in O(1) with derived
    // formula
    return (b / m - 1) * (b / m) / 2; a
}

// Driver program

    let b = 10, m = 2;
    document.write(maxSquare (b,m));

// This code is contributed by Mayank Tyagi
</script>

Output: 

10

Time complexity: O(1)

Auxiliary Space: O(1)
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