Determine the number of squares of unit area that a given line will pass through.

Given 2 endpoints (x1, y1) and (x2, y2) of a line, the task is to determine the number of squares of the unit area that line will pass through.

Examples:

Input: (x1 = 1, y1 = 1), (x2 = 4, y2 = 3) 
Output: 4 
In the diagram above the line is passing through 4 squares
Input: (x1 = 0, y1 = 0), (x2 = 2, y2 = 2) 
Output: 2 

Approach: 
Let, 

Dx = (x2 - x1)
Dy = (y2 - y1)

Therefore, 

x = x1 + Dx * t
y = y1 + Dy * t

We have to find (x, y) for t in (0, 1].
For, x and y to be integral Dx and Dy must be divisible by t. Also, t cannot be irrational since Dx and Dy are integers.
Therefore let t = p / q.
Dx and Dy must be divisible by q. So GCD of Dx and Dy must be q. 
Or, q = GCD(Dx, Dy).
There are only GCD(Dx, Dy) smallest subproblems.
Below is the implementation of the above approach: 

C++

#include<bits/stdc++.h>
using namespace std;

// Function to return the required position
int noOfSquares(int x1, int y1, int x2, int y2)
{
    int dx = abs(x2 - x1);
    int dy = abs(y2 - y1);

    int ans = dx + dy - __gcd(dx, dy);

    cout<<ans;
}

// Driver Code
int main()
{
    int x1 = 1, y1 = 1, x2 = 4, y2 = 3;

    noOfSquares(x1, y1, x2, y2);

    return 0;
}

Java

// Java program to determine the number
// of squares that line will pass through
class GFG
{
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);

}

// Function to return the required position
static void noOfSquares(int x1, int y1,
                        int x2, int y2)
{
    int dx = Math.abs(x2 - x1);
    int dy = Math.abs(y2 - y1);

    int ans = dx + dy - __gcd(dx, dy);

    System.out.println(ans);
}

// Driver Code
public static void main(String []args)
{
    int x1 = 1, y1 = 1, x2 = 4, y2 = 3;

    noOfSquares(x1, y1, x2, y2);
}
}

// This code contributed by Rajput-Ji

Python3

# Python3 program to determine the number
# of squares that line will pass through
from math import gcd

# Function to return the required position
def noOfSquares(x1, y1, x2, y2) :

    dx = abs(x2 - x1);
    dy = abs(y2 - y1);

    ans = dx + dy - gcd(dx, dy);

    print(ans);

# Driver Code
if __name__ == "__main__" :

    x1 = 1; y1 = 1; x2 = 4; y2 = 3;

    noOfSquares(x1, y1, x2, y2);

# This code is contributed by Ryuga

C#

using System;

class GFG
{
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);

}

// Function to return the required position
static void noOfSquares(int x1, int y1,
                        int x2, int y2)
{
    int dx = Math.Abs(x2 - x1);
    int dy = Math.Abs(y2 - y1);

    int ans = dx + dy - __gcd(dx, dy);

    Console.WriteLine(ans);
}

// Driver Code
static void Main()
{
    int x1 = 1, y1 = 1, x2 = 4, y2 = 3;

    noOfSquares(x1, y1, x2, y2);
}
}

// This code is contributed by mits

PHP

<?php
// PHP program to determine the number
// of squares that line will pass through

// Function to return the required position
function noOfSquares($x1, $y1, $x2, $y2)
{

    $dx = abs($x2 - $x1);
    $dy = abs($y2 - $y1);

    $ans = $dx + $dy - gcd($dx, $dy);

    echo($ans);
}

function gcd($a, $b)
{
    if ($b == 0)
        return $a;
    return gcd($b, $a % $b);

}

// Driver Code
$x1 = 1; $y1 = 1; $x2 = 4; $y2 = 3;
noOfSquares($x1, $y1, $x2, $y2);

// This code has been contributed
// by 29AjayKumar
?>

Javascript

<script>

function __gcd(a, b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);

}

// Function to return the required position
function noOfSquares(x1, y1, x2, y2)
{
    var dx = Math.abs(x2 - x1);
    var dy = Math.abs(y2 - y1);

    var ans = dx + dy - __gcd(dx, dy);

    document.write(ans);
}

// Driver Code
var x1 = 1, y1 = 1, x2 = 4, y2 = 3;
noOfSquares(x1, y1, x2, y2);

// This code is contributed by noob2000.
</script>

Output:

4

Time Complexity: O(logn)

Auxiliary Space: O(1)