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Determine the number of squares of unit area that a given line will pass through.
  • Difficulty Level : Expert
  • Last Updated : 26 Mar, 2021

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

 



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 numer
// 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 numer
// 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(1)
 

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