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 squaresInput:(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 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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