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Shortest path on a Square

  • Last Updated : 31 May, 2021

Given side of a square n and two points (x1, y1) and (x2, y2) on the boundaries of the given square. The task is to find the shortest path through the square sides between these two points where the corner coordinates of the square are (0, 0), (n, 0), (0, n), and (n, n).

Examples: 

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Input: n = 2, x1 = 0, y1 = 0, x2 = 1, y2 = 0 
Output:



Input: n = 26, x1 = 21, y1 = 0, x2 = 26, y2 = 14 
Output: 19 

Approach: 

  • If both the x and y coordinates of a point is greater than the other and the points are not on opposite sides of square then the shortest distance will be abs(x2 – x1) + abs(y2 – y1).
  • Else, the shortest distance will be equal to min((x1 + y1 + x2 + y2), (4 * n) – (x1 + y1 + x2 + y2))

Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the length
// of the minimum path between
// two points on a square of given side
int minPath(int n, int x1, int y1, int x2, int y2)
{
 
    // If both of the x and y coordinates
    // of one point is greater than the other
    if ((x1 <= x2 && y1 <= y2) || (x1 >= x2 && y1 >= y2))
    {
        // If the points are not on opposite sides
        if (!(abs(y2 - y1) == n || abs(x2 - x1) == n))
            return (abs(x1 - x2) + abs(y1 - y2));
    }
    return min(x1 + x2 + y1 + y2, (4 * n) - (x1 + x2 + y1 + y2));
}
 
// Driver code
int main()
{
    // Side of the square
    int n = 4;
    int x1 = 2, y1 = 0, x2 = 3, y2 = 4;
    cout << minPath(n, x1, y1, x2, y2);
 
    return 0;
}
 
// improved by Sonal Agrawal

Java




// Java implementation of the approach
class GFG{
 
// Function to return the length
// of the minimum path between
// two points on a square of given side
static int minPath(int n, int x1, int y1,
                          int x2, int y2)
{
     
    // If both of the x and y coordinates
    // of one point is greater than the other
    if ((x1 <= x2 && y1 <= y2) ||
        (x1 >= x2 && y1 >= y2))
    {
         
        // If the points are not on opposite sides
        if (!(Math.abs(y2 - y1) == n ||
              Math.abs(x2 - x1) == n))
            return (Math.abs(x1 - x2) +
                    Math.abs(y1 - y2));
    }
    return Math.min(x1 + x2 + y1 + y2, (4 * n) -
                   (x1 + x2 + y1 + y2));
}
 
// Driver code
public static void main(String[] args)
{
     
    // Side of the square
    int n = 4;
    int x1 = 2, y1 = 0, x2 = 3, y2 = 4;
     
    System.out.println(minPath(n, x1, y1, x2, y2));
}
}
 
// This code is contributed by sanjeev2552

Python3




# Python3 implementation of above approach
 
# Function to return the length of the
# minimum path between two points
# on a square of given side
def minPath(n, x1, y1, x2, y2):
 
    # If both of the x and y coordinates
    # of one point is greater than the other
    if (((x1 <= x2 and y1 <= y2) or
        (x1 >= x2 and y1 >= y2)) and
        not (abs(y2 - y1) == n or
             abs(x2 - x1) == n)):
        return (abs(x1 - x2) + abs(y1 - y2));
 
    return min(x1 + x2 + y1 + y2, (4 * n) -
              (x1 + x2 + y1 + y2));
 
# Driver code
 
# side of the square
n = 4; x1 = 2; y1 = 0
x2 = 3; y2 = 4
print(minPath(n, x1, y1, x2, y2))
 
# This code is contributed
# by Shashank_Sharma

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the length
// of the minimum path between
// two points on a square of given side
static int minPath(int n, int x1, int y1,
                        int x2, int y2)
{
 
    // If both of the x and y coordinates
    // of one point is greater than the other
    if ((x1 <= x2 && y1 <= y2) || (x1 >= x2 && y1 >= y2))
    {
        // If the points are not on opposite sides
        if (!(Math.Abs(y2 - y1) == n || Math.Abs(x2 - x1) == n))
            return (Math.Abs(x1 - x2) + Math.Abs(y1 - y2));
    }
    return Math.Min(x1 + x2 + y1 + y2, (4 * n) - (x1 + x2 + y1 + y2));
}
 
// Driver code
public static void Main()
{
    // Side of the square
    int n = 4;
    int x1 = 2, y1 = 0, x2 = 3, y2 = 4;
    Console.Write(minPath(n, x1, y1, x2, y2));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP




<?php
// Php implementation of the approach
 
// Function to return the length
// of the minimum path between
// two points on a square of given side
function minPath($n, $x1, $y1, $x2, $y2)
{
 
    // If both of the x and y coordinates
    // of one point is greater than the other
    if (($x1 <= $x2 && $y1 <= $y2) || ($x1 >= $x2 && $y1 >= $y2))
    {
        // If the points are not on opposite sides
        if (!(abs($y2 - $y1) == $n || abs($x2 - $x1) == $n))
            return (abs($x1 - $x2) + abs($y1 - $y2));
    }
    return min($x1 + $x2 + $y1 + $y2, (4 * $n) - ($x1 + $x2 + $y1 + $y2));
     
}
 
// Driver code
 
// Side of the square
$n = 4;
$x1 = 2 ;
$y1 = 0 ;
$x2 = 3 ;
$y2 = 4 ;
echo minPath($n, $x1, $y1, $x2, $y2);
     
// This code is contributed by Ryuga
?>

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to return the length
    // of the minimum path between
    // two points on a square of given side
    function minPath(n, x1, y1, x2, y2)
    {
 
        // If both of the x and y coordinates
        // of one point is greater than the other
        if ((x1 <= x2 && y1 <= y2) || (x1 >= x2 && y1 >= y2))
        {
         
            // If the points are not on opposite sides
            if (!(Math.abs(y2 - y1) == n || Math.abs(x2 - x1) == n))
                return (Math.abs(x1 - x2) + Math.abs(y1 - y2));
        }
        return Math.min(x1 + x2 + y1 + y2, (4 * n) - (x1 + x2 + y1 + y2));
    }
     
    // Side of the square
    let n = 4;
    let x1 = 2, y1 = 0, x2 = 3, y2 = 4;
    document.write(minPath(n, x1, y1, x2, y2));
     
    // This code is contributed by decode2207.
</script>
Output
7



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