Shortest path on a Square

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 are (0, 0), (n, 0), (0, n) and (n, n) .

Examples:

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

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 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++

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// 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))
        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 = 2;
    int x1 = 0, y1 = 0, x2 = 1, y2 = 0;
    cout << minPath(n, x1, y1, x2, y2);
  
    return 0;
}

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Python 3














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# 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)):  


        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 = 2; x1 = 0; y1 = 0


x2 = 1; y2 = 0


print(minPath(n, x1, y1, x2, y2)) 


  


# This code is contributed  


# by Shashank_Sharma 



















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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))

		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 = 2;

	int x1 = 0, y1 = 0, x2 = 1, y2 = 0;

	Console.Write(minPath(n, x1, y1, x2, y2));

}

}


// This code is contributed

// by Akanksha Rai



PHP














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<?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))  


        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 = 2;  


    $x1 = 0 ; 


    $y1 = 0 ; 


    $x2 = 1 ; 


    $y2 = 0 ;  


    echo minPath($n, $x1, $y1, $x2, $y2);  


      


// This code is contributed by Ryuga 


?> 



















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Output:


1






          
          
          
          

            

    

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