Reaching a point using clockwise or anticlockwise movements

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Given starting and ending position and a number N. Given that we are allowed to move in only four directions as shown in the image below. The directions of moves are U( $\uparrow$ ), R $\rightarrow$ , D $\downarrow$ and L $\leftarrow$ . We need to write a program to determine if starting from the given starting position we can reach the given end position in exactly N moves in moving about any direction(Clockwise or Anticlockwise).

Examples :

Input: start = U , end = L , N = 3 
Output: Clockwise 
Explanation: Step 1: move clockwise to reach R
             Step 2: move clockwise to reach D
             Step 3: move clockwise to reach L 
So we reach from U to L in 3 steps moving in 
clockwise direction.

Input: start = R , end = L , N = 3
Output: Not possible 
Explanation: It is not possible to start from 
R and end at L in 3 steps moving about in any 
direction. 

Input: start = D , end = R , N = 7 
Output: Clockwise
Explanation: Starting at D, we complete one 
complete clockwise round in 4 steps to reach D 
again, then it takes 3 step to reach R

The idea to solve this problem is to observe that we can complete one round in 4 steps by traveling in any direction (clockwise or anti-clockwise), so taking n%4 steps is equivalent to taking n steps from the starting point. Therefore n is reduced to n%4. Consider the values of ‘U’ as 0, ‘R’ as 1, ‘D’ as 2 and ‘L’ as 3. If the abs(value(a)-value(b)) is 2 and n is also 2, then we can move either in clockwise or anticlockwise direction to reach the end position from the start position. If moving k steps in clockwise direction take us to the end position from start position then we can say that the condition for clockwise move will be (value(a)+k)%4==value(b). Similarly, the condition for anticlockwise move will be (value(a)+k*3)%4==value(b) since taking k step from position a in clockwise direction is equivalent to taking (a + k*3)%4 steps in anticlockwise direction.
Below is the implementation of the above approach:

C++

// CPP program to determine if  
// starting from the starting  
// position we can reach the   
// end position in N moves  
// moving about any direction 
#include <bits/stdc++.h> 
using namespace std; 
  
// function that returns mark  
// up value of directions 
int value(char a) 
{ 
    if (a == 'U') 
        return 0; 
    if (a == 'R') 
        return 1; 
    if (a == 'D') 
        return 2; 
    if (a == 'L') 
        return 3; 
} 
  
// function to print 
// the possible move 
void printMove(char a, char b, int n) 
{ 
    // mod with 4 as completing  
    // 4 steps means completing  
    // one single round 
    n = n % 4; 
  
    // when n is 2 and the  
    // difference between moves is 2 
    if (n == 2 and abs(value(a) -  
                       value(b)) == 2) 
        cout << "Clockwise or Anticlockwise"; 
  
    // anticlockwise condition 
    else if ((value(a) + n * 3) % 4 == value(b)) 
        cout << "Anticlockwise"; 
  
    // clockwise condition 
    else if ((value(a) + n) % 4 == value(b)) 
        cout << "Clockwise"; 
    else
        cout << "Not Possible"; 
} 
  
// Driver Code 
int main() 
{ 
    char a = 'D', b = 'R'; 
    int n = 7; 
    printMove(a, b, n); 
  
    return 0; 
} 

Java

// Java program to determine if  
// starting from the starting  
// position we can reach the  
// end position in N moves  
// moving about any direction 
class GFG 
{ 
    // function that returns mark  
    // up value of directions 
    static int value(char a) 
    { 
        if (a == 'U') 
            return 0; 
        if (a == 'R') 
            return 1; 
        if (a == 'D') 
            return 2; 
        if (a == 'L') 
            return 3; 
              
            return -1; 
    } 
  
    // function to print 
    // the possible move 
    static void printMove(char a,  
                          char b,  
                          int n) 
    { 
        // mod with 4 as completing  
        // 4 steps means completing 
        // one single round 
        n = n % 4; 
      
        // when n is 2 and  
        // the difference 
        // between moves is 2 
        if (n == 2 && Math.abs(value(a) -  
                               value(b)) == 2) 
            System.out.println("Clockwise " + 
                        " or Anticlockwise"); 
      
        // anticlockwise condition 
        else if ((value(a) + n * 3) % 
                       4 == value(b)) 
            System.out.println("Anticlockwise"); 
      
        // clockwise condition 
        else if ((value(a) + n) % 4 == value(b)) 
            System.out.println("Clockwise"); 
        else
            System.out.println("Not Possible"); 
    } 
  
    // Driver Code 
    public static void main(String args[]) 
    { 
        char a = 'D', b = 'R'; 
        int n = 7; 
        printMove(a, b, n); 
    } 
} 
  
// This code is contributed by Sam007 

Python3

# python program to determine  
# if starting from the starting 
# position we can reach the end  
# position in N moves moving   
# any direction 
  
# function that returns mark 
# up value of directions 
def value(a): 
      
    if (a == 'U'): 
        return 0
    if (a == 'R'): 
        return 1
    if (a == 'D'): 
        return 2
    if (a == 'L'): 
        return 3
  
# function to print  
# the possible move 
def printMove(a, b, n): 
      
    # mod with 4 as completing 
    # 4 steps means completing 
    # one single round 
    n = n % 4; 
  
    # when n is 2 and  
    # the difference 
    # between moves is 2 
    if (n == 2 and
        abs(value(a) - value(b)) == 2): 
        print ("Clockwise or Anticlockwise") 
  
    # anticlockwise condition 
    elif ((value(a) + n * 3) % 4 == value(b)): 
        print ("Anticlockwise") 
  
    # clockwise condition 
    elif ((value(a) + n) % 4 == value(b)): 
        print ("Clockwise") 
    else: 
        print ("Not Possible") 
  
  
# Driver Code 
a = 'D'
b = 'R'
n = 7
printMove(a, b, n) 
  
# This code is contributed by Sam007. 

C#

// C# program to determine 
// if starting from the  
// starting position we  
// can reach the end position 
// in N moves moving about  
// any direction 
using System; 
  
class GFG 
{ 
    // function that returns mark  
    // up value of directions 
    static int value(char a) 
    { 
        if (a == 'U') 
            return 0; 
        if (a == 'R') 
            return 1; 
        if (a == 'D') 
            return 2; 
        if (a == 'L') 
            return 3; 
          
            return -1; 
    } 
  
    // function to print 
    // the possible move 
    static void printMove(char a,  
                          char b,  
                          int n) 
    { 
        // mod with 4 as completing  
        // 4 steps means completing 
        // one single round 
        n = n % 4; 
      
        // when n is 2 and  
        // the difference 
        // between moves is 2 
        if (n == 2 && Math.Abs(value(a) -  
                               value(b)) == 2) 
            Console.Write("Clockwise " + 
                    "or Anticlockwise"); 
      
        // anticlockwise condition 
        else if ((value(a) + n * 3) %  
                        4 == value(b)) 
            Console.Write("Anticlockwise"); 
      
        // clockwise condition 
        else if ((value(a) + n) %  
                    4 == value(b)) 
            Console.WriteLine("Clockwise"); 
        else
            Console.WriteLine("Not Possible"); 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
    char a = 'D', b = 'R'; 
    int n = 7; 
    printMove(a, b, n); 
    } 
} 
  
// This code is contributed by Sam007 

PHP

<?php 
// PHP program to determine  
// if starting from the  
// starting position we can 
// reach the end position in 
// N moves moving about  
// any direction 
  
// function that returns mark  
// up value of directions 
  
function value($a) 
{ 
    if ($a == 'U') 
        return 0; 
    if ($a == 'R') 
        return 1; 
    if ($a == 'D') 
        return 2; 
    if ($a == 'L') 
        return 3; 
} 
  
// function to print 
// the possible move 
function printMove($a, $b,$n) 
{ 
    // mod with 4 as completing  
    // 4 steps means completing  
    // one single round 
    $n = $n % 4; 
  
    // when n is 2 and the  
    // difference between 
    // moves is 2 
    if ($n == 2 and abs(value($a) -  
                        value($b)) == 2) 
        echo "Clockwise or Anticlockwise"; 
  
    // anticlockwise condition 
    else if ((value($a) + $n * 3) %  
                    4 == value($b)) 
        echo "Anticlockwise"; 
  
    // clockwise condition 
    else if ((value($a) + $n) %  
                4 == value($b)) 
        echo "Clockwise"; 
    else
        echo "Not Possible"; 
} 
  
// Driver Code 
$a = 'D'; $b = 'R'; 
$n = 7; 
printMove($a, $b, $n); 
  
// This code is contributed ajit. 
?> 

Javascript

<script> 
  
// JavaScript program to determine if  
// starting from the starting  
// position we can reach the  
// end position in N moves  
// moving about any direction 
  
    // function that returns mark  
    // up value of directions 
    function value(a) 
    { 
        if (a == 'U') 
            return 0; 
        if (a == 'R') 
            return 1; 
        if (a == 'D') 
            return 2; 
        if (a == 'L') 
            return 3; 
                
            return -1; 
    } 
    
    // function to print 
    // the possible move 
    function printMove(a, b, n) 
    { 
        // mod with 4 as completing  
        // 4 steps means completing 
        // one single round 
        n = n % 4; 
        
        // when n is 2 and  
        // the difference 
        // between moves is 2 
        if (n == 2 && Math.abs(value(a) -  
                               value(b)) == 2) 
            document.write("Clockwise " + 
                        " or Anticlockwise"); 
        
        // anticlockwise condition 
        else if ((value(a) + n * 3) % 
                       4 == value(b)) 
            document.write("Anticlockwise"); 
        
        // clockwise condition 
        else if ((value(a) + n) % 4 == value(b)) 
            document.write("Clockwise"); 
        else
            document.write("Not Possible"); 
    } 
    
  
// Driver code 
      
        let a = 'D', b = 'R'; 
        let n = 7; 
        printMove(a, b, n); 
      
    // This code is contributed by code_hunt. 
</script> 

Output :

Clockwise

Time Complexity: O(1), as we are not using any loop or recursion to traverse.

Auxiliary Space: O(1), as we are not using any extra space.

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