# Reaching a point using clockwise or anticlockwise movements

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( ), R , D and L . 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


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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  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 setps 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 setps 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 setps 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 setps 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

 

Output :

Clockwise


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