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
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 = 7printMove(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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