There is a robot starting at the position (0, 0), the origin, on a 2D plane. Given a sequence of its moves, judge if this robot ends up at (0, 0) after it completes its moves.

You are given a string moves that represents the move sequence of the robot where moves[i] represents its ith move. Valid moves are 'R' (right), 'L' (left), 'U' (up), and 'D' (down).

Return true if the robot returns to the origin after it finishes all of its moves, or false otherwise.

Note: The way that the robot is “facing” is irrelevant. 'R' will always make the robot move to the right once, 'L' will always make it move left, etc. Also, assume that the magnitude of the robot's movement is the same for each move.

Example 1:

Input: moves = "UD"
Output: true
Explanation: The robot moves up once, and then down once. All moves have the same magnitude, so it ended up at the origin where it started. Therefore, we return true.

Example 2:

Input: moves = "LL"
Output: false
Explanation: The robot moves left twice. It ends up two "moves" to the left of the origin. We return false because it is not at the origin at the end of its moves.

Constraints:

• 1 <= moves.length <= 2 * 104

• moves only contains the characters 'U', 'D', 'L' and 'R'.

Explanation:

Method 1:

By Using switch statement.

Step 1:

Convert the string to the character array.

moves.toCharArray();

Step 2:

For every character check the switch condition.

For every character check switch case if anyone matches done the appropriate increment and decrement operations.

Step 3:

If leftRight and upDown are zeros(0) then return true else false.

public class RobotReturnOrigin {
  public static void main(String[] args) {
    String s = "UD";
    System.out.println("Is Robot return to origin: "+judgeCircle(s));
  }

  public static boolean judgeCircle(String moves) {
    int upDown = 0;
    int leftRight = 0;

    for (char c : moves.toCharArray())
    {
      switch (c)
      {
        case 'U':
          upDown++;
          break;
        case 'D':
          upDown--;
          break;
        case 'L':
          leftRight++;
          break;
        case 'R':
          leftRight--;
          break;
      }
    }

    return (leftRight == 0 && upDown == 0); 
  }
}

Output:

Is Robot return to origin: true

Solution Link: https://leetcode.com/problems/robot-return-to-origin/submissions/1112162139/

Method 2:

Convert the string to character array and check the conditions using if else conditions.

CODE:

public class RobotReturnOrigin {
  public static void main(String[] args) {
    String s = "UD";
    System.out.println("Is Robot return to origin: "+judgeCircle(s));
  }

  public static boolean judgeCircle(String moves) {
    int leftRight=0,upDown=0;
    char[] c=moves.toCharArray();
    for(char ch:c){
      if(ch == 'U'){
        upDown++;
      }
      else if(ch == 'D'){
        upDown--;
      }
      else if(ch == 'L'){
        leftRight++;
      }
      else if(ch == 'R'){
        leftRight--;
      }
    }

    return (upDown==0 && leftRight==0);
  }
}

Output:

Is Robot return to origin: true

Solution Link: https://leetcode.com/problems/robot-return-to-origin/submissions/1112160616/

Method 3:

Without converting to character array using string check all the conditions.

public class RobotReturnOrigin {
  public static void main(String[] args) {
    String s = "UD";
    System.out.println("Is Robot return to origin: "+judgeCircle(s));
  }

  public static boolean judgeCircle(String moves) {

    int upDown=0, leftRight = 0;
    for(int i=0; i<moves.length(); i++){

      char move = moves.charAt(i);
      if(move == 'U')
        upDown++;
      else if(move == 'D')
        upDown--;
      else if(move == 'L')
        leftRight++;
      else if(move == 'R')
        leftRight--;
    }

    return (upDown==0 & leftRight==0);
  }
}

Output:

Is Robot return to origin: true

Solution Link: https://leetcode.com/problems/robot-return-to-origin/submissions/1053555665/

Thank you for reading.

Happy Coding😊…