Robot Return to Origin

The robot starts at the origin point (0,0). For the robot to return to the origin after all moves, it needs to make equal number of ‘U’ (Up) and ‘D’ (Down) moves, as well as equal number of ‘L’ (Left) and ‘R’ (Right) moves. This is because ‘U’ and ‘D’ moves cancel each other out in the vertical direction, while ‘L’ and ‘R’ moves cancel each other out in the horizontal direction.

So, we need to count the number of each type of move and check if the count of ‘U’ moves equals the count of ‘D’ moves, and the count of ‘L’ moves equals the count of ‘R’ moves.

Here’s the code:

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class Solution:
    def judgeCircle(self, moves: str) -> bool:
        # Initialize counters for all moves
        up_moves = down_moves = left_moves = right_moves = 0

        # Count each move
        for move in moves:
            if move == 'U':
                up_moves += 1
            elif move == 'D':
                down_moves += 1
            elif move == 'L':
                left_moves += 1
            elif move == 'R':
                right_moves += 1

        # Check if the robot returned to the origin
        return up_moves == down_moves and left_moves == right_moves

This solution is simple and clear, it directly implements the idea of counting moves and checking if the robot returned to the origin. It runs in O(n) time complexity, where n is the length of the moves string, because it iterates through the string once. The space complexity is O(1), because it only uses a constant amount of space to store the counters.

Identifying Problem Isomorphism

“Robot Return to Origin” asks you to determine if a sequence of movements given as a string brings a robot back to its starting point in a 2D grid.

A simpler problem is “Judge Route Circle”. This problem also asks you to determine if a sequence of moves makes a robot return to the origin. The key difference here is that “Judge Route Circle” does not specify the grid dimensions, whereas “Robot Return to Origin” does.

On the more complex side, “Robot Bounded in Circle” is isomorphic. This problem not only deals with the robot’s position but also its orientation. The task here is to determine if a series of moves leaves the robot in a circle.

The hierarchy based on complexity is as follows:

  1. “Judge Route Circle” - you need to check whether a series of moves returns the robot to the origin, but without the specifics of a grid.
  2. “Robot Return to Origin” - adds the dimension of a 2D grid to the previous problem.
  3. “Robot Bounded in Circle” - adds the complexity of the robot’s orientation in addition to its position.

While not an exact match, these problems share the theme of interpreting a sequence of moves to determine the robot’s position relative to its starting point.

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class Solution:
    def judgeCircle(self, moves: str) -> bool:
        return moves.count('L') == moves.count('R') and moves.count('U') == moves.count('D')

Problem Classification

This falls under the domain of Robotics and Algorithms. It’s a type of simulation problem, where the movements of a robot on a 2D plane are simulated based on a given sequence of moves.

‘What’ Components:

  1. Robot: We are given a robot that starts at the position (0,0) on a 2D plane.
  2. Move sequence: The robot follows a sequence of moves provided as a string, where each character represents a different direction of movement: ‘R’ for right, ‘L’ for left, ‘U’ for up, and ‘D’ for down.
  3. Robot’s final position: The robot moves according to the provided sequence. After completing all of its moves, we need to determine if the robot ends up at the origin (0,0).
  4. Return Value: We need to return a boolean value: True if the robot returns to the origin and False otherwise.

This problem is a simulation type problem with a clear deterministic result. The robot follows a series of instructions and we need to determine if these instructions lead it back to its starting point. This involves understanding the rules of the simulation (how each move affects the robot’s position) and then applying these rules to the sequence of moves. There is no need to find an optimal solution or make any decisions during the simulation, so it’s not a decision or optimization problem.

Language Agnostic Coding Drills

  1. Dissection of the code:

The code contains several distinct concepts:

a. Method Definition: The code defines a method judgeCircle which takes a string parameter moves and returns a boolean value.

b. String Manipulation: The method uses the count function on the string moves to count the occurrences of ‘L’, ‘R’, ‘U’, and ‘D’.

c. Comparisons: It uses comparison operators (==) to compare the counts of moves in opposing directions (‘L’ vs ‘R’ and ‘U’ vs ‘D’).

d. Logical Operators: It uses a logical AND operator (and) to combine the two comparisons.

  1. Coding drills in order of increasing difficulty:

    a. Method Definition: Define a method or function. This is one of the basic concepts in programming.

    b. String Manipulation: Use string functions, such as count, to manipulate or analyze strings. It requires understanding of built-in string functions and their usage.

    c. Comparisons: Use comparison operators to compare values. It involves understanding the different comparison operators and their behavior.

    d. Logical Operators: Use logical operators to combine boolean expressions. It requires an understanding of boolean logic and operator precedence.

  2. Problem-solving approach:

The problem-solving approach for this problem is relatively straightforward due to the problem’s nature.

a. First, we count the number of ‘L’ and ‘R’ moves. If the robot moves right and left the same number of times, it will end up on the same vertical line where it started.

b. Then we count the number of ‘U’ and ‘D’ moves. If the robot moves up and down the same number of times, it will end up on the same horizontal line where it started.

c. Using the AND operator, we combine these two conditions. If both conditions are True, the robot ends up at the origin, and we return True. If not, the robot doesn’t end up at the origin, and we return False.

Each of the coding drills contributes to the final solution: defining the function allows us to structure our code; string manipulation enables us to analyze the input; comparison operations let us compare the number of steps in each direction; and the logical operator lets us combine these comparisons to determine the final result.

Targeted Drills in Python

  1. Python Coding Drills:

    a. Method Definition:

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    def greet(name):
    print(f"Hello, {name}!")
greet("Python learner")

    This drill is about defining a method in Python, calling it, and passing an argument to it.

    b. String Manipulation:

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    s = "Hello, Python learner!"
print(s.count('l'))

    This drill shows how to use the count function to count the number of occurrences of a character in a string.

    c. Comparisons:

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    a = 5
b = 5
print(a == b)

    This drill shows how to use the == operator to compare two values in Python.

    d. Logical Operators:

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    a = 5
b = 5
c = 6
print(a == b and b == c)

    This drill demonstrates how to use the and operator to combine two comparisons.

  2. Problem-Specific Coding Drills:

    The problem-specific drill here would involve understanding how to navigate a 2D plane using specific movements and count them properly, as well as understanding how to check if the robot has returned to the origin.

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    moves = "UDLR"
up_down = moves.count('U') - moves.count('D')
left_right = moves.count('L') - moves.count('R')
print(up_down == 0 and left_right == 0)

    In this drill, we simulate the robot’s movements by counting the different types of moves it can make and checking whether it has returned to its original position. It’s crucial for this problem as it directly relates to understanding the robot’s movements and how to verify if it has returned to the origin.

  3. Assembling the Drills into a Solution:

    We can combine these drills to build the solution to our problem. We start by defining our function judgeCircle that takes in the robot’s movements. Inside this function, we perform our string manipulation and counting for each type of move the robot can make. We then use comparisons to check if the robot moved the same distance in opposite directions, which would mean it returned to the origin. Lastly, we use the logical operator and to combine these comparisons and return the result.

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    def judgeCircle(moves):
    up_down = moves.count('U') - moves.count('D')
    left_right = moves.count('L') - moves.count('R')
    return up_down == 0 and left_right == 0
print(judgeCircle("UDLRUDLR"))

    This code will return True if the robot returns to the origin and False otherwise.