LeetCode 2061: Number of Spaces Cleaning Robot Cleaned Solution

Master LeetCode problem 2061 (Number of Spaces Cleaning Robot Cleaned), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2061. Number of Spaces Cleaning Robot Cleaned

Medium

Array Matrix Simulation

Problem Explanation

Explanation:

The problem requires us to determine the number of spaces cleaned by a cleaning robot based on its movement instructions. The cleaning robot starts at position (0,0) in a grid and moves according to the given instructions. The robot can move up, down, left, or right by one unit at a time.

To solve this problem, we can simulate the movement of the cleaning robot and keep track of the spaces cleaned. We will maintain a set to store the coordinates of the spaces cleaned during the robot's movement.

Algorithm:

Initialize a set to store the coordinates of the spaces cleaned.
Initialize the robot's position as (0,0).
Iterate through each movement instruction:
- Update the robot's position based on the instruction.
- Add the new position to the set of cleaned spaces.
Return the size of the set, which represents the number of spaces cleaned.

Time Complexity:

The time complexity of this algorithm is O(N), where N is the number of movement instructions given.

Space Complexity:

The space complexity of this algorithm is O(N), where N is the number of unique spaces cleaned during the robot's movement.

Solution Code

class Solution {
    public int numberofSpacesCleaned(String instructions) {
        Set<String> cleanedSpaces = new HashSet<>();
        int x = 0, y = 0;
        
        cleanedSpaces.add("0,0");
        
        for (char instruction : instructions.toCharArray()) {
            switch(instruction) {
                case 'U':
                    y++;
                    break;
                case 'D':
                    y--;
                    break;
                case 'L':
                    x--;
                    break;
                case 'R':
                    x++;
                    break;
            }
            cleanedSpaces.add(x + "," + y);
        }
        
        return cleanedSpaces.size();
    }
}

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