README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3248.Snake%20in%20Matrix/README_EN.md	1171	Weekly Contest 410 Q1	Array String Simulation

3248. Snake in Matrix

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Description

There is a snake in an n x n matrix grid and can move in four possible directions. Each cell in the grid is identified by the position: grid[i][j] = (i * n) + j.

The snake starts at cell 0 and follows a sequence of commands.

You are given an integer n representing the size of the grid and an array of strings commands where each command[i] is either "UP", "RIGHT", "DOWN", and "LEFT". It's guaranteed that the snake will remain within the grid boundaries throughout its movement.

Return the position of the final cell where the snake ends up after executing commands.

 

Example 1:

Input: n = 2, commands = ["RIGHT","DOWN"]

Output: 3

Explanation:

0	1
2	3

0	1
2	3

0	1
2	3

Example 2:

Input: n = 3, commands = ["DOWN","RIGHT","UP"]

Output: 1

Explanation:

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

0	1	2
3	4	5
6	7	8

 

Constraints:

• 2 <= n <= 10
• 1 <= commands.length <= 100
• commands consists only of "UP", "RIGHT", "DOWN", and "LEFT".
• The input is generated such the snake will not move outside of the boundaries.

Solutions

Solution 1: Simulation

We can use two variables $x$ and $y$ to represent the position of the snake. Initially, $x = y = 0$. Then, we traverse $\textit{commands}$ and update the values of $x$ and $y$ based on the current command. Finally, we return $x \times n + y$.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{commands}$. The space complexity is $O(1)$.

Python3

class Solution:
    def finalPositionOfSnake(self, n: int, commands: List[str]) -> int:
        x = y = 0
        for c in commands:
            match c[0]:
                case "U":
                    x -= 1
                case "D":
                    x += 1
                case "L":
                    y -= 1
                case "R":
                    y += 1
        return x * n + y

Java

class Solution {
    public int finalPositionOfSnake(int n, List<String> commands) {
        int x = 0, y = 0;
        for (var c : commands) {
            switch (c.charAt(0)) {
                case 'U' -> x--;
                case 'D' -> x++;
                case 'L' -> y--;
                case 'R' -> y++;
            }
        }
        return x * n + y;
    }
}

C++

class Solution {
public:
    int finalPositionOfSnake(int n, vector<string>& commands) {
        int x = 0, y = 0;
        for (const auto& c : commands) {
            switch (c[0]) {
            case 'U': x--; break;
            case 'D': x++; break;
            case 'L': y--; break;
            case 'R': y++; break;
            }
        }
        return x * n + y;
    }
};

Go

func finalPositionOfSnake(n int, commands []string) int {
	x, y := 0, 0
	for _, c := range commands {
		switch c[0] {
		case 'U':
			x--
		case 'D':
			x++
		case 'L':
			y--
		case 'R':
			y++
		}
	}
	return x*n + y
}

TypeScript

function finalPositionOfSnake(n: number, commands: string[]): number {
    let [x, y] = [0, 0];
    for (const c of commands) {
        c[0] === 'U' && x--;
        c[0] === 'D' && x++;
        c[0] === 'L' && y--;
        c[0] === 'R' && y++;
    }
    return x * n + y;
}