LeetCode #3248 — EASY

Snake in Matrix

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

2
["RIGHT","DOWN"]

Example 2

3
["DOWN","RIGHT","UP"]

Step 02

Core Insight

What unlocks the optimal approach

Try to update the row and column of the snake after each command.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3248: Snake in Matrix
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;
    }
}

// Accepted solution for LeetCode #3248: Snake in Matrix
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
}

# Accepted solution for LeetCode #3248: Snake in Matrix
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

// Accepted solution for LeetCode #3248: Snake in Matrix
/**
 * [3248] Snake in Matrix
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn final_position_of_snake(n: i32, commands: Vec<String>) -> i32 {
        let (mut x, mut y) = (0, 0);

        commands.iter().for_each(|str| match str.as_str() {
            "UP" => x -= 1,
            "DOWN" => x += 1,
            "LEFT" => y -= 1,
            "RIGHT" => y += 1,
            _ => {}
        });

        x * n + y
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3248() {
        assert_eq!(
            3,
            Solution::final_position_of_snake(2, vec_string!("RIGHT", "DOWN"))
        );
        assert_eq!(
            1,
            Solution::final_position_of_snake(3, vec_string!("DOWN", "RIGHT", "UP"))
        );
    }
}

// Accepted solution for LeetCode #3248: Snake in Matrix
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.