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.
Move from brute-force thinking to an efficient approach using array strategy.
You are given an n x n integer matrix board where the cells are labeled from 1 to n2 in a Boustrophedon style starting from the bottom left of the board (i.e. board[n - 1][0]) and alternating direction each row.
You start on square 1 of the board. In each move, starting from square curr, do the following:
next with a label in the range [curr + 1, min(curr + 6, n2)].
next has a snake or ladder, you must move to the destination of that snake or ladder. Otherwise, you move to next.n2.A board square on row r and column c has a snake or ladder if board[r][c] != -1. The destination of that snake or ladder is board[r][c]. Squares 1 and n2 are not the starting points of any snake or ladder.
Note that you only take a snake or ladder at most once per dice roll. If the destination to a snake or ladder is the start of another snake or ladder, you do not follow the subsequent snake or ladder.
[[-1,4],[-1,3]], and on the first move, your destination square is 2. You follow the ladder to square 3, but do not follow the subsequent ladder to 4.Return the least number of dice rolls required to reach the square n2. If it is not possible to reach the square, return -1.
Example 1:
Input: board = [[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,35,-1,-1,13,-1],[-1,-1,-1,-1,-1,-1],[-1,15,-1,-1,-1,-1]] Output: 4 Explanation: In the beginning, you start at square 1 (at row 5, column 0). You decide to move to square 2 and must take the ladder to square 15. You then decide to move to square 17 and must take the snake to square 13. You then decide to move to square 14 and must take the ladder to square 35. You then decide to move to square 36, ending the game. This is the lowest possible number of moves to reach the last square, so return 4.
Example 2:
Input: board = [[-1,-1],[-1,3]] Output: 1
Constraints:
n == board.length == board[i].length2 <= n <= 20board[i][j] is either -1 or in the range [1, n2].1 and n2 are not the starting points of any snake or ladder.Problem summary: You are given an n x n integer matrix board where the cells are labeled from 1 to n2 in a Boustrophedon style starting from the bottom left of the board (i.e. board[n - 1][0]) and alternating direction each row. You start on square 1 of the board. In each move, starting from square curr, do the following: Choose a destination square next with a label in the range [curr + 1, min(curr + 6, n2)]. This choice simulates the result of a standard 6-sided die roll: i.e., there are always at most 6 destinations, regardless of the size of the board. If next has a snake or ladder, you must move to the destination of that snake or ladder. Otherwise, you move to next. The game ends when you reach the square n2. A board square on row r and column c has a snake or ladder if board[r][c] != -1. The destination of that snake or ladder is board[r][c]. Squares 1 and n2 are not the starting points of any
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,-1,-1,-1,-1,-1],[-1,35,-1,-1,13,-1],[-1,-1,-1,-1,-1,-1],[-1,15,-1,-1,-1,-1]]
[[-1,-1],[-1,3]]
most-profitable-path-in-a-tree)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #909: Snakes and Ladders
class Solution {
public int snakesAndLadders(int[][] board) {
int n = board.length;
Deque<Integer> q = new ArrayDeque<>();
q.offer(1);
int m = n * n;
boolean[] vis = new boolean[m + 1];
vis[1] = true;
for (int ans = 0; !q.isEmpty(); ++ans) {
for (int k = q.size(); k > 0; --k) {
int x = q.poll();
if (x == m) {
return ans;
}
for (int y = x + 1; y <= Math.min(x + 6, m); ++y) {
int i = (y - 1) / n, j = (y - 1) % n;
if (i % 2 == 1) {
j = n - j - 1;
}
i = n - i - 1;
int z = board[i][j] == -1 ? y : board[i][j];
if (!vis[z]) {
vis[z] = true;
q.offer(z);
}
}
}
}
return -1;
}
}
// Accepted solution for LeetCode #909: Snakes and Ladders
func snakesAndLadders(board [][]int) int {
n := len(board)
q := []int{1}
m := n * n
vis := make([]bool, m+1)
vis[1] = true
for ans := 0; len(q) > 0; ans++ {
for k := len(q); k > 0; k-- {
x := q[0]
q = q[1:]
if x == m {
return ans
}
for y := x + 1; y <= min(x+6, m); y++ {
i, j := (y-1)/n, (y-1)%n
if i%2 == 1 {
j = n - j - 1
}
i = n - i - 1
z := y
if board[i][j] != -1 {
z = board[i][j]
}
if !vis[z] {
vis[z] = true
q = append(q, z)
}
}
}
}
return -1
}
# Accepted solution for LeetCode #909: Snakes and Ladders
class Solution:
def snakesAndLadders(self, board: List[List[int]]) -> int:
n = len(board)
q = deque([1])
vis = {1}
ans = 0
m = n * n
while q:
for _ in range(len(q)):
x = q.popleft()
if x == m:
return ans
for y in range(x + 1, min(x + 6, m) + 1):
i, j = divmod(y - 1, n)
if i & 1:
j = n - j - 1
i = n - i - 1
z = y if board[i][j] == -1 else board[i][j]
if z not in vis:
vis.add(z)
q.append(z)
ans += 1
return -1
// Accepted solution for LeetCode #909: Snakes and Ladders
use std::collections::{HashSet, VecDeque};
impl Solution {
pub fn snakes_and_ladders(board: Vec<Vec<i32>>) -> i32 {
let n = board.len();
let m = (n * n) as i32;
let mut q = VecDeque::new();
q.push_back(1);
let mut vis = HashSet::new();
vis.insert(1);
let mut ans = 0;
while !q.is_empty() {
for _ in 0..q.len() {
let x = q.pop_front().unwrap();
if x == m {
return ans;
}
for y in x + 1..=i32::min(x + 6, m) {
let (mut i, mut j) = ((y - 1) / n as i32, (y - 1) % n as i32);
if i % 2 == 1 {
j = (n as i32 - 1) - j;
}
i = (n as i32 - 1) - i;
let z = if board[i as usize][j as usize] == -1 {
y
} else {
board[i as usize][j as usize]
};
if !vis.contains(&z) {
vis.insert(z);
q.push_back(z);
}
}
}
ans += 1;
}
-1
}
}
// Accepted solution for LeetCode #909: Snakes and Ladders
function snakesAndLadders(board: number[][]): number {
const n = board.length;
const q: number[] = [1];
const m = n * n;
const vis: boolean[] = Array(m + 1).fill(false);
vis[1] = true;
for (let ans = 0; q.length > 0; ans++) {
const nq: number[] = [];
for (const x of q) {
if (x === m) {
return ans;
}
for (let y = x + 1; y <= Math.min(x + 6, m); y++) {
let i = Math.floor((y - 1) / n);
let j = (y - 1) % n;
if (i % 2 === 1) {
j = n - j - 1;
}
i = n - i - 1;
const z = board[i][j] === -1 ? y : board[i][j];
if (!vis[z]) {
vis[z] = true;
nq.push(z);
}
}
}
q.length = 0;
for (const x of nq) {
q.push(x);
}
}
return -1;
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.