LeetCode #773 — HARD

Sliding Puzzle

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

On an 2 x 3 board, there are five tiles labeled from 1 to 5, and an empty square represented by 0. A move consists of choosing 0 and a 4-directionally adjacent number and swapping it.

The state of the board is solved if and only if the board is [[1,2,3],[4,5,0]].

Given the puzzle board board, return the least number of moves required so that the state of the board is solved. If it is impossible for the state of the board to be solved, return -1.

Example 1:

Input: board = [[1,2,3],[4,0,5]]
Output: 1
Explanation: Swap the 0 and the 5 in one move.

Example 2:

Input: board = [[1,2,3],[5,4,0]]
Output: -1
Explanation: No number of moves will make the board solved.

Example 3:

Input: board = [[4,1,2],[5,0,3]]
Output: 5
Explanation: 5 is the smallest number of moves that solves the board.
An example path:
After move 0: [[4,1,2],[5,0,3]]
After move 1: [[4,1,2],[0,5,3]]
After move 2: [[0,1,2],[4,5,3]]
After move 3: [[1,0,2],[4,5,3]]
After move 4: [[1,2,0],[4,5,3]]
After move 5: [[1,2,3],[4,5,0]]

Constraints:

  • board.length == 2
  • board[i].length == 3
  • 0 <= board[i][j] <= 5
  • Each value board[i][j] is unique.
Patterns Used
📊Dynamic Programming→🔀Backtracking→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: On an 2 x 3 board, there are five tiles labeled from 1 to 5, and an empty square represented by 0. A move consists of choosing 0 and a 4-directionally adjacent number and swapping it. The state of the board is solved if and only if the board is [[1,2,3],[4,5,0]]. Given the puzzle board board, return the least number of moves required so that the state of the board is solved. If it is impossible for the state of the board to be solved, return -1.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Backtracking

Example 1

[[1,2,3],[4,0,5]]

Example 2

[[1,2,3],[5,4,0]]

Example 3

[[4,1,2],[5,0,3]]
Step 02

Core Insight

What unlocks the optimal approach

  • Perform a breadth-first-search, where the nodes are the puzzle boards and edges are if two puzzle boards can be transformed into one another with one move.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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
Largest constraint values
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 #773: Sliding Puzzle
class Solution {
    private String[] t = new String[6];
    private int[][] board;

    public int slidingPuzzle(int[][] board) {
        this.board = board;
        String start = gets();
        String end = "123450";
        if (end.equals(start)) {
            return 0;
        }
        Set<String> vis = new HashSet<>();
        Deque<String> q = new ArrayDeque<>();
        q.offer(start);
        vis.add(start);
        int ans = 0;
        while (!q.isEmpty()) {
            ++ans;
            for (int n = q.size(); n > 0; --n) {
                String x = q.poll();
                setb(x);
                for (String y : next()) {
                    if (y.equals(end)) {
                        return ans;
                    }
                    if (!vis.contains(y)) {
                        vis.add(y);
                        q.offer(y);
                    }
                }
            }
        }
        return -1;
    }

    private String gets() {
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 3; ++j) {
                t[i * 3 + j] = String.valueOf(board[i][j]);
            }
        }
        return String.join("", t);
    }

    private void setb(String s) {
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 3; ++j) {
                board[i][j] = s.charAt(i * 3 + j) - '0';
            }
        }
    }

    private int[] find0() {
        for (int i = 0; i < 2; ++i) {
            for (int j = 0; j < 3; ++j) {
                if (board[i][j] == 0) {
                    return new int[] {i, j};
                }
            }
        }
        return new int[] {0, 0};
    }

    private List<String> next() {
        int[] p = find0();
        int i = p[0], j = p[1];
        int[] dirs = {-1, 0, 1, 0, -1};
        List<String> res = new ArrayList<>();
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k];
            int y = j + dirs[k + 1];
            if (x >= 0 && x < 2 && y >= 0 && y < 3) {
                swap(i, j, x, y);
                res.add(gets());
                swap(i, j, x, y);
            }
        }
        return res;
    }

    private void swap(int i, int j, int x, int y) {
        int t = board[i][j];
        board[i][j] = board[x][y];
        board[x][y] = t;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.

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