LeetCode #3311 — HARD

Construct 2D Grid Matching Graph Layout

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

Solve on LeetCode
The Problem

Problem Statement

You are given a 2D integer array edges representing an undirected graph having n nodes, where edges[i] = [ui, vi] denotes an edge between nodes ui and vi.

Construct a 2D grid that satisfies these conditions:

  • The grid contains all nodes from 0 to n - 1 in its cells, with each node appearing exactly once.
  • Two nodes should be in adjacent grid cells (horizontally or vertically) if and only if there is an edge between them in edges.

It is guaranteed that edges can form a 2D grid that satisfies the conditions.

Return a 2D integer array satisfying the conditions above. If there are multiple solutions, return any of them.

Example 1:

Input: n = 4, edges = [[0,1],[0,2],[1,3],[2,3]]

Output: [[3,1],[2,0]]

Explanation:

Example 2:

Input: n = 5, edges = [[0,1],[1,3],[2,3],[2,4]]

Output: [[4,2,3,1,0]]

Explanation:

Example 3:

Input: n = 9, edges = [[0,1],[0,4],[0,5],[1,7],[2,3],[2,4],[2,5],[3,6],[4,6],[4,7],[6,8],[7,8]]

Output: [[8,6,3],[7,4,2],[1,0,5]]

Explanation:

Constraints:

  • 2 <= n <= 5 * 104
  • 1 <= edges.length <= 105
  • edges[i] = [ui, vi]
  • 0 <= ui < vi < n
  • All the edges are distinct.
  • The input is generated such that edges can form a 2D grid that satisfies the conditions.

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: You are given a 2D integer array edges representing an undirected graph having n nodes, where edges[i] = [ui, vi] denotes an edge between nodes ui and vi. Construct a 2D grid that satisfies these conditions: The grid contains all nodes from 0 to n - 1 in its cells, with each node appearing exactly once. Two nodes should be in adjacent grid cells (horizontally or vertically) if and only if there is an edge between them in edges. It is guaranteed that edges can form a 2D grid that satisfies the conditions. Return a 2D integer array satisfying the conditions above. If there are multiple solutions, return any of them.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

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

Example 2

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

Example 3

9
[[0,1],[0,4],[0,5],[1,7],[2,3],[2,4],[2,5],[3,6],[4,6],[4,7],[6,8],[7,8]]
Step 02

Core Insight

What unlocks the optimal approach

  • Observe the indegrees of the nodes.
  • The case where there are two nodes with an indegree of 1, and all the others have an indegree of 2 can be handled separately.
  • The nodes with the smallest degrees are the corners.
  • You can simulate the grid creation process using BFS or a similar approach after making some observations on the indegrees.
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 #3311: Construct 2D Grid Matching Graph Layout
class Solution {
    public int[][] constructGridLayout(int n, int[][] edges) {
        List<Integer>[] g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int[] e : edges) {
            int u = e[0], v = e[1];
            g[u].add(v);
            g[v].add(u);
        }

        int[] deg = new int[5];
        Arrays.fill(deg, -1);

        for (int x = 0; x < n; x++) {
            deg[g[x].size()] = x;
        }

        List<Integer> row = new ArrayList<>();
        if (deg[1] != -1) {
            row.add(deg[1]);
        } else if (deg[4] == -1) {
            int x = deg[2];
            for (int y : g[x]) {
                if (g[y].size() == 2) {
                    row.add(x);
                    row.add(y);
                    break;
                }
            }
        } else {
            int x = deg[2];
            row.add(x);
            int pre = x;
            x = g[x].get(0);
            while (g[x].size() > 2) {
                row.add(x);
                for (int y : g[x]) {
                    if (y != pre && g[y].size() < 4) {
                        pre = x;
                        x = y;
                        break;
                    }
                }
            }
            row.add(x);
        }

        List<List<Integer>> res = new ArrayList<>();
        res.add(new ArrayList<>(row));

        boolean[] vis = new boolean[n];
        int rowSize = row.size();
        for (int i = 0; i < n / rowSize - 1; i++) {
            for (int x : row) {
                vis[x] = true;
            }
            List<Integer> nxt = new ArrayList<>();
            for (int x : row) {
                for (int y : g[x]) {
                    if (!vis[y]) {
                        nxt.add(y);
                        break;
                    }
                }
            }
            res.add(new ArrayList<>(nxt));
            row = nxt;
        }

        int[][] ans = new int[res.size()][rowSize];
        for (int i = 0; i < res.size(); i++) {
            for (int j = 0; j < rowSize; j++) {
                ans[i][j] = res.get(i).get(j);
            }
        }
        return ans;
    }
}
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)
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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

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