LeetCode #1743 — MEDIUM

Restore the Array From Adjacent Pairs

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums.

You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i] = [u_i, v_i] indicates that the elements u_i and v_i are adjacent in nums.

It is guaranteed that every adjacent pair of elements nums[i] and nums[i+1] will exist in adjacentPairs, either as [nums[i], nums[i+1]] or [nums[i+1], nums[i]]. The pairs can appear in any order.

Return the original array nums. If there are multiple solutions, return any of them.

Example 1:

Input: adjacentPairs = [[2,1],[3,4],[3,2]]
Output: [1,2,3,4]
Explanation: This array has all its adjacent pairs in adjacentPairs.
Notice that adjacentPairs[i] may not be in left-to-right order.

Example 2:

Input: adjacentPairs = [[4,-2],[1,4],[-3,1]]
Output: [-2,4,1,-3]
Explanation: There can be negative numbers.
Another solution is [-3,1,4,-2], which would also be accepted.

Example 3:

Input: adjacentPairs = [[100000,-100000]]
Output: [100000,-100000]

Constraints:

nums.length == n
adjacentPairs.length == n - 1
adjacentPairs[i].length == 2
2 <= n <= 10⁵
-10⁵ <= nums[i], u_i, v_i <= 10⁵
There exists some nums that has adjacentPairs as its pairs.

Step 01

Brute Force Baseline

Problem summary: There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums. You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i] = [ui, vi] indicates that the elements ui and vi are adjacent in nums. It is guaranteed that every adjacent pair of elements nums[i] and nums[i+1] will exist in adjacentPairs, either as [nums[i], nums[i+1]] or [nums[i+1], nums[i]]. The pairs can appear in any order. Return the original array nums. 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

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

Example 2

[[4,-2],[1,4],[-3,1]]

Example 3

[[100000,-100000]]

Step 02

Core Insight

What unlocks the optimal approach

Find the first element of nums - it will only appear once in adjacentPairs.
The adjacent pairs are like edges of a graph. Perform a depth-first search from the first element.

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 #1743: Restore the Array From Adjacent Pairs
class Solution {
    public int[] restoreArray(int[][] adjacentPairs) {
        int n = adjacentPairs.length + 1;
        Map<Integer, List<Integer>> g = new HashMap<>();
        for (int[] e : adjacentPairs) {
            int a = e[0], b = e[1];
            g.computeIfAbsent(a, k -> new ArrayList<>()).add(b);
            g.computeIfAbsent(b, k -> new ArrayList<>()).add(a);
        }
        int[] ans = new int[n];
        for (Map.Entry<Integer, List<Integer>> entry : g.entrySet()) {
            if (entry.getValue().size() == 1) {
                ans[0] = entry.getKey();
                ans[1] = entry.getValue().get(0);
                break;
            }
        }
        for (int i = 2; i < n; ++i) {
            List<Integer> v = g.get(ans[i - 1]);
            ans[i] = v.get(1) == ans[i - 2] ? v.get(0) : v.get(1);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1743: Restore the Array From Adjacent Pairs
func restoreArray(adjacentPairs [][]int) []int {
	n := len(adjacentPairs) + 1
	g := map[int][]int{}
	for _, e := range adjacentPairs {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	ans := make([]int, n)
	for k, v := range g {
		if len(v) == 1 {
			ans[0] = k
			ans[1] = v[0]
			break
		}
	}
	for i := 2; i < n; i++ {
		v := g[ans[i-1]]
		ans[i] = v[0]
		if v[0] == ans[i-2] {
			ans[i] = v[1]
		}
	}
	return ans
}

# Accepted solution for LeetCode #1743: Restore the Array From Adjacent Pairs
class Solution:
    def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
        g = defaultdict(list)
        for a, b in adjacentPairs:
            g[a].append(b)
            g[b].append(a)
        n = len(adjacentPairs) + 1
        ans = [0] * n
        for i, v in g.items():
            if len(v) == 1:
                ans[0] = i
                ans[1] = v[0]
                break
        for i in range(2, n):
            v = g[ans[i - 1]]
            ans[i] = v[0] if v[1] == ans[i - 2] else v[1]
        return ans

// Accepted solution for LeetCode #1743: Restore the Array From Adjacent Pairs
struct Solution;

use std::collections::HashMap;
use std::collections::HashSet;

impl Solution {
    fn restore_array(adjacent_pairs: Vec<Vec<i32>>) -> Vec<i32> {
        let mut adj: HashMap<i32, HashSet<i32>> = HashMap::new();
        let mut nodes: HashSet<i32> = HashSet::new();
        for pair in adjacent_pairs {
            let u = pair[0];
            let v = pair[1];
            adj.entry(u).or_default().insert(v);
            adj.entry(v).or_default().insert(u);
            nodes.insert(u);
            nodes.insert(v);
        }
        let mut res = vec![];
        let mut u = *adj
            .iter()
            .filter(|(_, edges)| edges.len() == 1)
            .map(|(u, _)| u)
            .min()
            .unwrap();
        res.push(u);
        while adj[&u].len() == 1 {
            let v = *adj[&u].iter().next().unwrap();
            adj.get_mut(&u).unwrap().remove(&v);
            adj.get_mut(&v).unwrap().remove(&u);
            res.push(v);
            u = v;
        }
        res
    }
}

#[test]
fn test() {
    let adjacent_pairs = vec_vec_i32![[2, 1], [3, 4], [3, 2]];
    let res = vec![1, 2, 3, 4];
    assert_eq!(Solution::restore_array(adjacent_pairs), res);
    let adjacent_pairs = vec_vec_i32![[4, -2], [1, 4], [-3, 1]];
    let res = vec![-3, 1, 4, -2];
    assert_eq!(Solution::restore_array(adjacent_pairs), res);
    let adjacent_pairs = vec_vec_i32![[100000, -100000]];
    let res = vec![-100000, 100000];
    assert_eq!(Solution::restore_array(adjacent_pairs), res);
}

// Accepted solution for LeetCode #1743: Restore the Array From Adjacent Pairs
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1743: Restore the Array From Adjacent Pairs
// class Solution {
//     public int[] restoreArray(int[][] adjacentPairs) {
//         int n = adjacentPairs.length + 1;
//         Map<Integer, List<Integer>> g = new HashMap<>();
//         for (int[] e : adjacentPairs) {
//             int a = e[0], b = e[1];
//             g.computeIfAbsent(a, k -> new ArrayList<>()).add(b);
//             g.computeIfAbsent(b, k -> new ArrayList<>()).add(a);
//         }
//         int[] ans = new int[n];
//         for (Map.Entry<Integer, List<Integer>> entry : g.entrySet()) {
//             if (entry.getValue().size() == 1) {
//                 ans[0] = entry.getKey();
//                 ans[1] = entry.getValue().get(0);
//                 break;
//             }
//         }
//         for (int i = 2; i < n; ++i) {
//             List<Integer> v = g.get(ans[i - 1]);
//             ans[i] = v.get(1) == ans[i - 2] ? v.get(0) : v.get(1);
//         }
//         return ans;
//     }
// }

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.

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.