LeetCode #2508 — HARD

Add Edges to Make Degrees of All Nodes Even

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

Solve on LeetCode
The Problem

Problem Statement

There is an undirected graph consisting of n nodes numbered from 1 to n. You are given the integer n and a 2D array edges where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi. The graph can be disconnected.

You can add at most two additional edges (possibly none) to this graph so that there are no repeated edges and no self-loops.

Return true if it is possible to make the degree of each node in the graph even, otherwise return false.

The degree of a node is the number of edges connected to it.

Example 1:

Input: n = 5, edges = [[1,2],[2,3],[3,4],[4,2],[1,4],[2,5]]
Output: true
Explanation: The above diagram shows a valid way of adding an edge.
Every node in the resulting graph is connected to an even number of edges.

Example 2:

Input: n = 4, edges = [[1,2],[3,4]]
Output: true
Explanation: The above diagram shows a valid way of adding two edges.

Example 3:

Input: n = 4, edges = [[1,2],[1,3],[1,4]]
Output: false
Explanation: It is not possible to obtain a valid graph with adding at most 2 edges.

Constraints:

  • 3 <= n <= 105
  • 2 <= edges.length <= 105
  • edges[i].length == 2
  • 1 <= ai, bi <= n
  • ai != bi
  • There are no repeated edges.

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: There is an undirected graph consisting of n nodes numbered from 1 to n. You are given the integer n and a 2D array edges where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi. The graph can be disconnected. You can add at most two additional edges (possibly none) to this graph so that there are no repeated edges and no self-loops. Return true if it is possible to make the degree of each node in the graph even, otherwise return false. The degree of a node is the number of edges connected to it.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

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

Example 2

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

Example 3

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

Related Problems

  • Minimum Degree of a Connected Trio in a Graph (minimum-degree-of-a-connected-trio-in-a-graph)
Step 02

Core Insight

What unlocks the optimal approach

  • Notice that each edge that we add changes the degree of exactly 2 nodes.
  • The number of nodes with an odd degree in the original graph should be either 0, 2, or 4. Try to work on each of these cases.
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 #2508: Add Edges to Make Degrees of All Nodes Even
class Solution {
    public boolean isPossible(int n, List<List<Integer>> edges) {
        Set<Integer>[] g = new Set[n + 1];
        Arrays.setAll(g, k -> new HashSet<>());
        for (var e : edges) {
            int a = e.get(0), b = e.get(1);
            g[a].add(b);
            g[b].add(a);
        }
        List<Integer> vs = new ArrayList<>();
        for (int i = 1; i <= n; ++i) {
            if (g[i].size() % 2 == 1) {
                vs.add(i);
            }
        }
        if (vs.size() == 0) {
            return true;
        }
        if (vs.size() == 2) {
            int a = vs.get(0), b = vs.get(1);
            if (!g[a].contains(b)) {
                return true;
            }
            for (int c = 1; c <= n; ++c) {
                if (a != c && b != c && !g[a].contains(c) && !g[c].contains(b)) {
                    return true;
                }
            }
            return false;
        }
        if (vs.size() == 4) {
            int a = vs.get(0), b = vs.get(1), c = vs.get(2), d = vs.get(3);
            if (!g[a].contains(b) && !g[c].contains(d)) {
                return true;
            }
            if (!g[a].contains(c) && !g[b].contains(d)) {
                return true;
            }
            if (!g[a].contains(d) && !g[b].contains(c)) {
                return true;
            }
            return false;
        }
        return false;
    }
}
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

BRUTE FORCE
O(n²) time
O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP
O(n) time
O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Related Problems
#1Two Sumeasy→#13Roman to Integereasy→#30Substring with Concat of All Wordshard→#36Valid Sudokumedium→#41First Missing Positivehard→