Leetcode #2508: Add Edges to Make Degrees of All Nodes Even
In this guide, we solve Leetcode #2508 Add Edges to Make Degrees of All Nodes Even in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
class Solution:
def isPossible(self, n: int, edges: List[List[int]]) -> bool:
g = defaultdict(set)
for a, b in edges:
g[a].add(b)
g[b].add(a)
vs = [i for i, v in g.items() if len(v) & 1]
if len(vs) == 0:
return True
if len(vs) == 2:
a, b = vs
if a not in g[b]:
return True
return any(a not in g[c] and c not in g[b] for c in range(1, n + 1))
if len(vs) == 4:
a, b, c, d = vs
if a not in g[b] and c not in g[d]:
return True
if a not in g[c] and b not in g[d]:
return True
if a not in g[d] and b not in g[c]:
return True
return False
return False
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.