Recently in our oavoservice interview assistance services, TikTok interviews have frequently featured a classic "division equation" problem — LeetCode 399.
The problem seems simple but hides subtle traps. Many candidates think at first glance: "This is just mathematical computation, right?" Wrong. This problem tests graph theory modeling ability and DFS traversal techniques, and 90% of mathematical approaches will crash during Follow-up questions.
Let's see how oavoservice real-time interview assistance helps candidates identify the interviewer's traps.
01 TikTok Interview Original Question (Complete English Version)
Title: Division
You are given an array of variable pairs equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] represent the equation Ai / Bi = values[i].
Each Ai or Bi is a string that represents a single variable.
You are also given some queries, where queries[j] = [Cj, Dj] represents the jth query where you must find the answer for Cj / Dj = ?. Return the answers to all queries. If a single answer cannot be determined, return -1.0.
Note:
- The input is always valid
- You may assume that evaluating the queries will not result in division by zero
- There is no contradiction
- Variables that do not occur in the list of equations are undefined
Example Input:
equations = [["a","b"],["b","c"]]
values = [2.0,3.0]
queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
Output: [6.00000,0.50000,-1.00000,1.00000,-1.00000]
02 Common Candidate Mistake: Falling into "Mathematical Computation" Mindset
Most candidates see this problem and their first reaction is:
"I need to solve this using algebraic equation systems, build matrices, then use Gaussian elimination..."
Stop! The interviewer thinks: "Another programmer brainwashed by math class."
This approach might work for simple cases, but when the interviewer asks:
- "What if there are 10,000 variables?"
- "What if there are 50,000 queries?"
- "Can you support dynamically adding new equations?"
The mathematical approach instantly crashes with exploding time complexity.
03 oavoservice Key Guidance: Graph Theory Modeling Approach
In our real-time assistance, we quickly point candidates in the right direction:
Core Insight: Treat variables as nodes and division relationships as directed weighted edges!
- Nodes: Each variable (like "a", "b", "c")
- Edge weights:
a/b = 2.0means edgea → bhas weight2.0, and edgeb → ahas weight1/2.0 = 0.5 - Queries: DFS from start to end, multiply all edge weights along the path
One-sentence solution:
Treat variables as nodes, division as directed weighted edges, use DFS to find paths and multiply weights.
04 oavoservice Optimized Solution (Live Implementation)
from collections import defaultdict
def calcEquation(equations, values, queries):
# 1. Build graph: g[X][Y] = value of X/Y; also build reverse edge g[Y][X] = 1/value
graph = defaultdict(dict)
for (x, y), val in zip(equations, values):
graph[x][y] = val
graph[y][x] = 1.0 / val
# 2. DFS: from start to end, multiply weights along the way
def dfs(start, end, visited):
# Handle boundary cases
if start not in graph or end not in graph:
return -1.0
if start == end:
return 1.0
visited.add(start)
for neighbor, weight in graph[start].items():
if neighbor in visited:
continue
result = dfs(neighbor, end, visited)
if result != -1.0:
return weight * result
return -1.0
# 3. Process all queries
return [dfs(c, d, set()) for c, d in queries]
# Verify results
equations = [["a","b"],["b","c"]]
values = [2.0, 3.0]
queries = [["a","c"],["b","a"],["a","e"],["a","a"],["x","x"]]
print(calcEquation(equations, values, queries))
# Output: [6.0, 0.5, -1.0, 1.0, -1.0]
05 Interviewer Follow-ups & Optimization Points
After candidates write the basic version, TikTok interviewers typically ask:
Q1: "How can you optimize time complexity?"
oavoservice guidance approach:
- Memoization: After DFS finds a path, directly store
graph[start][end] = result - Union-Find optimization: Use weighted Union-Find for nearly O(1) queries
Q2: "If there are many queries, how would you preprocess?"
Advanced solution: Floyd-Warshall all-pairs shortest path
# Preprocessing: O(V³) to calculate paths between any two nodes
def preprocess_all_paths(graph):
nodes = list(graph.keys())
for k in nodes:
for i in nodes:
for j in nodes:
if k in graph[i] and j in graph[k]:
graph[i][j] = graph[i][k] * graph[k][j]
Q3: "Does it support dynamic updates?"
Real-time update strategy:
- Incremental graph updates
- Cache invalidation mechanism
- Lazy recomputation
06 Why Do 70% of Candidates Fail?
- Wrong approach: Obsessing over mathematical formulas, ignoring graph theory essence
- Missing boundaries: Forgetting to handle
x/x = 1and undefined variables-1 - No cycle detection: DFS without
visitedset leads to infinite loops - Poor complexity awareness: Unable to optimize for large datasets
What interviewers really want to see: The ability to abstract real-world problems into graph theory models.
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❌ Unclear graph theory modeling approaches
❌ Weak grasp of DFS/BFS algorithms
❌ Unable to quickly optimize during interviewer follow-ups
❌ Lack of big tech interview experience and techniques
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