How do you solve Evaluate Division on LeetCode?

Evaluate Division (LeetCode #399) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * α(n)) time complexity and O(n) space complexity. Key insight: Understanding the problem as a graph allows for efficient traversal and relationship mapping.

What is the brute force solution for Evaluate Division?

The brute force approach for Evaluate Division is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each query individually by traversing the equations to find a path from the numerator to the denominator. This is akin to searching for a route on a map without using any shortcuts. The optimal Optimal Solution approach improves this to O(n * α(n)).

What are the interview tips for Evaluate Division?

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can represent the problem using data structures like graphs or union-find. Practice explaining your thought process as you code, as communication is key in interviews.

What are common mistakes when solving Evaluate Division?

Failing to handle cases where variables are not present in the equations. Not considering the direction of relationships when calculating values.

#399

Evaluate Division

Medium

ArrayStringDepth-First SearchBreadth-First SearchUnion-FindGraph TheoryShortest PathGraph TraversalUnion-FindDepth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * α(n))
Space	O(1)	O(n)

💡

Intuition

Time O(n * α(n))Space O(n)

The optimal solution leverages graph traversal with a union-find structure to efficiently find the relationships between variables. This is like having a map with direct routes, allowing us to quickly find the shortest path between two points.

⚙️

Algorithm

3 steps

1Step 1: Build a graph using a union-find data structure to connect variables based on the equations.
2Step 2: For each query, check if both variables are in the same connected component.
3Step 3: If they are connected, calculate the value using the stored products; otherwise, return -1.0.

solution.py36 lines

1# Full working Python code
2class UnionFind:
3    def __init__(self):
4        self.parent = {}
5        self.value = {}
6
7    def find(self, x):
8        if self.parent[x] != x:
9            origin = self.find(self.parent[x])
10            self.value[x] *= self.value[self.parent[x]]
11            self.parent[x] = origin
12        return self.parent[x]
13
14    def union(self, x, y, value):
15        if x not in self.parent:
16            self.parent[x] = x
17            self.value[x] = 1.0
18        if y not in self.parent:
19            self.parent[y] = y
20            self.value[y] = 1.0
21        rootX = self.find(x)
22        rootY = self.find(y)
23        self.parent[rootX] = rootY
24        self.value[rootX] = value * self.value[y] / self.value[x]
25
26def calcEquation(equations, values, queries):
27    uf = UnionFind()
28    for (a, b), value in zip(equations, values):
29        uf.union(a, b, value)
30    results = []
31    for c, d in queries:
32        if c in uf.parent and d in uf.parent and uf.find(c) == uf.find(d):
33            results.append(uf.value[c] / uf.value[d])
34        else:
35            results.append(-1.0)
36    return results

ℹ

Complexity note: The time complexity is O(n * α(n)) where α(n) is the inverse Ackermann function, which grows very slowly, making this approach efficient. The space complexity is O(n) due to the storage of parent and value mappings.

1Understanding the problem as a graph allows for efficient traversal and relationship mapping.
2Union-Find is particularly useful for dynamic connectivity problems, where we need to group elements based on relationships.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.