How do you solve Unit Conversion I on LeetCode?

Unit Conversion I (LeetCode #3528) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Graph traversal is key to solving unit conversions efficiently.

What is the time complexity of Unit Conversion I?

The optimal solution for Unit Conversion I runs in O(n) time and uses O(n) space. Each unit is processed once, leading to linear complexity.

What is the brute force solution for Unit Conversion I?

The brute force approach for Unit Conversion I is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach involves repeatedly converting units using all available conversions until all units are calculated. It’s inefficient as it checks every possible path. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Unit Conversion I?

Unit Conversion I uses the following concepts: Depth-First Search, Breadth-First Search, Graph Theory. The recommended patterns to study are: Graph Traversal, Dynamic Programming.

What are the interview tips for Unit Conversion I?

Clarify the input format and constraints. Think about edge cases, such as no conversions. Explain your thought process while coding.

What are common mistakes when solving Unit Conversion I?

Not handling cycles in the conversion graph. Forgetting to apply modulo operation.

#3528

Unit Conversion I

Medium

Depth-First SearchBreadth-First SearchGraph TheoryGraph TraversalDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Using BFS or DFS, we can efficiently traverse the conversion graph from the base unit, multiplying conversion factors along the way to compute equivalent units.

⚙️

Algorithm

3 steps

1Step 1: Build a graph from the conversions array where each unit points to its conversions.
2Step 2: Perform a BFS starting from unit 0, updating the equivalent units using the conversion factors.
3Step 3: Return the results modulo 10^9 + 7.

solution.py17 lines

1from collections import defaultdict, deque
2
3def unitConversion(conversions):
4    graph = defaultdict(list)
5    for src, tgt, factor in conversions:
6        graph[src].append((tgt, factor))
7    n = len(conversions) + 1
8    baseUnitConversion = [0] * n
9    baseUnitConversion[0] = 1
10    queue = deque([0])
11    while queue:
12        unit = queue.popleft()
13        for tgt, factor in graph[unit]:
14            if baseUnitConversion[tgt] == 0:
15                queue.append(tgt)
16            baseUnitConversion[tgt] = (baseUnitConversion[tgt] + baseUnitConversion[unit] * factor) % (10**9 + 7)
17    return baseUnitConversion

ℹ

Complexity note: Each unit is processed once, leading to linear complexity.

1Graph traversal is key to solving unit conversions efficiently.
2Understanding the relationships between units helps in structuring the problem.

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