How do you solve Difference Between Maximum and Minimum Price Sum on LeetCode?

Difference Between Maximum and Minimum Price Sum (LeetCode #2538) 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: The minimum price sum is always the price of the root node.

What is the time complexity of Difference Between Maximum and Minimum Price Sum?

The optimal solution for Difference Between Maximum and Minimum Price Sum runs in O(n) time and uses O(n) space. This complexity is efficient because we only traverse each node once during the DFS, leading to linear time complexity.

What is the brute force solution for Difference Between Maximum and Minimum Price Sum?

The brute force approach for Difference Between Maximum and Minimum Price Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. We can calculate the price sums for all paths starting from each node by performing a depth-first search (DFS) for each possible root. This will help us find the maximum and minimum price sums for each root and then compute their difference. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Difference Between Maximum and Minimum Price Sum?

Difference Between Maximum and Minimum Price Sum uses the following concepts: Array, Dynamic Programming, Tree, Depth-First Search. The recommended patterns to study are: Depth-First Search, Tree Traversal.

What are the interview tips for Difference Between Maximum and Minimum Price Sum?

Always clarify the tree structure and how nodes are connected before starting your solution. Discuss your thought process and approach before diving into coding to ensure you have a clear plan. Consider edge cases, such as trees with only one node or all nodes having the same price.

What are common mistakes when solving Difference Between Maximum and Minimum Price Sum?

Not considering that the tree can be rooted at any node, leading to incorrect calculations. Failing to properly handle the parent-child relationship during DFS, which can cause infinite loops.

#2538

Difference Between Maximum and Minimum Price Sum

Hard

ArrayDynamic ProgrammingTreeDepth-First SearchDepth-First SearchTree Traversal

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)

We can use a single DFS traversal to calculate the maximum and minimum price sums for paths starting from each node. By leveraging the tree structure, we can efficiently compute the required values without needing to re-root the tree multiple times.

⚙️

Algorithm

4 steps

1Step 1: Construct the graph from the edges.
2Step 2: Use a DFS to calculate the maximum and minimum price sums for paths starting from an arbitrary root.
3Step 3: Store the results in a way that allows us to compute the maximum difference efficiently.
4Step 4: Return the maximum difference found.

solution.py21 lines

1# Full working Python code
2from collections import defaultdict
3
4def maxDifference(n, edges, price):
5    graph = defaultdict(list)
6    for a, b in edges:
7        graph[a].append(b)
8        graph[b].append(a)
9
10    def dfs(node, parent):
11        max_sum = price[node]
12        min_sum = price[node]
13        for neighbor in graph[node]:
14            if neighbor != parent:
15                child_max, child_min = dfs(neighbor, node)
16                max_sum = max(max_sum, child_max + price[node])
17                min_sum = min(min_sum, child_min + price[node])
18        return max_sum, min_sum
19
20    max_sum, min_sum = dfs(0, -1)
21    return max_sum - min_sum

ℹ

Complexity note: This complexity is efficient because we only traverse each node once during the DFS, leading to linear time complexity.

1The minimum price sum is always the price of the root node.
2Using DFS allows us to efficiently calculate the maximum and minimum price sums in a single traversal.

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