Minimum Increments to Equalize Leaf Paths

To solve the problem, we need to find the minimum number of increments required to make all root-to-leaf paths have equal scores. The approach focuses on traversing the tree to compute the cost of each path and adjusting the costs to match the highest path score.

Problem Statement

You are given a tree with n nodes, rooted at node 0. The tree's structure is represented by a list of edges where each edge connects two nodes. Each node has an associated cost, which represents the cost of traversing that node. The goal is to make the sum of costs for all root-to-leaf paths equal by increasing the cost of the nodes along those paths. The number of increments required to achieve this is the desired output.

For each root-to-leaf path, you must adjust the costs so that all paths have the same total score. Specifically, each path’s cost must be raised to the maximum root-to-leaf path score found in the tree. The task is to find the minimum number of increments to achieve this uniformity.

Examples

Example 1

Input: n = 3, edges = [[0,1],[0,2]], cost = [2,1,3]

Output: 1

There are two root-to-leaf paths: To make all root-to-leaf path scores equal to 5, increase the cost of node 1 by 2. Only one node is increased, so the output is 1.

Example 2

Input: n = 3, edges = [[0,1],[1,2]], cost = [5,1,4]

Output: 0

There is only one root-to-leaf path: Path 0 → 1 → 2 has a score of 5 + 1 + 4 = 10 . Since only one root-to-leaf path exists, all path costs are trivially equal, and the output is 0.

Example 3

Input: n = 5, edges = [[0,4],[0,1],[1,2],[1,3]], cost = [3,4,1,1,7]

Output: 1

There are three root-to-leaf paths: To make all root-to-leaf path scores equal to 10, increase the cost of node 1 by 2. Thus, the output is 1.

Constraints

• 2 <= n <= 105
• edges.length == n - 1
• edges[i] == [ui, vi]
• 0 <= ui, vi < n
• cost.length == n
• 1 <= cost[i] <= 109
• The input is generated such that edges represents a valid tree.

Solution Approach

Tree Traversal

The problem involves traversing the tree, calculating the score for each root-to-leaf path. We use DFS or BFS to explore all paths from the root to the leaves.

State Tracking

During traversal, we track the score of each path, and for each node, we track its cost. The key idea is to determine the maximum score of any path and calculate how much each other path needs to be incremented.

Increment Calculation

Once the maximum path score is found, we compute how much each path needs to be incremented to match this maximum score. This step involves accumulating the required increments for each path.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the tree traversal approach, typically O(n) where n is the number of nodes in the tree. Each node is visited once during the DFS/BFS traversal, and space complexity is proportional to the height of the tree or the number of nodes in the tree.

What Interviewers Usually Probe

• The candidate demonstrates understanding of binary-tree traversal.
• They properly account for incremental adjustments required to equalize path scores.
• The candidate leverages state tracking effectively to minimize the number of increments.

Common Pitfalls or Variants

Common pitfalls

• Forgetting to consider all root-to-leaf paths when calculating the total cost.
• Misunderstanding how to propagate increments along the paths.
• Not efficiently managing the tree's traversal, which could lead to higher time complexity.

Follow-up variants

• Handling cases with very deep trees, requiring more complex state tracking.
• Optimizing the approach for large trees with millions of nodes.
• Adjusting the problem to allow path score reduction as well as increment.

FAQ

What is the primary pattern for solving the Minimum Increments to Equalize Leaf Paths problem?

The primary pattern involves binary-tree traversal and state tracking to compute the required increments to equalize path scores.

How can I optimize the solution for large inputs?

Optimizing the solution involves efficient tree traversal methods like DFS or BFS and minimizing redundant calculations during state tracking.

What happens if multiple paths have the same score?

If multiple paths have the same score, no increment is required for those paths. Only the other paths will need to be adjusted.

Can the approach handle trees with millions of nodes?

Yes, the approach can handle large trees as long as the traversal and state tracking are implemented efficiently to maintain linear time complexity.

Why is DFS/BFS recommended for this problem?

DFS/BFS is recommended because it allows us to explore each root-to-leaf path in the tree efficiently while tracking costs and calculating the necessary increments.