3593. Minimum Increments to Equalize Leaf Paths

Problem Link

Description

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You are given an integer n and an undirected tree rooted at node 0 with n nodes numbered from 0 to n - 1. This is represented by a 2D array edges of length n - 1, where edges[i] = [ui, vi] indicates an edge from node ui to vi .

Each node i has an associated cost given by cost[i], representing the cost to traverse that node.

The score of a path is defined as the sum of the costs of all nodes along the path.

Your goal is to make the scores of all root-to-leaf paths equal by increasing the cost of any number of nodes by any non-negative amount.

Return the minimum number of nodes whose cost must be increased to make all root-to-leaf path scores equal.

 

Example 1:

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

Output: 1

Explanation:

There are two root-to-leaf paths:

• Path 0 → 1 has a score of 2 + 1 = 3.
• Path 0 → 2 has a score of 2 + 3 = 5.

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

Explanation:

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

Explanation:

There are three root-to-leaf paths:

• Path 0 → 4 has a score of 3 + 7 = 10.
• Path 0 → 1 → 2 has a score of 3 + 4 + 1 = 8.
• Path 0 → 1 → 3 has a score of 3 + 4 + 1 = 8.

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

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Python3

class Solution:
    def minIncrease(self, n: int, edges: List[List[int]], cost: List[int]) -> int:
        graph = defaultdict(list)
        res = 0
 
        for a, b in edges:
            graph[a].append(b)
            graph[b].append(a)
 
        def dfs(node, prev):
            nonlocal res
 
            children = []
            for adj in graph[node]:
                if adj != prev:
                    children.append(dfs(adj, node))
            
            if not children:
                return cost[node]
            
            maxCost = max(children)
            res += sum(int(maxCost > c) for c in children)
            return maxCost + cost[node]
 
        dfs(0, -1)
        return res