Asked in: SHARECHAT
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All Testcases Passed ✔
Solution
from collections import defaultdict
from functools import lru_cache
def minDifference(N, towns, roads):
total_sum = sum(towns)
// ... rest of solution available after purchaseExplanation
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To solve this problem of dividing a tree of N regions into two parts by removing exactly one road so that the absolute difference in the total number of towns between the two parts is minimized, it’s important to carefully understand the problem structure and leverage properties of trees and recursion.
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### Understanding the problem
You have a tree with N nodes (regions), each node having a weight (the number of towns in that region). The tree edges represent bidirectional roads connecting these regions. Removing one edge will split the tree into exactly two disconnected subtrees (two new nations). The goal is to find which edge removal yields the minimum absolute difference between the sum of towns in these two subtrees.
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### Key Insights
- The tree is connected and has N-1 edges. Removing exactly one edge breaks it into two subtrees.
- The sum of towns across all regions is fixed.
- For each edge removal, you want to compute the sum of towns in one of the resulting subtrees, because the other subtree’s sum is the total sum minus the first subtree’s sum.
- The difference for that edge removal is then |totalSum - 2 * subtreeSum|.
- We want to find the edge removal with the minimal such difference.
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### Step 1: Compute the total number of towns
First, sum up the number of towns across all N regions. This total sum will be used to compute the difference for each edge removal.
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### Step 2: Tree structure and traversal
Because the regions are connected as a tree, you can use tree traversal techniques such as Depth-First Search (DFS) to explore the structure.
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### Step 3: Use DFS to compute subtree sums
- Choose an arbitrary node as the root (usually node 1).
- Run a DFS starting from the root to compute the sum of towns for each subtree rooted at a node.
- The subtree sum at any node includes the towns in that node and all nodes reachable from it in its subtree.
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### Step 4: Evaluate the effect of removing each edge
- During DFS, when you move from a parent node to a child node, consider the edge between them as a candidate for removal.
- Removing this edge splits the tree into two parts:
- The subtree rooted at the child (with sum = subtreeSum[child]).
- The rest of the tree, whose sum is totalSum - subtreeSum[child].
- The absolute difference for this split is |totalSum - 2 * subtreeSum[child]|.
- Keep track of the minimum such difference found.
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### Step 5: Optimize traversal and storage
- Since the tree has N nodes and N-1 edges, a single DFS traversal can compute subtree sums and simultaneously check differences for all edges.
- You only need to compute subtree sums once.
- No need to explicitly remove edges or rebuild trees; instead, think of the split logically by subtree sums.
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### Step 6: Edge cases and validation
- When a subtree has very few towns, the difference will be large because the other subtree is almost the whole tree.
- When the subtree sum is close to half the total sum, the difference is minimized.
- The best split often occurs at edges connecting large and balanced subtrees.
- Validate that the DFS is correctly marking visited nodes to avoid cycles (even though the problem states it’s a tree, good practice).
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### Example to conceptualize
Suppose you have regions with towns: [10, 20, 30, 40] and roads connecting them forming a tree.
- Total towns = 100.
- If a subtree has sum 40, then difference = |100 - 2 * 40| = 20.
- Another subtree with sum 50 gives difference = |100 - 2 * 50| = 0 (perfectly balanced).
- Your algorithm should find this edge to cut.
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### Summary of approach:
1. Calculate total towns sum.
2. Use DFS to calculate subtree sums for each node.
3. For each subtree sum (computed during DFS), calculate difference = |totalSum - 2 * subtreeSum|.
4. Track and return the minimal difference found.
5. This leverages tree structure and recursion efficiently, avoiding redundant calculations.
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By focusing on subtree sums and their relationship to the total sum, you convert the problem into a single-pass DFS traversal problem with O(N) complexity, suitable for large N. This approach intuitively balances mathematical insight with efficient graph traversal.
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