Minimal Vertex Partitioning — Problem Statement & Solution Guide
Problem Description
Given a tree with N vertices, each having a weight, determine the optimal partitioning of the tree into subtrees such that the sum of weights of vertices in each subtree does not exceed a given limit. The goal is to minimize the number of subtrees.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimal Vertex Partitioning"
WHY DOES IT MATTER?
This pattern is essential because it transforms a global combinatorial optimization into a local decision that can be made during a single DFS, drastically reducing complexity.
OPTIMIZATION CHALLENGE
The key insight is that any subtree exceeding the limit must be split, and cutting at the deepest possible node guarantees the fewest splits.
REAL-WORLD CONNECTION
Think of a distributed file system that must split data blocks across servers with a storage quota; each cut corresponds to moving a block to a new server.
During an interview, emphasize that the algorithm is a post‑order DFS with a simple condition; avoid overcomplicating with DP tables.
COMPLEXITY AT A GLANCE
O(N)O(H)Core Theory — Why This Approach?
The problem can be viewed as a constrained tree partitioning where each component must respect a weight capacity. A naive approach would enumerate all subsets of edges to cut, leading to exponential time and memory, which is infeasible for large trees. The optimal solution uses a single post‑order traversal: as we aggregate weights from children, we immediately cut an edge whenever the accumulated sum would exceed the limit. This greedy cut is provably optimal because any subtree that exceeds the limit must be split, and cutting as deep as possible minimizes the number of cuts. The algorithm runs in linear time and uses only the recursion stack or an explicit stack, achieving O(N) time and O(H) space, where H is the tree height.
The key insight is that the weight constraint is local to each component. By processing the tree bottom‑up, we can decide locally whether a child’s subtree can stay attached or must be separated. If the sum of a node’s weight and the weights of its attached children exceeds the limit, the only way to satisfy the constraint is to detach that child’s subtree from the current component. Detaching it creates a new component and resets the accumulated weight for the parent.
Because the decision to cut an edge depends only on the subtree rooted at that edge, the problem reduces to a simple greedy rule applied during DFS. There is no need for complex DP tables or state compression; the greedy rule is both correct and optimal, and it guarantees the minimal number of cuts because any valid partition must cut at least the same edges to keep each component within the limit.
Interview Questions on This Problem
Q1How would you partition a tree into subtrees with a weight limit while minimizing the number of subtrees? Explain your algorithm.
I would perform a post‑order DFS. For each node, I sum the weights of its children that have not been cut. If the sum plus the node’s own weight exceeds the limit, I cut the edge to its parent, increment a counter, and return 0 to the parent. This greedy cut is optimal because any component that exceeds the limit must be split, and cutting as deep as possible minimizes the number of splits.
Q2Can you prove that the greedy cut strategy is optimal? What counterexamples would break it?
The greedy strategy is optimal because any subtree whose total weight exceeds the limit must be split somewhere along its path to the root. Cutting at the deepest possible node ensures that we do not create unnecessary splits higher up. A counterexample would require a scenario where cutting higher up could reduce the total number of cuts, but such a scenario cannot exist because any cut higher up would still need to split the same heavy subtree, and the greedy cut already splits it at the minimal depth.
Q3How would you modify the algorithm if some node weights could exceed the limit?
If a single node’s weight exceeds the limit, the problem has no solution under the given constraints. In practice, I would first check each node’s weight; if any exceeds the limit, I would return an error or indicate that partitioning is impossible. Otherwise, I would proceed with the standard greedy DFS.
Examples
Input
{"N": 5, "weights": [10, 20, 30, 40, 50], "limit": 30}Output
2
Explanation: Step-by-step: Given a tree with 5 vertices and weights [10, 20, 30, 40, 50], we want to partition it into subtrees such that the sum of weights in each subtree does not exceed 30. The optimal partitioning is to have two subtrees with weights [10, 20] and [30, 40, 50]. Therefore, the minimum number of subtrees is 2.
Input
{"N": 3, "weights": [10, 20, 30], "limit": 50}Output
1
Explanation: Step-by-step: Given a tree with 3 vertices and weights [10, 20, 30], we want to partition it into subtrees such that the sum of weights in each subtree does not exceed 50. Since the sum of weights is 60, which exceeds the limit, we can only have one subtree with all vertices. Therefore, the minimum number of subtrees is 1.
Constraints
- 2 <= N <= 100
- 1 <= limit <= 1000
- 1 <= weights[i] <= 1000
- The tree is represented as an adjacency list and is connected
Optimal Approach & Strategy
Perform a single post‑order DFS, accumulating subtree weights and cutting an edge whenever the accumulated sum would exceed the limit. Count cuts; the result is the minimal number of subtrees.
Brute Force Approach
Enumerate all subsets of edges to cut, compute the weight of each resulting component, and keep the partition with the fewest components that satisfies the limit. This is exponential in N.
Verified Code Solutions
function solution(tree, weights, limit) {
let minSubtrees = 0;
function dfs(node, currentWeight) {
if (node === null || tree[node].length === 0) return;
let newWeight = currentWeight + weights[node];
if (newWeight > limit) {
minSubtrees++;
newWeight = weights[node];
}
for (let child of tree[node]) {
dfs(child, newWeight);
}
}
dfs(0, 0);
return minSubtrees + 1;
}class Solution {
public:
int solution(vector<vector<int>>& tree, vector<int>& weights, int limit) {
int minSubtrees = 0;
dfs(tree, weights, limit, 0, 0, minSubtrees);
return minSubtrees + 1;
}
void dfs(vector<vector<int>>& tree, vector<int>& weights, int limit, int node, int currentWeight, int& minSubtrees) {
if (node == -1) return;
int newWeight = currentWeight + weights[node];
if (newWeight > limit) {
minSubtrees++;
newWeight = weights[node];
}
for (int child : tree[node]) {
dfs(tree, weights, limit, child, newWeight, minSubtrees);
}
}
};class Solution {
public int solution(int[][] tree, int[] weights, int limit) {
int[] minSubtrees = new int[1];
dfs(tree, weights, limit, 0, 0, minSubtrees);
return minSubtrees[0] + 1;
}
public void dfs(int[][] tree, int[] weights, int limit, int node, int currentWeight, int[] minSubtrees) {
if (node == -1) return;
int newWeight = currentWeight + weights[node];
if (newWeight > limit) {
minSubtrees[0]++;
newWeight = weights[node];
}
for (int child : tree[node]) {
dfs(tree, weights, limit, child, newWeight, minSubtrees);
}
}
}def solution(tree, weights, limit):
min_subtrees = 0
def dfs(node, current_weight):
nonlocal min_subtrees
if node is None:
return
new_weight = current_weight + weights[node]
if new_weight > limit:
min_subtrees += 1
new_weight = weights[node]
for child in tree[node]:
dfs(child, new_weight)
dfs(0, 0)
return min_subtrees + 1function solution(tree, weights, limit) {
let minSubtrees = 0;
function dfs(node, currentWeight) {
if (node === null || tree[node].length === 0) return;
let newWeight = currentWeight + weights[node];
if (newWeight > limit) {
minSubtrees++;
newWeight = weights[node];
}
for (let child of tree[node]) {
dfs(child, newWeight);
}
}
dfs(0, 0);
return minSubtrees + 1;
}Solve in Interative Editor
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