LeetCode 3812 - Minimum Edge Toggles on a Tree

We are given a tree with n nodes and n - 1 edges. Every node has a binary color represented by the strings start and target. An operation consists of selecting an edge and toggling both of its endpoints.

leetcodehardtreedepth-first-searchgraph-theorytopological-sortsorting

LeetCode Problem 3812

Difficulty: 🔴 Hard
Topics: Tree, Depth-First Search, Graph Theory, Topological Sort, Sorting

Solution

LeetCode 3812 - Minimum Edge Toggles on a Tree

Problem Understanding

We are given a tree with n nodes and n - 1 edges. Every node has a binary color represented by the strings start and target.

An operation consists of selecting an edge and toggling both of its endpoints. If a node currently has color 0, it becomes 1; if it has color 1, it becomes 0.

The goal is to transform the coloring described by start into the coloring described by target.

Instead of returning the sequence of operations itself, we must return the set of edge indices that should be toggled. Since toggling the same edge twice cancels itself out, every edge is either used zero times or one time in an optimal solution.

Among all valid solutions with the minimum number of edge toggles, we must return the chosen edge indices in increasing order.

If no sequence of edge toggles can transform start into target, we return [-1].

A useful way to view the problem is to define, for every node:

$$d[v] = start[v] \oplus target[v]$$

where d[v] = 1 means node v must be flipped an odd number of times overall, and d[v] = 0 means node v must be flipped an even number of times.

Each selected edge contributes exactly one flip to each endpoint. Therefore, if we define:

$$x_e \in {0,1}$$

to indicate whether edge e is selected, then every node must satisfy:

$$\sum_{e \text{ incident to } v} x_e \equiv d[v] \pmod 2$$

The problem becomes a parity system on a tree.

Constraints Analysis

The constraints allow:

  • n ≤ 100000
  • Tree with exactly n - 1 edges

This immediately rules out any exponential search over edge subsets.

We need an algorithm that is close to linear time, ideally O(n).

Important Edge Cases

A few cases require special attention:

  • The transformation may already be complete (start == target).
  • The parity requirements may be impossible to satisfy.
  • The tree may be highly unbalanced, forming a chain of length 100000.
  • Multiple valid solutions may exist in general graphs, but a tree structure often forces uniqueness.
  • The root node requires special handling because it has no parent edge.

Approaches

Brute Force

A brute-force solution would consider every subset of edges.

Since there are n - 1 edges, there are:

$$2^{n-1}$$

possible subsets.

For each subset, we could simulate all toggles, compute the resulting coloring, and check whether it matches target.

The smallest valid subset would be the answer.

This is correct because it explicitly examines every possible solution. However, it is completely infeasible for n = 100000.

Key Insight

The crucial observation is that only parity matters.

Define:

$$d[v] = start[v] \oplus target[v]$$

Each node requires either an odd number (1) or even number (0) of incident selected edges.

Because the graph is a tree, the parity equations have a very special structure.

Root the tree arbitrarily at node 0.

Consider a non-root node u.

The only edge connecting the subtree of u to the rest of the tree is the parent edge (parent[u], u).

After all descendants have been processed, the parity requirement at node u determines uniquely whether the parent edge must be selected.

Specifically:

  • If the current parity inside the subtree equals d[u], the parent edge is not needed.
  • Otherwise, the parent edge must be selected.

This allows a bottom-up DFS.

Even better, every selected edge is forced. There is no choice.

The algorithm computes the unique solution satisfying all non-root nodes. Afterward, the root parity is checked.

If the root parity is also satisfied, the solution exists and is automatically minimum because every edge value is uniquely determined.

Comparison of Approaches

Approach Time Complexity Space Complexity Notes
Brute Force O(2^(n-1) · n) O(n) Tries every subset of edges
Optimal O(n) O(n) Bottom-up parity propagation on the tree

Algorithm Walkthrough

  1. Compute the required parity array.

For every node:

$$d[v] = (start[v] \ne target[v])$$

This tells us whether node v needs an odd number of flips. 2. Build the tree.

Store for each node its neighbors and the corresponding edge indices. 3. Root the tree at node 0.

Using DFS, compute:

  • parent of every node
  • edge index connecting it to its parent
  • traversal order
  1. Process nodes in reverse DFS order.

This guarantees every child is processed before its parent. 5. Maintain a value need[v].

Initially:

$$need[v] = d[v]$$

During processing, children contribute their parity requirements upward. 6. For every non-root node u:

If need[u] = 1, the parent edge must be selected.

Record that edge index.

Selecting the parent edge contributes one incident selected edge to both endpoints.

Therefore the parent's requirement is toggled:

$$need[parent[u]] \mathrel{\oplus}= 1$$ 7. Continue until all non-root nodes are processed. 8. Check the root.

After all propagation is complete:

  • If need[0] = 0, a valid solution exists.
  • If need[0] = 1, no solution exists.
  1. Sort the chosen edge indices and return them.

Why it Works

For every non-root node, the parent edge is the only edge leaving its subtree. Therefore once all child decisions are fixed, the parity requirement at that node uniquely determines whether the parent edge must be selected.

The algorithm enforces the parity condition for every non-root node exactly once. The only remaining constraint is the root equation. If the root equation is satisfied, all parity equations are satisfied simultaneously.

Since every edge decision is uniquely forced, any valid solution is unique. Consequently, that solution is automatically the minimum-length solution.

Python Solution

from typing import List

class Solution:
    def minimumFlips(self, n: int, edges: List[List[int]], start: str, target: str) -> List[int]:
        graph = [[] for _ in range(n)]
        
        for idx, (u, v) in enumerate(edges):
            graph[u].append((v, idx))
            graph[v].append((u, idx))
        
        need = [0] * n
        for i in range(n):
            need[i] = (ord(start[i]) - ord('0')) ^ (ord(target[i]) - ord('0'))
        
        parent = [-1] * n
        parent_edge = [-1] * n
        
        order = [0]
        parent[0] = 0
        
        for node in order:
            for nxt, edge_idx in graph[node]:
                if parent[nxt] != -1:
                    continue
                parent[nxt] = node
                parent_edge[nxt] = edge_idx
                order.append(nxt)
        
        answer = []
        
        for node in reversed(order[1:]):
            if need[node] == 1:
                answer.append(parent_edge[node])
                need[parent[node]] ^= 1
        
        if need[0] != 0:
            return [-1]
        
        answer.sort()
        return answer

Implementation Explanation

The adjacency list stores both neighboring nodes and edge indices because the final answer must contain edge indices rather than endpoints.

The array need stores the parity requirement for every node. A value of 1 means the node must be flipped an odd number of times.

A DFS traversal rooted at node 0 records each node's parent and the edge connecting it to the parent.

Processing occurs in reverse traversal order. When a node still has parity requirement 1, its parent edge must be selected. That edge satisfies the node's remaining parity requirement and simultaneously toggles the parent's requirement.

Every selected edge index is stored in answer.

After all non-root nodes are handled, only the root remains. If the root parity is nonzero, no solution exists and we return [-1].

Otherwise the collected edge indices form the unique valid solution.

Go Solution

func minimumFlips(n int, edges [][]int, start string, target string) []int {
	graph := make([][][2]int, n)

	for i, e := range edges {
		u, v := e[0], e[1]
		graph[u] = append(graph[u], [2]int{v, i})
		graph[v] = append(graph[v], [2]int{u, i})
	}

	need := make([]int, n)
	for i := 0; i < n; i++ {
		need[i] = int(start[i]-'0') ^ int(target[i]-'0')
	}

	parent := make([]int, n)
	parentEdge := make([]int, n)

	for i := 0; i < n; i++ {
		parent[i] = -1
		parentEdge[i] = -1
	}

	order := make([]int, 0, n)
	order = append(order, 0)
	parent[0] = 0

	for i := 0; i < len(order); i++ {
		node := order[i]

		for _, item := range graph[node] {
			nextNode := item[0]
			edgeIdx := item[1]

			if parent[nextNode] != -1 {
				continue
			}

			parent[nextNode] = node
			parentEdge[nextNode] = edgeIdx
			order = append(order, nextNode)
		}
	}

	answer := make([]int, 0)

	for i := len(order) - 1; i >= 1; i-- {
		node := order[i]

		if need[node] == 1 {
			answer = append(answer, parentEdge[node])
			need[parent[node]] ^= 1
		}
	}

	if need[0] != 0 {
		return []int{-1}
	}

	for i := 1; i < len(answer); i++ {
		key := answer[i]
		j := i - 1

		for j >= 0 && answer[j] > key {
			answer[j+1] = answer[j]
			j--
		}

		answer[j+1] = key
	}

	return answer
}

Go-Specific Notes

The Go version uses slices for adjacency lists and stores (neighbor, edgeIndex) as a fixed-size array [2]int.

An iterative traversal is used instead of recursive DFS to avoid stack overflow on a path-shaped tree containing 100000 nodes.

The logic is otherwise identical to the Python implementation.

Worked Examples

Example 1

Input:

n = 3
edges = [[0,1],[1,2]]
start = "010"
target = "100"

Compute parity requirements:

Node Start Target Need
0 0 1 1
1 1 0 1
2 0 0 0

Reverse processing order:

Node Need Action
2 0 Nothing
1 1 Select edge 0

Selecting edge 0 toggles parent requirement:

Node New Need
0 0

Root becomes 0, so the solution is valid.

Result:

[0]

Example 2

Input:

n = 7
edges = [[0,1],[1,2],[2,3],[3,4],[3,5],[1,6]]
start = "0011000"
target = "0010001"

Parity requirements:

Node Need
0 0
1 0
2 0
3 1
4 0
5 0
6 1

Reverse processing:

Node Need Selected Edge
6 1 5
5 0 None
4 0 None
3 1 2
2 1 1
1 0 None

Chosen edges:

[5, 2, 1]

Sorted:

[1, 2, 5]

Example 3

Input:

n = 2
edges = [[0,1]]
start = "00"
target = "01"

Parity requirements:

Node Need
0 0
1 1

Process node 1:

Select edge 0.

Root parity becomes:

need[0] = 1

The root constraint fails.

Result:

[-1]

Complexity Analysis

Measure Complexity Explanation
Time O(n) Building the graph, traversal, and reverse processing each visit every node and edge once
Space O(n) Adjacency list, parent arrays, traversal order, and answer storage

The algorithm performs only a constant amount of work per node and per edge. Since a tree contains exactly n - 1 edges, the overall running time is linear. The memory usage is also linear because all auxiliary structures store information proportional to the number of nodes or edges.

Test Cases

sol = Solution()

assert sol.minimumFlips(
    3,
    [[0, 1], [1, 2]],
    "010",
    "100"
) == [0]  # Example 1

assert sol.minimumFlips(
    7,
    [[0,1],[1,2],[2,3],[3,4],[3,5],[1,6]],
    "0011000",
    "0010001"
) == [1, 2, 5]  # Example 2

assert sol.minimumFlips(
    2,
    [[0, 1]],
    "00",
    "01"
) == [-1]  # Example 3

assert sol.minimumFlips(
    2,
    [[0, 1]],
    "00",
    "00"
) == []  # Already equal

assert sol.minimumFlips(
    2,
    [[0, 1]],
    "00",
    "11"
) == [0]  # Single edge solves

assert sol.minimumFlips(
    3,
    [[0,1],[1,2]],
    "000",
    "111"
) == [-1]  # Odd parity impossible

assert sol.minimumFlips(
    4,
    [[0,1],[1,2],[2,3]],
    "0000",
    "1001"
) == [0, 2]  # Path structure

assert sol.minimumFlips(
    5,
    [[0,1],[0,2],[0,3],[0,4]],
    "00000",
    "11110"
) == [-1]  # Star with impossible parity

assert sol.minimumFlips(
    5,
    [[0,1],[0,2],[0,3],[0,4]],
    "00000",
    "11000"
) == [0]  # Star with one selected edge

assert sol.minimumFlips(
    6,
    [[0,1],[1,2],[2,3],[3,4],[4,5]],
    "000000",
    "101010"
) == [0, 1, 2, 3, 4]  # Long chain propagation

Test Summary

Test Why
Example 1 Basic valid transformation
Example 2 Multiple selected edges in different subtrees
Example 3 Impossible configuration
Already equal Empty answer case
Two-node valid Smallest nontrivial valid tree
All ones target on path Global parity inconsistency
Path tree Bottom-up propagation check
Impossible star Root parity failure
Valid star Multiple leaves attached to root
Long chain Stresses propagation through many levels

Edge Cases

Transformation Already Completed

If start and target are identical, every node has need[v] = 0. No edge needs to be selected, so the correct answer is an empty array.

A buggy implementation might unnecessarily select edges or fail to handle an empty result. The presented algorithm naturally returns an empty list because no node ever triggers an edge selection.

Impossible Global Parity

Every selected edge contributes two flips, one at each endpoint. Therefore the total parity of required flips across all nodes must be even.

For example:

start = "00"
target = "01"

Only one node needs changing. This is impossible because every operation affects two nodes.

The algorithm detects this automatically when the final root parity remains nonzero.

Deep Path-Shaped Tree

A tree can be a chain containing 100000 nodes.

A recursive DFS may overflow the call stack in Python or Go.

The implementation uses an iterative traversal and stores nodes in an explicit order array, ensuring that it remains safe even for the largest possible input.

Root Node Handling

Every non-root node has exactly one parent edge that can satisfy any remaining parity requirement.

The root has no parent edge.

As a result, after all propagation is complete, the root becomes the final consistency check. If its parity requirement is nonzero, no solution exists. Correct handling of this special case is essential for determining feasibility.