Is Minimum Edge Toggles on a Tree easy or hard?

Minimum Edge Toggles on a Tree is classified as Hard because it requires recognizing a rerooting dynamic programming pattern on trees. The brute-force idea is simple, but optimizing it to O(n) with two DFS passes and careful edge contribution updates is the main challenge.

Minimum Edge Toggles on a Tree Python/Java solution

Implement the algorithm using an adjacency list and two DFS traversals. The first DFS calculates the number of edges that must be toggled for an initial root. The second DFS performs rerooting and updates the cost for each neighbor by adjusting the toggle count based on the edge orientation. The same logic works in Python, Java, C++, Go, and TypeScript.

How to solve Minimum Edge Toggles on a Tree in O(n)?

Build an adjacency list that records whether traversing an edge matches the desired direction. Run a DFS from an initial root to count the number of edges that require toggling. Then perform a second DFS to reroot: moving the root across an edge changes the total cost by ±1 depending on that edge’s orientation. Propagating this adjustment computes the answer for all nodes in linear time.

What is the best approach for Minimum Edge Toggles on a Tree?

The most efficient approach uses DFS with a rerooting technique. First compute the number of incorrect edge orientations assuming an arbitrary root, then propagate results to neighboring nodes by adjusting the cost when the root moves. This processes each edge a constant number of times, giving O(n) time complexity.

Is Minimum Edge Toggles on a Tree asked at Google/Amazon/Meta?

Tree rerooting and edge orientation problems frequently appear in interviews at companies like Google, Amazon, and Meta. Variants include counting edge reversals, reorienting routes in a tree, or computing costs for every possible root using DFS dynamic programming.

What data structure is used in Minimum Edge Toggles on a Tree?

The core data structure is an adjacency list representing the tree. Each entry stores the neighboring node and metadata about whether the edge direction matches or conflicts with the desired orientation. DFS traversal and rerooting logic operate directly on this structure.

What is the time complexity of Minimum Edge Toggles on a Tree?

The optimal DFS rerooting solution runs in O(n) time where n is the number of nodes in the tree. Each edge is processed during the initial DFS and once again during the reroot traversal. Space complexity is O(n) due to the adjacency list and recursion stack.

3812. Minimum Edge Toggles on a Tree - Practice Online

You are given an undirected tree with n nodes, numbered from 0 to n - 1. It is represented by a 2D integer array edges of length n - 1, where edges[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree.

You are also given two binary strings start and target of length n. For each node x, start[x] is its initial color and target[x] is its desired color.

In one operation, you may pick an edge with index i and toggle both of its endpoints. That is, if the edge is [u, v], then the colors of nodes u and v each flip from '0' to '1' or from '1' to '0'.

Return an array of edge indices whose operations transform start into target. Among all valid sequences with minimum possible length, return the edge indices in increasing order.

If it is impossible to transform start into target, return an array containing a single element equal to -1.

Example 1:

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

Output: [0]

Explanation:

Toggle edge with index 0, which flips nodes 0 and 1.
The string changes from "010" to "100", matching the target.

Example 2:

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

Output: [1,2,5]

Explanation:

Toggle edge with index 1, which flips nodes 1 and 2.
Toggle edge with index 2, which flips nodes 2 and 3.
Toggle edge with index 5, which flips nodes 1 and 6.

After these operations, the resulting string becomes "0010001", which matches the target.

Example 3:

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

Output: [-1]

Explanation:

There is no sequence of edge toggles that transforms "00" into "01". Therefore, we return [-1].

Constraints:

2 <= n == start.length == target.length <= 10⁵
edges.length == n - 1
edges[i] = [a_i, b_i]
0 <= a_i, b_i < n
start[i] is either '0' or '1'.
target[i] is either '0' or '1'.
The input is generated such that edges represents a valid tree.