How do you solve Minimum Edge Toggles on a Tree on LeetCode?

Minimum Edge Toggles on a Tree (LeetCode #3812) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: DFS is efficient for tree traversal.

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

The optimal solution for Minimum Edge Toggles on a Tree runs in O(n) time and uses O(n) space. DFS visits each node once, making it linear in terms of the number of nodes.

What is the brute force solution for Minimum Edge Toggles on a Tree?

The brute force approach for Minimum Edge Toggles on a Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. Try every possible combination of edge toggles to see if we can match the target colors. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Edge Toggles on a Tree?

Explain your thought process clearly. Use examples to illustrate your approach. Practice DFS problems to build confidence.

What are common mistakes when solving Minimum Edge Toggles on a Tree?

Not considering the toggle state correctly. Failing to check if the result has an even number of toggles.

#3812

Minimum Edge Toggles on a Tree

Hard

TreeDepth-First SearchGraph TheoryTopological SortSortingGraph TraversalBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use DFS to traverse the tree and determine which edges need toggling based on the difference between start and target colors.

⚙️

Algorithm

3 steps

1Step 1: Perform DFS from any root node, tracking the current toggle state.
2Step 2: For each node, if its color differs from the target, toggle the edge to its parent.
3Step 3: Collect the indices of toggled edges and return them.

solution.py18 lines

1def minEdgeToggles(n, edges, start, target):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for i, (u, v) in enumerate(edges):
5        graph[u].append((v, i))
6        graph[v].append((u, i))
7    result = []
8    def dfs(node, parent, toggle):
9        current_color = start[node] ^ toggle
10        if current_color != target[node]:
11            toggle ^= 1
12            if parent != -1:
13                result.append(edges[parent][1])
14        for neighbor, edge_index in graph[node]:
15            if neighbor != parent:
16                dfs(neighbor, edge_index, toggle)
17    dfs(0, -1, 0)
18    return sorted(result) if len(result) % 2 == 0 else [-1]

ℹ

Complexity note: DFS visits each node once, making it linear in terms of the number of nodes.

1DFS is efficient for tree traversal.
2Toggling affects both connected nodes, so track the toggle state.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.