Asked in: No companies listed
Image of the Question
All Testcases Passed ✔
Solution
from functools import lru_cache
from sys import setrecursionlimit
from collections import defaultdict
setrecursionlimit(10**6)
// ... rest of solution available after purchaseExplanation
```
To approach this problem effectively, it's crucial to first develop a clear understanding of the tree structure and the implications of the flip operation. The problem involves a rooted tree where each node holds a binary value (0 or 1), and we are allowed to perform a specific type of flip operation on any node. The flip operation affects the selected node and also impacts certain nodes in its subtree, specifically those whose indices have the same parity (i.e., even or odd index) as the selected node. The goal is to convert the initial binary values of all nodes into their corresponding expected values using the minimum number of operations.
We start by recognizing that the structure of the tree is important. Since it's rooted at node 0, and the edges are undirected, a Depth-First Search (DFS) traversal will help us understand parent-child relationships and also define the subtree for each node. In a DFS, when we visit a node, we can recursively explore all its children and build up our logic based on the traversal path.
The flip operation is subtle but critical: flipping a node flips all nodes in its subtree that share the same parity. This means that any flip not only affects the current node but can have cascading effects on other nodes down the tree depending on their index parity. Because of this, a naïve approach that greedily flips nodes without considering the parity and depth relationships will likely cause more flips than necessary.
To design a good strategy, we must simulate the impact of each flip operation as we traverse the tree. Since each node is either even or odd indexed, we need to keep track of how many flips have been applied so far to even-indexed and odd-indexed nodes, respectively. This can be done using two variables: one to track the current flip count for even parity and another for odd parity. These counters will help determine the effective value of a node when we visit it during DFS.
Here’s the idea: as we perform a DFS from the root (node 0), we carry along the current state of the flip counters. When visiting a node, we compute its effective current value by checking how many times its parity has been flipped before. If the number of flips affecting this node is odd, its value is considered flipped; if even, it's as it was in the original `initial` array.
Once we know the node’s effective value, we compare it to the expected value. If they don’t match, it means we must perform a flip operation at this node. After performing the flip, we must update the corresponding parity counter so that all future children with matching parity are affected accordingly.
This approach ensures that we only flip when necessary and that we don’t double-flip or mistakenly flip nodes that already have the correct value. The strategy boils down to using DFS traversal to propagate the effect of flips and intelligently decide where to apply them based on the mismatch between the effective current value and the expected value.
Throughout this traversal, we can maintain a list or count of nodes where we apply the flip operation, and by the end, the size of this list gives us the minimum number of operations required.
To summarize, the key insights are:
1. Understand the tree structure using DFS.
2. Track the number of flips affecting nodes of even and odd parity separately.
3. At each node, determine the effective current value after accounting for flips.
4. Compare the effective value to the expected value and decide whether to flip.
5. Propagate the flip effect by updating the parity-specific flip counter for the subtree traversal.
By following this disciplined, parity-aware DFS approach, the minimum number of required operations can be determined efficiently, while also ensuring correctness.
```