Скачать или смотреть 💥 LeetCode 2322 Minimum Score After Removals on a Tree | LeetCode 2322 | DFS + XOR

In this video, we solve LeetCode 2322 - Minimum Score After Removals on a Tree using a clever DFS approach with XOR bit manipulation.

We explore how removing two edges from the tree partitions it into three components, and how to minimise the difference between the maximum and minimum XOR of these three parts.

🔥 Topics Covered:

DFS Traversal with XOR

Subtree XOR computation

How to choose two edges optimally

Time and space complexity breakdown

💡 Logic Explained Simply:

Compute the total XOR of the tree.

Perform DFS to simulate first cut and compute cut1.

Use another DFS to explore valid second cuts and compute cut2.

Calculate the third cut using XOR properties: cut3 = total ^ cut1 ^ cut2

Track the minimum difference between max and min of cut1, cut2, and cut3.

✅ Efficient Java implementation shown step-by-step.

If you’re preparing for coding interviews at FAANG or other top companies, understanding this problem will boost your graph + bit manipulation skills.