The Minimum Score After Removals on a Tree problem gives you a tree (graph with n nodes and n-1 edges) where each node has a value. You remove exactly two edges, splitting the tree into three components. The score is the difference between the maximum and minimum XOR values of the three components' node-value XORs. Find the minimum possible score. This Minimum Score After Removals on a Tree interview question involves tree traversal, XOR properties, and bit manipulation.
Microsoft, Meta, and Amazon ask this hard problem because it tests XOR reasoning on trees — a combination rarely seen but highly valued. It requires understanding XOR prefix sums on trees (using DFS timestamps), ancestor-descendant relationships, and enumeration over edge pairs. The array, DFS, bit manipulation, and tree interview pattern is comprehensively tested.
Use DFS with Euler tour / DFS timestamps to compute subtree XOR values. For each node, compute the XOR of all values in its subtree. When you remove an edge, the XOR of the cut component is just the subtree XOR. Enumerate all pairs of edges. For each pair, determine the three component XORs using subtree XORs and the total XOR. The three cases depend on whether one removed subtree is an ancestor of the other. Check all O(n²) pairs efficiently.
Tree: 1-2-3-4 (path). Values: [1,2,4,8]. Total XOR = 1^2^4^8 = 15. Remove edge (1-2) and (2-3): components are {1}=1, {2}=2, {3,4}=4^8=12. Score = max(12,2,1) - min(12,2,1) = 12 - 1 = 11. Try other pairs to find the minimum score.
Tree problems involving path or subtree queries often become elegant with DFS timestamp precomputation. Learn how entry and exit times in a DFS define ancestor-descendant relationships: node u is an ancestor of v if in[u] ≤ in[v] and out[v] ≤ out[u]. Practicing XOR problems on trees separately before combining them with removal enumeration will build the confidence needed to tackle this type of problem in an interview.
