Minimize Inversions in a Prefix Tree - Solution & Explanation

Problem Overview: You get a prefix tree (trie) where each node's children can be reordered. Each reordering changes the sequence of values read off the tree, and you want the arrangement that produces the fewest inversions, meaning pairs that appear out of order.

Approach 1: Brute Force Over All Child Orderings (Exponential)

Enumerate every permutation of children at every node, build the resulting sequence, and count inversions with a nested loop. The number of arrangements grows as the product of factorials of child counts, so this is only useful for checking small inputs. It runs in O(k! * n^2) time and O(n) space, where k is the largest branching factor. Use it to validate a faster solution on tiny test cases, not as a submission.

Approach 2: Recursive Greedy with Pairwise Inversion Counts (O(n^2))

Process the tree bottom-up with DFS. For each node, compute the sequence produced by each child subtree and the cross-inversions between every pair of children, in both possible orders. Then pick the order that yields fewer cross-inversions. The key insight is that choices at different nodes are independent, so a local greedy decision at each node gives a global optimum. Count cross-inversions with nested loops over the child sequences. This is simpler to reason about than the optimal version, and it is fast enough for moderate inputs. Time is O(n^2) in the worst case and space is O(n) for the recursion and stored sequences.

Approach 3: Greedy DFS with Merge-Based Inversion Counting (Optimal, O(n log n))

Keep the same bottom-up greedy idea, but replace the nested loops with a merge step from merge sort. Each subtree returns its sorted sequence along with an inversion count. When merging two child sequences, one merge pass counts how many elements from the left child exceed elements from the right child, which gives both cross-inversion totals in linear time. Swap the children only when the swapped order gives fewer cross-inversions, then merge. Total time is O(n log n) and space is O(n). Prefer this approach whenever the input is large, because the nested-loop version hits its limit quickly.

Recommended for interviews: Interviewers expect you to state the brute force, then justify the greedy per-node choice, and finally implement the merge-based counting. The brute force shows you understand the search space. The greedy argument and the merge-based count show you can cut it down to O(n log n). Explain why independent subtree choices compose correctly before you write any code.

Related topics: tree, divide and conquer, sorting.