Is Minimize Inversions in a Prefix Tree easy or hard?

It is rated Hard, with a 33.4% acceptance rate. The difficulty comes from proving that local per-node greedy choices give a global optimum and from implementing the merge-based counting correctly.

How to solve Minimize Inversions in a Prefix Tree in O(n log n)?

Process the tree with DFS and return a sorted sequence plus an inversion count from each subtree. At each node, count cross-inversions between children during a merge step, swap children if the swapped order has fewer inversions, then merge. The per-node choices are independent, so the greedy result is globally optimal.

What is the best approach for Minimize Inversions in a Prefix Tree?

The best approach is a bottom-up greedy DFS that uses merge-based inversion counting. At each node, you compare the cross-inversions for both child orders and keep the cheaper one, merging sorted child sequences in the same pass. This runs in O(n log n) time with O(n) space.

What data structure is used in Minimize Inversions in a Prefix Tree?

The problem uses a tree, specifically a prefix tree, as the input structure. The solution relies on recursion or an explicit stack for DFS, and on arrays or lists to hold sorted child sequences during the merge step.

What is the time complexity of Minimize Inversions in a Prefix Tree?

The optimal solution runs in O(n log n) time, where n is the number of nodes. The naive version that counts cross-inversions with nested loops takes O(n^2), and enumerating all child orderings takes factorial time.

Minimize Inversions in a Prefix Tree Python solution: how do I write it?

In Python, write a recursive DFS that returns a (sorted list, inversion count) pair for each node. Merge the children's lists with a two-pointer pass that counts cross-inversions, and choose the child order that gives the smaller count. Use an iterative stack if recursion depth could exceed Python's default limit.

Is Minimize Inversions in a Prefix Tree asked at Google, Amazon, or Meta?

This problem is tagged as General on FleetCode, so company-specific frequency is not listed. Problems that combine tree traversal with merge sort style counting appear often in senior-level interviews at large tech companies, so practice the merge-based version.

4077. Minimize Inversions in a Prefix Tree - Practice Online

You are given an array words of n unique strings containing only lowercase English letters from 'a' to 'j'. You are also given an integer array target containing each index from 0 to n - 1 exactly once.

Build a prefix tree from words:

The root represents the empty string "".
Each other node represents a different non-empty prefix of a word. The parent of a node is the prefix obtained by removing its last character.
The node representing the complete word words[i] is marked with index i. A marked node may also have children.

You may choose the order of the children of each node independently. You cannot change which node is a child's parent.

Traverse the tree starting at the root. When visiting a node, first record its index if it is marked. Then visit its children in the chosen order, completing the traversal of each child before moving to the next. This is a pre-order traversal.

The recorded indices form an array traversal of length n. Let pos[i] be the position of index i in target.

An inversion is a pair of positions (a, b) such that 0 <= a < b < n and pos[traversal[a]] > pos[traversal[b]].

Return the minimum possible number of inversions over all choices of child orderings.

Example 1:

Input: words = ["bad","bag","fig","fed"], target = [3,1,0,2]

Output: 2

Explanation:

Visit the root's "f" child before its "b" child. Within the "f" branch, visit "fed" before "fig". Within the "b" branch, visit "bag" before "bad".

This produces traversal = [3,2,1,0]:

Traversal position	Word index	Word	Position in target
0	3	`"fed"`	0
1	2	`"fig"`	3
2	1	`"bag"`	1
3	0	`"bad"`	2

The positions in target are [0,3,1,2]. The inversions are the pairs of traversal positions (1,2) and (1,3), since 3 > 1 and 3 > 2.

No child ordering produces fewer than 2 inversions.

Example 2:

Input: words = ["ab","ac","ba","bc"], target = [0,3,1,2]

Output: 1

Explanation:

Visit the root's "a" child before its "b" child. Within these branches, visit "ab" before "ac", and "bc" before "ba".

This produces traversal = [0,1,3,2]:

Traversal position	Word index	Word	Position in target
0	0	`"ab"`	0
1	1	`"ac"`	2
2	3	`"bc"`	1
3	2	`"ba"`	3

The positions in target are [0,2,1,3]. The only inversion is the pair of traversal positions (1,2), since 2 > 1.

No child ordering produces zero inversions.

Constraints:

1 <= n == words.length <= 10⁴
1 <= words[i].length <= 20
words[i] contains only lowercase English letters from 'a' to 'j'
All words[i] are unique
0 <= target[i] <= n - 1
target is a permutation of integers from 0 to n - 1