Longest Common Prefix of K Strings After Removal

To solve the problem, use a trie to store all strings and compute the longest common prefix for each element. Efficient management of the trie is critical. This involves keeping track of how the removal of each word affects the common prefix among the remaining strings.

Problem Statement

You are given an array of strings and an integer k. For each index i, find the length of the longest common prefix among any k strings selected from distinct indices in the remaining array after removing the ith element. Return an array of the results where the value for each index corresponds to the result for that element's removal. If the removal results in fewer than k strings, return 0.

In other words, for each string in the array, you must identify the longest common prefix of any k remaining strings after removing that specific string. Efficiently managing the data structure to store and compute the prefixes for the remaining strings is key to solving this problem. If fewer than k strings are left after removal, return 0 for that index.

Examples

Example 1

Input: words = ["jump","run","run","jump","run"], k = 2

Output: [3,4,4,3,4]

Example 2

Input: words = ["dog","racer","car"], k = 2

Output: [0,0,0]

Constraints

• 1 <= k <= words.length <= 105
• 1 <= words[i].length <= 104
• words[i] consists of lowercase English letters.
• The sum of words[i].length is smaller than or equal 105.

Solution Approach

Use a Trie for Prefix Storage

The key to solving this problem efficiently is utilizing a trie to store all the strings. A trie allows us to easily compute the longest common prefix between any set of strings. By storing all strings initially in the trie, we can calculate the common prefix for any k strings quickly, while managing string removals.

Track Frequency of Prefixes

As we process each string, we can track the frequency of each prefix in the trie. This enables us to quickly determine if a prefix is common to at least k remaining strings after a string is removed.

Handle Edge Cases of Removal

When a string is removed, the trie must be adjusted efficiently. If fewer than k strings remain after removal, the result for that index should be 0. Proper management of the trie structure when a string is removed is crucial to solving the problem correctly.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is dependent on the trie operations, which are influenced by both the number of strings and their lengths. Space complexity is also affected by the size of the trie, which grows with the total length of all strings in the input array.

What Interviewers Usually Probe

• Ensure the candidate uses a trie or similar efficient data structure to solve the problem.
• Check if the candidate considers edge cases where fewer than k strings remain after removal.
• Look for an understanding of how to efficiently update the trie after each string removal.

Common Pitfalls or Variants

Common pitfalls

• Failing to efficiently manage the trie when strings are removed, leading to high time complexity.
• Not handling cases where fewer than k strings remain after removal, returning incorrect results.
• Overcomplicating the solution without leveraging the trie structure to its full potential.

Follow-up variants

• Consider variations where k can be larger than the number of strings in the array.
• What if strings are very short, or there are multiple duplicate strings in the array?
• Can you solve this problem with a different data structure, and how would the complexity change?

FAQ

How does a trie help in solving this problem efficiently?

A trie allows for fast computation of the longest common prefix by storing strings in a tree-like structure. This enables efficient prefix lookup and management, especially when considering the removal of elements.

What happens when fewer than k strings remain after removal?

In such cases, the result for that index should be 0, as it is impossible to compute a common prefix from fewer than k strings.

Why is this problem considered hard?

The difficulty arises from the need to efficiently compute the longest common prefix among k strings after removing each element, which requires careful data structure management to avoid excessive time complexity.

What is the time complexity of the solution?

The time complexity depends on the size of the trie and the number of strings. Efficient trie operations are key to maintaining optimal performance.

What are common mistakes when solving this problem?

Common mistakes include failing to update the trie correctly after string removal or not handling edge cases where fewer than k strings remain after removal.