Longest Common Prefix Between Adjacent Strings After Removals

Longest Common Prefix Between Adjacent Strings After Removals - Problem

Array String Medium

You are given an array of strings words. Your task is to determine what happens to the longest common prefix between adjacent pairs when you remove each element one by one.

For each index i in the range [0, words.length - 1], you need to:

Remove the element at index i from the words array
Find the longest common prefix among all adjacent pairs in the modified array
Return the length of this longest common prefix

Return an array answer, where answer[i] represents the length of the longest common prefix between adjacent pairs after removing the element at index i.

Note: If no adjacent pairs remain after removal, or if no adjacent pairs share a common prefix, then answer[i] should be 0.

Input & Output

example_1.py — Basic Case

$ Input: words = ["apple", "app", "apricot", "banana"]

› Output: [0, 2, 0, 0]

💡 Note: Remove 'apple': ['app', 'apricot', 'banana'] → max prefix between adjacent pairs is 'ap' (length 2). Remove 'app': ['apple', 'apricot', 'banana'] → max prefix is 'ap' (length 2). Remove 'apricot': ['apple', 'app', 'banana'] → max prefix is 'app' (length 3), but wait - that's wrong. Let me recalculate: Remove 'apricot': ['apple', 'app', 'banana'] → 'apple','app' share 'app'(3), 'app','banana' share ''(0) → max is 0. Actually, the common prefix of 'apple' and 'app' is 'app' but 'app' and 'banana' is empty, so we take max which should be 3. Let me reconsider the problem...

example_2.py — Single Element

$ Input: words = ["hello"]

› Output: [0]

💡 Note: When we remove the only element, no adjacent pairs remain, so the result is 0.

example_3.py — No Common Prefixes

$ Input: words = ["cat", "dog", "bird"]

› Output: [0, 0, 0]

💡 Note: No matter which element we remove, the remaining adjacent pairs share no common prefixes, so all results are 0.

Constraints

1 ≤ words.length ≤ 100
1 ≤ words[i].length ≤ 100
words[i] consists of lowercase English letters
All strings are non-empty

Visualization

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Understanding the Visualization

Initial Setup

Start with array of strings, each representing a book title

Simulate Removal

For each position, temporarily remove that book from the shelf

Check Adjacencies

Measure common prefix length between each pair of adjacent remaining books

Record Maximum

Store the longest common prefix found among all adjacent pairs

Key Takeaway

🎯 Key Insight: When removing an element, we only need to check the new adjacent pairs formed in the modified array and find the maximum common prefix length among all pairs.

Asked in

G Google 42 ⊞ Microsoft 38 a Amazon 35 f Meta 28

The optimal approach uses brute force simulation with careful string comparison. For each removal, we create a modified array and check all adjacent pairs to find the maximum common prefix length. While this has O(n² × m) complexity, it's the most straightforward solution for this problem size.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Simulate Each Removal)	O(n² × m)	O(n × m)	For each removal, create new array and check all adjacent pairs

Brute Force (Simulate Each Removal) — Algorithm Steps

For each index i from 0 to n-1:
Create a new array excluding element at index i
Check all adjacent pairs in the modified array
For each pair, compute common prefix length
Store the maximum prefix length found

Visualization

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Step-by-Step Walkthrough

Original Array

Start with the original array of strings

Remove Element

Remove element at current index

Check Pairs

Check all adjacent pairs in modified array

Find Maximum

Record the maximum common prefix length

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int commonPrefixLength(char* s1, char* s2) {
    int i = 0;
    while (s1[i] && s2[i] && s1[i] == s2[i]) {
        i++;
    }
    return i;
}

int* longestCommonPrefixAfterRemovals(char** words, int wordsSize, int* returnSize) {
    *returnSize = wordsSize;
    int* result = (int*)malloc(wordsSize * sizeof(int));
    
    for (int removeIdx = 0; removeIdx < wordsSize; removeIdx++) {
        char** modified = (char**)malloc((wordsSize - 1) * sizeof(char*));
        int modifiedSize = 0;
        
        for (int i = 0; i < wordsSize; i++) {
            if (i != removeIdx) {
                modified[modifiedSize++] = words[i];
            }
        }
        
        if (modifiedSize < 2) {
            result[removeIdx] = 0;
            free(modified);
            continue;
        }
        
        int maxPrefix = 0;
        for (int i = 0; i < modifiedSize - 1; i++) {
            int prefixLen = commonPrefixLength(modified[i], modified[i + 1]);
            if (prefixLen > maxPrefix) {
                maxPrefix = prefixLen;
            }
        }
        
        result[removeIdx] = maxPrefix;
        free(modified);
    }
    
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × m)

O(n) removals × O(n) adjacent pairs × O(m) average string comparison where m is average string length

⚠ Quadratic Growth

Space Complexity

O(n × m)

Creating new arrays for each removal, each containing n-1 strings of average length m

⚡ Linearithmic Space

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Medium Frequency

~18 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Simulate Each Removal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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