What is the time complexity of Longest Common Prefix Between Adjacent Strings After Removals?

The optimal solution for Longest Common Prefix Between Adjacent Strings After Removals runs in O(n) time and uses O(n) space. Precomputing LCPs takes O(n), and each removal check is O(1).

What is the brute force solution for Longest Common Prefix Between Adjacent Strings After Removals?

The brute force approach for Longest Common Prefix Between Adjacent Strings After Removals is Brute Force, with O(n²) time complexity and O(1) space complexity. For each string removal, we check all adjacent pairs to find the longest common prefix. This is straightforward but inefficient as it recalculates for every removal. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Common Prefix Between Adjacent Strings After Removals?

Longest Common Prefix Between Adjacent Strings After Removals uses the following concepts: Array, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Common Prefix Between Adjacent Strings After Removals?

Explain your thought process clearly. Consider edge cases upfront. Optimize after understanding the brute force solution.

What are common mistakes when solving Longest Common Prefix Between Adjacent Strings After Removals?

Not considering edge cases with few strings. Incorrectly calculating the longest common prefix.

#3598

Longest Common Prefix Between Adjacent Strings After Removals

Medium

ArrayStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Precompute the longest common prefixes for adjacent pairs. This allows quick access to LCP lengths after each removal without recalculating.

⚙️

Algorithm

3 steps

1Step 1: Precompute the longest common prefix lengths for all adjacent pairs in the original array.
2Step 2: For each index i, check the LCP of pairs that would be adjacent after removing words[i].
3Step 3: Store the maximum LCP length for each removal in the answer array.

solution.py17 lines

1def longestCommonPrefix(words):
2    def lcp(s1, s2):
3        count = 0
4        while count < min(len(s1), len(s2)) and s1[count] == s2[count]:
5            count += 1
6        return count
7    n = len(words)
8    lcp_pairs = [lcp(words[i], words[i + 1]) for i in range(n - 1)]
9    answer = []
10    for i in range(n):
11        if i == 0:
12            answer.append(lcp_pairs[1])
13        elif i == n - 1:
14            answer.append(lcp_pairs[n - 2])
15        else:
16            answer.append(max(lcp_pairs[i - 1], lcp_pairs[i]))
17    return answer

ℹ

Complexity note: Precomputing LCPs takes O(n), and each removal check is O(1).

1Adjacent pairs determine the common prefix length.
2Precomputation can save time during repeated checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.