How do you solve Find the Length of the Longest Common Prefix on LeetCode?

Find the Length of the Longest Common Prefix (LeetCode #3043) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Using a HashSet allows for O(1) average time complexity for lookups.

What is the brute force solution for Find the Length of the Longest Common Prefix?

The brute force approach for Find the Length of the Longest Common Prefix is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible pair of integers from both arrays and finds the common prefix for each pair. It keeps track of the longest common prefix found during these comparisons. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Find the Length of the Longest Common Prefix?

Find the Length of the Longest Common Prefix uses the following concepts: Array, Hash Table, String, Trie. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Length of the Longest Common Prefix?

Clarify the problem statement and constraints before jumping to code. Think about edge cases, such as arrays with single elements or no common prefixes. Practice writing clean and efficient code with proper variable naming.

What are common mistakes when solving Find the Length of the Longest Common Prefix?

Not considering all prefixes of integers. Confusing the length of the prefix with the actual prefix value.

#3043

Find the Length of the Longest Common Prefix

Medium

ArrayHash TableStringTrieHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n + m)Space O(n)

The optimal solution uses a HashSet to store all possible prefixes of integers in arr1. Then, it checks each integer in arr2 to see if any of its prefixes exist in the HashSet, allowing for efficient lookup.

⚙️

Algorithm

3 steps

1Step 1: Create a HashSet to store all prefixes of each integer in arr1.
2Step 2: For each integer in arr2, generate its prefixes and check if they exist in the HashSet.
3Step 3: Keep track of the maximum length of the common prefixes found.

solution.py21 lines

1# Full working Python code
2arr1 = [1, 10, 100]
3arr2 = [1000]
4
5def longest_common_prefix(arr1, arr2):
6    prefixes = set()
7    for x in arr1:
8        str_x = str(x)
9        for i in range(1, len(str_x) + 1):
10            prefixes.add(str_x[:i])
11
12    max_length = 0
13    for y in arr2:
14        str_y = str(y)
15        for i in range(1, len(str_y) + 1):
16            if str_y[:i] in prefixes:
17                max_length = max(max_length, i)
18
19    return max_length
20
21print(longest_common_prefix(arr1, arr2))

ℹ

Complexity note: The time complexity is O(n + m) where n is the total number of digits in arr1 and m is the total number of digits in arr2. The space complexity is O(n) due to the HashSet storing prefixes from arr1.

1Using a HashSet allows for O(1) average time complexity for lookups.
2Generating prefixes efficiently reduces the number of comparisons needed.

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