How do you solve Longest Consecutive Sequence on LeetCode?

Longest Consecutive Sequence (LeetCode #128) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashSet allows for O(1) average time complexity for lookups, which is crucial for efficiently finding consecutive sequences.

What is the brute force solution for Longest Consecutive Sequence?

The brute force approach for Longest Consecutive Sequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible sequence of numbers in the array. It does this by sorting the array and then counting consecutive numbers, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Consecutive Sequence?

Longest Consecutive Sequence uses the following concepts: Array, Hash Table, Union-Find. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Consecutive Sequence?

Always consider edge cases, such as empty arrays or arrays with all duplicate values. Think about the data structures that can optimize your solution, like HashSets for fast lookups. Explain your thought process clearly during the interview, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Longest Consecutive Sequence?

Failing to handle duplicates correctly, which can lead to incorrect streak counts. Not using a HashSet for lookups, which can lead to inefficient solutions.

#128

Longest Consecutive Sequence

Medium

ArrayHash TableUnion-FindHash 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)

The optimal approach uses a HashSet to store the numbers, allowing for O(1) average time complexity for lookups. This way, we can efficiently find the start of each sequence and count its length without sorting.

⚙️

Algorithm

5 steps

1Step 1: Insert all numbers into a HashSet for O(1) access.
2Step 2: Initialize a variable to keep track of the longest sequence length.
3Step 3: Iterate through each number in the array. For each number, check if it is the start of a sequence (i.e., number - 1 is not in the set).
4Step 4: If it is the start, count the length of the sequence by checking for consecutive numbers in the set.
5Step 5: Update the longest sequence length if the current count is greater.

solution.py14 lines

1# Full working Python code
2
3def longest_consecutive(nums):
4    num_set = set(nums)
5    longest = 0
6    for num in num_set:
7        if num - 1 not in num_set:  # start of a sequence
8            current_num = num
9            current_streak = 1
10            while current_num + 1 in num_set:
11                current_num += 1
12                current_streak += 1
13            longest = max(longest, current_streak)
14    return longest

ℹ

Complexity note: The time complexity is O(n) because we traverse the array and perform constant time operations with the HashSet. The space complexity is O(n) due to storing the numbers in the HashSet.

1Using a HashSet allows for O(1) average time complexity for lookups, which is crucial for efficiently finding consecutive sequences.
2Identifying the start of a sequence by checking if the previous number is absent helps avoid unnecessary checks.

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