Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
Example 1:
Input: [100, 4, 200, 1, 3, 2]
Output: 4
Explanation: Longest consecutive sequence is [1, 2, 3, 4]
.
Therefore its length is 4.
Example 2:
Input: [0, 2, 0, 1, 2, 3, 1]
Output: 4
Explanation: Longest consecutive sequence is [0, 1, 2, 3]
.
Therefore its length is 4.
Note that we count each value once, even tho values 0, 1 and 2 appear 2 times each in nums
For this lesson, your algorithm should run in O(n log n) time and use O(1) extra space.
(There are faster solutions which we will discuss in future lessons)
The problem requires finding the length of the longest consecutive sequence in an unsorted array of integers.
The core challenge is to identify the longest sequence of consecutive integers in an unsorted array. This problem is significant in various applications such as data analysis, where identifying trends or patterns in data is crucial. A common pitfall is to assume that sorting the array and then finding the sequence is the only way, but there are more efficient methods.
To solve this problem, we can start with a naive approach and then optimize it:
The naive solution involves sorting the array and then finding the longest consecutive sequence. This approach is straightforward but not optimal in terms of time complexity.
We can optimize the solution by using a set to keep track of the elements and then iterate through the array to find the longest sequence. This approach ensures that we only traverse the array once, achieving O(n) time complexity.
Here is a step-by-step breakdown of the optimized algorithm:
def longest_consecutive(nums):
# Sort the array
nums.sort()
# Initialize variables
longest_streak = 0
current_streak = 1
# Iterate through the sorted array
for i in range(1, len(nums)):
# Check if the current element is consecutive to the previous element
if nums[i] != nums[i - 1]: # Skip duplicates
if nums[i] == nums[i - 1] + 1:
current_streak += 1
else:
longest_streak = max(longest_streak, current_streak)
current_streak = 1
# Return the longest sequence length
return max(longest_streak, current_streak)
# Example usage
print(longest_consecutive([100, 4, 200, 1, 3, 2])) # Output: 4
print(longest_consecutive([0, 2, 0, 1, 2, 3, 1])) # Output: 4
The time complexity of the optimized solution is O(n log n) due to the sorting step. The space complexity is O(1) as we are not using any extra space apart from a few variables.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
When approaching such problems, it is essential to:
In this blog post, we discussed how to find the longest consecutive sequence in an unsorted array of integers. We explored a naive solution and then optimized it to achieve O(n log n) time complexity. Understanding and solving such problems is crucial for developing problem-solving skills and improving algorithmic thinking.
For further reading and practice, consider the following resources: