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
Your algorithm should run in O(n) time and use O(n) extra space.
The problem requires finding the length of the longest consecutive elements sequence in an unsorted array of integers. The algorithm should run in O(n) time and use O(n) extra space.
Input: [100, 4, 200, 1, 3, 2] Output: 4 Explanation: Longest consecutive sequence is [1, 2, 3, 4]. Therefore its length is 4.
The core challenge is to identify the longest sequence of consecutive integers in an unsorted array efficiently. This problem is significant in various applications such as data analysis, where finding patterns in data is crucial. A common pitfall is to sort the array first, which would result in a time complexity of O(n log n), not meeting the O(n) requirement.
To solve this problem efficiently, we can use a set to store the elements of the array. This allows O(1) average time complexity for lookups. The idea is to iterate through the array and for each element, check if it is the start of a sequence (i.e., the previous element is not in the set). If it is, we then count the length of the sequence by checking consecutive elements.
A naive solution would involve sorting the array and then finding the longest consecutive sequence. However, this approach has a time complexity of O(n log n) due to the sorting step, which is not optimal.
The optimized solution involves using a set to achieve O(n) time complexity. Here’s the thought process:
Here is a step-by-step breakdown of the algorithm:
def longest_consecutive(nums):
# Create a set of the array elements
num_set = set(nums)
max_length = 0
# Iterate through the array
for num in nums:
# Check if it is the start of a sequence
if num - 1 not in num_set:
current_num = num
current_length = 1
# Count the length of the sequence
while current_num + 1 in num_set:
current_num += 1
current_length += 1
# Update the maximum sequence length
max_length = max(max_length, current_length)
return max_length
# 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 this approach is O(n) because each element is processed at most twice (once when checking if it is the start of a sequence and once when counting the length of the sequence). The space complexity is O(n) due to the set used to store the elements.
Potential edge cases include:
Examples:
Input: [] Output: 0 Input: [1] Output: 1 Input: [2, 2, 2] Output: 1
To test the solution comprehensively, consider a variety of test cases:
Example test cases:
assert longest_consecutive([100, 4, 200, 1, 3, 2]) == 4
assert longest_consecutive([0, 2, 0, 1, 2, 3, 1]) == 4
assert longest_consecutive([]) == 0
assert longest_consecutive([1]) == 1
assert longest_consecutive([2, 2, 2]) == 1
assert longest_consecutive([-1, -2, -3, -4, -5]) == 5
When approaching such problems:
To improve problem-solving skills, practice similar problems, study algorithms, and understand their trade-offs.
In this blog post, we discussed how to find the length of the longest consecutive sequence in an unsorted array of integers efficiently. We explored the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.
For further reading and practice, consider the following resources: