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 elements sequence in an unsorted array of integers.
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. This problem is significant in various applications such as data analysis, where identifying trends or patterns in data is crucial.
Potential pitfalls include handling duplicates and ensuring that the solution is efficient in terms of time complexity.
To solve this problem, we can start with a naive approach and then move to more optimized solutions.
The naive solution involves 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.
We can optimize the solution by using a set to store the elements and then iterate through the array to find the longest consecutive sequence. This approach can achieve O(n) time complexity, but for this lesson, we will focus on the O(n log n) solution.
Here is a step-by-step breakdown of the O(n log n) algorithm:
import java.util.Arrays;
public class LongestConsecutiveSequence {
public static int longestConsecutive(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
// Sort the array
Arrays.sort(nums);
int longestStreak = 1;
int currentStreak = 1;
for (int i = 1; i < nums.length; i++) {
// Skip duplicates
if (nums[i] != nums[i - 1]) {
// Check if the current number is consecutive
if (nums[i] == nums[i - 1] + 1) {
currentStreak += 1;
} else {
// Update the longest streak
longestStreak = Math.max(longestStreak, currentStreak);
currentStreak = 1;
}
}
}
// Return the maximum of the longest streak and the current streak
return Math.max(longestStreak, currentStreak);
}
public static void main(String[] args) {
int[] nums1 = {100, 4, 200, 1, 3, 2};
int[] nums2 = {0, 2, 0, 1, 2, 3, 1};
System.out.println("Longest consecutive sequence length (Example 1): " + longestConsecutive(nums1)); // Output: 4
System.out.println("Longest consecutive sequence length (Example 2): " + longestConsecutive(nums2)); // Output: 4
}
}
The time complexity of the above 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 that scales with the input size.
Potential edge cases include:
To test the solution comprehensively, consider the following test cases:
When approaching such problems, consider the following tips:
In this blog post, we discussed how to find the longest consecutive sequence in an unsorted array of integers using an O(n log n) approach. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving problem-solving skills and preparing for coding interviews.
For further reading and practice, consider the following resources: