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 duplicate elements 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. This approach is not optimal because sorting takes O(n log n) time, and we need to traverse the sorted array to find the sequence.
We can use a set to achieve a more efficient solution. The idea is to use a set to store the elements of the array, and then iterate through the array to find the start of a sequence. For each start, we count the length of the sequence.
Here is a step-by-step breakdown of the algorithm:
longestStreak
to keep track of the longest sequence length.longestStreak
whenever a longer sequence is found.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int longestConsecutive(vector<int>& nums) {
if (nums.empty()) return 0;
sort(nums.begin(), nums.end());
int longestStreak = 1;
int currentStreak = 1;
for (int i = 1; i < nums.size(); ++i) {
if (nums[i] != nums[i - 1]) {
if (nums[i] == nums[i - 1] + 1) {
currentStreak += 1;
} else {
longestStreak = max(longestStreak, currentStreak);
currentStreak = 1;
}
}
}
return max(longestStreak, currentStreak);
}
int main() {
vector<int> nums = {100, 4, 200, 1, 3, 2};
cout << "Longest consecutive sequence length: " << longestConsecutive(nums) << endl;
return 0;
}
The time complexity of the above approach 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, consider the following tips:
In this blog post, we discussed how to find the length of the longest consecutive sequence in an unsorted array of integers. We explored a naive solution and then optimized it using sorting. Understanding and solving such problems is crucial for developing strong problem-solving skills.
For further reading and practice, consider the following resources: