Longest Consecutive Sequence
MediumOnly start counting from sequence heads
Put all numbers in a set, then from each number that starts a run, count upward as far as you can.
The idea
Put everything in a set. A number starts a run only if num−1 is absent; from each such start, walk upward counting. Every number is visited at most twice, so it's O(n).
100
4
200
1
3
2
0
1
2
3
4
5
Step 1 of 8. Brute force: from each number, keep searching for the next consecutive one. Values: 100, 4, 200, 1, 3, 2.
1/8
Brute force
timeO(n²)spaceO(n)
Search each run.
1for (const x of nums) {2 let len = 1, cur = x;3 while (set.has(cur + 1)) { cur++; len++; }4 best = Math.max(best, len);5}Input
- array
- [100, 4, 200, 1, 3, 2]
Memory
- i
- —
Output
- best
- —
- answer
- —
Check yourself
3 quick questions about this walkthrough. A wrong answer costs nothing.
Examples
Example 1
- Input:
- nums = [100, 4, 200, 1, 3, 2]
- Output:
- 4
- Explanation:
- 1,2,3,4 form a run of length 4.
Example 2
- Input:
- nums = [0, 3, 7, 2, 5, 8, 4, 6, 0, 1]
- Output:
- 9
- Explanation:
- 0..8 is a run of length 9.
Example 3
- Input:
- nums = [10]
- Output:
- 1
- Explanation:
- A single number is a run of length 1.
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