• Computer Science
• Python
• TypeScript

LeetCode Meditations: Longest Consecutive Sequence

The description for Longest Consecutive Sequence states:

Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.

You must write an algorithm that runs in O(n) time.

For example:

longestConsecutive([100, 4, 200, 1, 3, 2]);
// -> 4
// The longest consecutive elements sequence is `[1, 2, 3, 4]`, the length is 4.

longestConsecutive([0, 3, 7, 2, 5, 8, 4, 6, 0, 1]);
// -> 9

longestConsecutive([100, 4, 200, 1, 3, 2]);
// -> 4
// The longest consecutive elements sequence is `[1, 2, 3, 4]`, the length is 4.

longestConsecutive([0, 3, 7, 2, 5, 8, 4, 6, 0, 1]);
// -> 9

The first idea is to get the non-duplicate elements, sort them, and then count how many of them follows a consecutive series. However, it won't be $O(n)$ time as the description says: any sorting can't be better than $O(n \ log \ n)$. But let's see how we can go about it nonetheless.

Getting the non-duplicate ones is easy: we can use a Set. Sorting them is no problem either, but how can we count the consecutive ones? Looping through each element, we can't just check if the current element + 1 is in the set and update the count, because that doesn't mean that there is a consecutive order.

So instead, we can keep track of multiple counts, rather than holding just one count variable. In order to do that, we need to keep multiple starting points for each sequence that potentially exists.

The tricky part is when the sequences change. For example, in a sorted array like [1, 2, 3, 4, 100, 150], it is obvious that the first sequence is of length $4$, but when it comes to 100, we need to reset our count to start a new sequence.

In TypeScript, it might look like this:

function longestConsecutive(nums: number[]): number {
  if (!nums.length) { 
    return 0; 
  }

  let counts: number[] = [];
  let count = 0;
  let numsSorted = [...new Set(nums)].sort((a, b) => a - b);

  for (let i = 0; i < numsSorted.length; i++) {
    counts.push(++count);
    if (numsSorted[i + 1] !== numsSorted[i] + 1) {
      count = 0;
    }
  }

  return Math.max(...counts);
};

function longestConsecutive(nums: number[]): number {
  if (!nums.length) { 
    return 0; 
  }

  let counts: number[] = [];
  let count = 0;
  let numsSorted = [...new Set(nums)].sort((a, b) => a - b);

  for (let i = 0; i < numsSorted.length; i++) {
    counts.push(++count);
    if (numsSorted[i + 1] !== numsSorted[i] + 1) {
      count = 0;
    }
  }

  return Math.max(...counts);
};

So, as we loop through each element, we keep count, and add it to our counts array, and only reset it when the next element is not the next one in sequence.

This solution passes the tests, but note that when we get to the last element, numsSorted[i + 1] is just undefined, so checking for inequality is meaningless.

Time and space complexity

Since we are sorting nums, time complexity can't be better than $O(n \ log \ n)$. The space complexity will be $O(n)$ because of the additional storage for numsSorted and counts arrays, which will grow linearly as the length nums increases.

In fact, there is a much better way of doing this, so let's take a deep breath, and see how we can improve.

When you notice that we use a Set anyway, why not use it for what it's already good at, checking if an element is in it, instead of just pruning the duplicate elements? The good part is that we don't even need to sort them.

function longestConsecutive(nums: number[]): number {
  if (!nums.length) { 
    return 0;
  }

  let count: number;
  let nums_ = new Set(nums);
  let longestSeq = 0;
    
  for (let n of nums) {
    if (!nums_.has(n - 1)) {
      count = 0;
      while (nums_.has(n + count)) {
        count++;
      }

      if (count > longestSeq) {
        longestSeq = count;
      }
    }
  }

  return longestSeq;
};

function longestConsecutive(nums: number[]): number {
  if (!nums.length) { 
    return 0;
  }

  let count: number;
  let nums_ = new Set(nums);
  let longestSeq = 0;
    
  for (let n of nums) {
    if (!nums_.has(n - 1)) {
      count = 0;
      while (nums_.has(n + count)) {
        count++;
      }

      if (count > longestSeq) {
        longestSeq = count;
      }
    }
  }

  return longestSeq;
};

This time we check if an element has a previous one that comes before it; if not, we reset count and continue incrementing it while there is a consecutive sequence from that element onward. We update the longest sequence accordingly if the current count is greater than the previous longest sequence.

Here is the Python version:

class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        if not nums:
            return 0

        nums_ = set(nums)
        longest_seq = 0

        for n in nums:
            if n - 1 not in nums_:
                count = 0
                while n + count in nums_:
                    count += 1

                if count > longest_seq:
                    longest_seq = count

        return longest_seq

class Solution:
    def longestConsecutive(self, nums: List[int]) -> int:
        if not nums:
            return 0

        nums_ = set(nums)
        longest_seq = 0

        for n in nums:
            if n - 1 not in nums_:
                count = 0
                while n + count in nums_:
                    count += 1

                if count > longest_seq:
                    longest_seq = count

        return longest_seq

Time and space complexity

The time complexity is just $O(n)$ this time, as we only iterate through the nums array.
The space complexity is again $O(n)$ though, because we need to allocate space for nums_.

This was the last problem in Arrays & Hashing section in Blind 75. Next up, we'll look at the Two Pointers technique. Until then, happy coding.