The task

is taken from leetcode

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

Your algorithm should run in O(n) complexity.

Example:

Input: [100, 4, 200, 1, 3, 2]

Output: 4

Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.

My solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var longestConsecutive = function(nums) {
  const set = new Set(nums);
  let longestStreak = 1;
  let max = 0;
  set.forEach(x => {
    if (!set.has(x - 1)) {
      let num = x;
      while(set.has(++num)) { ++longestStreak; }
      max = Math.max(max, longestStreak);
      longestStreak = 1;      
    }
  });
  return max;
};

Time complexity is O(2n) which is still in the range of O(n) time complexity

This task is taken from Leetcode -

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

Your algorithm should run in \$O(n)\$ complexity.

Example:

Input: [100, 4, 200, 1, 3, 2]

Output: 4

/** Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4. */

My solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var longestConsecutive = function(nums) {
  const set = new Set(nums);
  let longestStreak = 1;
  let max = 0;
  set.forEach(x => {
    if (!set.has(x - 1)) {
      let num = x;
      while(set.has(++num)) { ++longestStreak; }
      max = Math.max(max, longestStreak);
      longestStreak = 1;      
    }
  });
  return max;
};

Time complexity is \$O(2n)\$ which is still in the range of \$O(n)\$ time complexity.