128 Longest Consecutive Sequence

The problem requires us to find the longest consecutive sequence from a given array. But we need to come up with a $O(N)$ solution: that means sorting the array is out of question since it takes $O(n\ast logN)$ time.

The approach in the code below, contains a set with all elements from input. Then once we iterate over the input, we check if any number is the starting point of a consecutive sequence by checking if $set.contains(current-1)$. Since this element is the starting point, we don’t need to worry about the consecutive elements greater than this - there won’t be any.

Once we find this element, maintain a variable to count until we don’t find $(current-1)$ in the set. Also, ensure to update current by $current=current-1$.

Note: Ensure to remove a element that you once visited from the set, as it would cause revisits and the overall complexity will bump to $O(N\ast K)$ where K is consecutive elements for each element in input array.

Code:

Java

class Solution {
    public int longestConsecutive(int[] nums) {
       Set<Integer> check = new HashSet<>();
       for (int a : nums) {
           check.add(a);
       }
 
       int max = 0;
       for (Integer a : nums) {
           if (!check.contains(a-1)) {
               int cur = a + 1;
 
               while (check.contains(cur)) {
                   check.remove(cur);
                   cur += 1;
               }
               max = Math.max(max,cur - a);
           }
       }
 
       return max;
 
    }
}

Big O Analysis

• Runtime The runtime complexity here is $O(N)$
• Memory The memory usage is $O(N)$ since we use a set of input size N.

— A

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