Longest Increasing Subsequence

Find the longest increasing subsequence of a given array of integers, A.

In other words, find a subsequence of array in which the subsequence’s elements are in strictly increasing order, and in which the subsequence is as long as possible.
This subsequence is not necessarily contiguous, or unique.
In this case, we only care about the length of the longest increasing subsequence.

Input Format:

The first and the only argument is an integer array A.

Output Format:

Return an integer representing the length of the longest increasing subsequence.

Constraints:

1 <= length(A) <= 2500
1 <= A[i] <= 2000

Example :

Input 1:
    A = [1, 2, 1, 5]

Output 1:
    3
    
Explanation 1:
    The sequence : [1, 2, 5]

Input 2:
    A = [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]
    
Output 2:
    6

Explanation 2:
    The sequence : [0, 2, 6, 9, 13, 15] or [0, 4, 6, 9, 11, 15] or [0, 4, 6, 9, 13, 15]

Solution 1:

Time: O(n^2)
Space: O(n)

public class Solution {
    // DO NOT MODIFY THE LIST. IT IS READ ONLY
    public int lis(final List<Integer> A) {
        int[] inc = new int[A.size()];
        int max = 0;
        for (int i = 0; i < A.size(); i ++) {
            inc[i] = 1;
            for (int j = 0; j < i; j ++) {
                if (A.get(j) < A.get(i)) {
                    inc[i] = Math.max(inc[i], inc[j] + 1);
                }
            }
            max = Math.max(max, inc[i]);
        }
        return max;
    }
}

Method 2:

Use a TreeSet (Or list) to keep track of current ascending subsequence, replace number with current element if current element is smaller than any of existing number, add new number to the set if the number is greater than all existing numbers, in which case increase the size of the set

Solution 2:

Time: O(nlogn)
Space: O(n)

public class Solution {
    // DO NOT MODIFY THE LIST. IT IS READ ONLY
    public int lis(final List<Integer> A) {
        TreeSet<Integer> set = new TreeSet<>();
        int max = 0;
        for (int num : A) {
            Integer largerOrEuqal = set.ceiling(num);
            if (largerOrEuqal != null) {
                set.remove(largerOrEuqal);
            }
            set.add(num);
            max = Math.max(max, set.size());
        }
        return max;
    }
}