300. Longest Increasing Subsequence

Given an unsorted array of integers, find the length of longest increasing subsequence.

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.

Your algorithm should run in O(n2) complexity.

Follow up: Could you improve it to O(n log n) time complexity?

Solution:

Method 1: Dynamic Programming

We use LIS[I] to denote the max LIS that end at nums[i] including nums[i].

It is the max of all (LIS[j] + 1), where j < i, if nums[i] > nums[j].

The time complexity is O(n²) and the space complexity is O(n).



Code:

public class Solution {
    public int lengthOfLIS(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int max = 1;
        int[] LIS = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            LIS[i] = 1;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    LIS[i] = Math.max(LIS[i], LIS[j] + 1);
                }
            }
            max = Math.max(max, LIS[i]);
        }
        return max;
    }
}

Method 2: DP + Binary Search

The idea is to maintain an array that array[i] means the smallest number that can form an increasing sequence with length i.

For example,  [4, 5, 6, 3]

len = 1   :      [4], [5], [6], [3]    => tails[1] = 3
len = 2   :      [4, 5], [5, 6]         => tails[2] = 5
len = 3   :      [4, 5, 6]                => tails[3] = 6

Therefore, we create an array tails with length + 1.

Traverse all numbers from the input array.

For each number num, use binary search to find the first index in tails where tails[i] >= num.

Replace tails[i] with num.

After traversing, go backward and find the first valid index, which is the longest length.

The time complexity is O(nlogn) and the space complexity is O(n).



Code:

public class Solution {
    public int lengthOfLIS(int[] nums) {
        int[] tails = new int[nums.length + 1];
        Arrays.fill(tails, Integer.MAX_VALUE);
        tails[0] = Integer.MIN_VALUE;
        
        int max = 0; // input can be [].
        for (int num : nums) {
            int index = findFirst(tails, num);
            tails[index] = num;
            max = Math.max(max, index);
        }
        return max;
    }
    
    public int findFirst(int[] nums, int target) {
        int start = 1;
        int end = nums.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] < target) {
                start = mid;
            }
            else {
                end = mid;
            }
        }
        if (nums[start] >= target) {
            return start;
        }
        return end;
    }
}