Longest Increasing Subsequence

Given an integer array nums, return the length of the longest strictly increasing subsequence.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Example 1:

Input: nums = [10,9,2,5,3,7,101,18]

Output: 4

Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.

Example 2:

Input: nums = [0,1,0,3,2,3]

Output: 4

Example 3:

Input: nums = [7,7,7,7,7,7,7]

Output: 1

 

Constraints:

1 <= nums.length <= 2500

-104 <= nums[i] <= 104

 

Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity? 

Solution :

C++ :

class Solution {
public:
    // There's a typical DP solution with O(N^2) Time and O(N) space 
    // DP[i] means the result ends at i
    // So for dp[i], dp[i] is max(dp[j]+1), for all j < i and nums[j] < nums[i]
    int lengthOfLIS(vector<int>& nums) {
        const int size = nums.size();
        if (size == 0) { return 0; } 
        vector<int> dp(size, 1);
        int res = 1;
        for (int i = 1; i < size; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[j] < nums[i]) {
                    dp[i] = max(dp[i], dp[j]+1);
                }
            }
            res = max (res, dp[i]);
        }
        return res;
    }
};

PYTHON :

def lengthOfLIS(self, nums):
    tails = [0] * len(nums)
    size = 0
    for x in nums:
        i, j = 0, size
        while i != j:
            m = (i + j) / 2
            if tails[m] < x:
                i = m + 1
            else:
                j = m
        tails[i] = x
        size = max(i + 1, size)
    return size