LeetCode 300. Longest Increasing Subsequence

Dynamic Programming | Binary Search

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:

Could you come up with the O(n2) solution?
Could you improve it to O(n log(n)) time complexity?

Solution:

Solution 1: DP

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp(n);
        
        int ans = 0;
        for (int i = 0; i < n; i++) {
            int tmp = 1;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    tmp = max(tmp, dp[j]+1);
                }
            }
            
            ans = max(ans, tmp);
            dp[i] = tmp;
        }
        
        return ans;
    }
};

Time Complexity: O(n^2)

Space Complexity: O(n)

Solution 2: Binary Search with DP

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> optimal_structure;
        
        for (int num : nums) {
            auto it = std::lower_bound(optimal_structure.begin(), optimal_structure.end(), num);
            if (it != optimal_structure.end()) {
                *it = num;
            } else {
                optimal_structure.push_back(num);
            }
        }
        
        return optimal_structure.size();
    }
};

Time Complexity: O(nlog(n))

Space Complexity: O(n)

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Solution:

Solution 1: DP

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp(n);
        
        int ans = 0;
        for (int i = 0; i < n; i++) {
            int tmp = 1;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    tmp = max(tmp, dp[j]+1);
                }
            }
            
            ans = max(ans, tmp);
            dp[i] = tmp;
        }
        
        return ans;
    }
};

Time Complexity: O(n^2)

Space Complexity: O(n)

Solution 2: Binary Search with DP

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> optimal_structure;
        
        for (int num : nums) {
            auto it = std::lower_bound(optimal_structure.begin(), optimal_structure.end(), num);
            if (it != optimal_structure.end()) {
                *it = num;
            } else {
                optimal_structure.push_back(num);
            }
        }
        
        return optimal_structure.size();
    }
};

Time Complexity: O(nlog(n))

Space Complexity: O(n)