---
description: Dynamic Programming | Binary Search
---

# LeetCode 300. 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:**

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

## Solution:

[English Version in Youtube](https://youtu.be/MGoFqowsxbU)

[中文版解答Youtube Link](https://youtu.be/7g3uoteFKug)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1Vh411C7if/)

### 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)
