300. Longest Increasing Subsequence

📝 Problem Details

• Title: 300. Longest Increasing Subsequence
• Link: https://leetcode.com/problems/longest-increasing-subsequence/
• Difficulty: Medium
• Tags/Categories: Dynamic-Programming

input array nums return the length of the longest strictly increasing subsequence subsequence: can remove elements so long as the order of the remaining elements stay the same

💡 Explanation of Solution

1. subproblem identification
- whats the longest subsequence ending at index i

2. use a bottom up approach as it will inform the result of future index values

3. dp state
- let dp[i] be the length of the longest increasing sequence that ends at index i
- base case: every element by itself has an increasing subsequence is 1 --> dp[i] = 1

4. transition formula
- to build dp[i] check all previous indicies j < i
- if nums[j] < nums[i] (increasing), then update dp[i]
	- dp[i] = max(dp[i], dp[j]+1)
	- final answer is max(dp[i]) for all i

⌛ Complexity Analysis

Time Complexity: O(n^2)
Space Complexity: O(n) for dp table

💻 Implementation of Solution

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        if(n == 0) return 0;
 
        vector<int> dp(n, 1); // base case: every element is a subsequence of length 1
 
        int maxLength = 1;
        for(int i=1; i < n; i++) {
            for(int j=0; j < i; j++) {
                if(nums[j] < nums[i]) { // valid increasing sequence
                    dp[i] = max(dp[i], dp[j] + 1);
                }
            }
            maxLength = max(maxLength, dp[i]); // track max LIS found
        }
        return maxLength;
    }
};