This approach uses dynamic programming to solve the problem in O(n^2) time complexity. We maintain a dp array where dp[i] represents the length of the longest increasing subsequence that ends with nums[i]. For each element, we iterate over all previous elements to see if they can be included in the subsequence ending at the current element, and update dp[i] accordingly.
Time Complexity: O(n^2) where n is the length of the input array. Space Complexity: O(n) for storing the dp array.
1#include <vector>
2#include <algorithm>
3#include <iostream>
4using namespace std;
5
6int lengthOfLIS(vector<int>& nums) {
7 int n = nums.size();
8 if (n == 0) return 0;
9 vector<int> dp(n, 1);
10 int maxLen = 1;
11 for (int i = 1; i < n; ++i) {
12 for (int j = 0; j < i; ++j) {
13 if (nums[i] > nums[j]) {
14 dp[i] = max(dp[i], dp[j] + 1);
15 }
16 }
17 maxLen = max(maxLen, dp[i]);
18 }
19 return maxLen;
20}
21
22int main() {
23 vector<int> nums = {10,9,2,5,3,7,101,18};
24 cout << lengthOfLIS(nums) << endl;
25 return 0;
26}This solution also uses a dynamic programming array to store the length of the longest subsequence ending at each position. The dp array is iteratively built by comparing elements and updating dp values based on smaller indices.
In this approach, we use a combination of dynamic programming and binary search to solve the problem in O(n log n) time. This is often called the 'patience sorting' technique where we maintain a list, 'ends', to store the smallest ending value of any increasing subsequence with length i+1 in 'ends[i]'. We use binary search to find the position where each element in nums can be placed in 'ends'.
Time Complexity: O(n log n). Space Complexity: O(n).
1from bisect import bisect_left
2
3class Solution:
4 def lengthOfLIS(self, nums):
5 ends = []
6 for num in nums:
7 pos = bisect_left(ends, num)
8 if pos < len(ends):
9 ends[pos] = num
10 else:
11 ends.append(num)
12 return len(ends)
13
14# Example usage:
15nums = [10,9,2,5,3,7,101,18]
16sol = Solution()
17print(sol.lengthOfLIS(nums))Python solution makes use of the bisect module to apply binary search on the list 'ends'. This approach ensures finding the right position for each element in logarithmic time.