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).
1import java.util.*;
2
3public class LongestIncreasingSubsequence {
4 public int lengthOfLIS(int[] nums) {
5 if (nums == null || nums.length == 0)
6 return 0;
7 List<Integer> ends = new ArrayList<>();
8 for (int num : nums) {
9 int pos = Collections.binarySearch(ends, num);
10 if (pos < 0) pos = -(pos + 1);
11 if (pos < ends.size()) {
12 ends.set(pos, num);
13 } else {
14 ends.add(num);
15 }
16 }
17 return ends.size();
18 }
19 public static void main(String[] args) {
20 LongestIncreasingSubsequence lis = new LongestIncreasingSubsequence();
21 int[] nums = {10,9,2,5,3,7,101,18};
22 System.out.println(lis.lengthOfLIS(nums));
23 }
24}Java's Collections.binarySearch is used to find the position to insert each element in the ends list. The position determines whether to extend or replace an element in the list.