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.
1class Solution:
2 def lengthOfLIS(self, nums):
3 if not nums:
4 return 0
5 dp = [1] * len(nums)
6 max_len = 1
7 for i in range(1, len(nums)):
8 for j in range(i):
9 if nums[i] > nums[j]:
10 dp[i] = max(dp[i], dp[j] + 1)
11 max_len = max(max_len, dp[i])
12 return max_len
13
14# Example usage:
15nums = [10,9,2,5,3,7,101,18]
16sol = Solution()
17print(sol.lengthOfLIS(nums))In Python, we use a list comprehension to initialize the dp array, then apply the nested loops to update the dp as in other examples.
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).
1#include <stdio.h>
2#include <stdlib.h>
3
4int comparator(const void *a, const void *b) {
5 return (*(int *)a - *(int *)b);
6}
7
8int lowerBound(int *arr, int size, int value) {
9 int low = 0, high = size;
10 while (low < high) {
11 int mid = low + (high - low) / 2;
12 if (arr[mid] < value) {
13 low = mid + 1;
14 } else {
15 high = mid;
16 }
17 }
18 return low;
19}
20
21int lengthOfLIS(int* nums, int numsSize) {
22 if (numsSize == 0) return 0;
23 int* ends = (int*)malloc(numsSize * sizeof(int));
24 int length = 0;
25 for (int i = 0; i < numsSize; ++i) {
26 int pos = lowerBound(ends, length, nums[i]);
27 ends[pos] = nums[i];
28 if (pos == length) {
29 length++;
30 }
31 }
32 free(ends);
33 return length;
34}
35
36int main() {
37 int nums[] = {10,9,2,5,3,7,101,18};
38 int length = sizeof(nums)/sizeof(nums[0]);
39 printf("%d\n", lengthOfLIS(nums, length));
40 return 0;
41}This C solution uses a lowerBound function which acts similarly to std::lower_bound in C++. It finds the position to insert (or replace) each element of nums in the 'ends' list, effectively forming the required subsequences.