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.
1import java.util.*;
2
3public class LongestIncreasingSubsequence {
4 public int lengthOfLIS(int[] nums) {
5 if (nums == null || nums.length == 0)
6 return 0;
7 int n = nums.length;
8 int[] dp = new int[n];
9 Arrays.fill(dp, 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] = Math.max(dp[i], dp[j] + 1);
15 }
16 }
17 maxLen = Math.max(maxLen, dp[i]);
18 }
19 return maxLen;
20 }
21 public static void main(String[] args) {
22 LongestIncreasingSubsequence lis = new LongestIncreasingSubsequence();
23 int[] nums = {10,9,2,5,3,7,101,18};
24 System.out.println(lis.lengthOfLIS(nums));
25 }
26}The Java solution follows the same logic as C and C++ solutions, using a dp array initialized to 1. It iterates over pairs of elements, updating the array to store the longest subsequence lengths found.
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.