#### Longest Increasing Subsequence in an Array

Given an array of integers, this problem aims to find the length of the longest increasing subsequence in the array. For example, given the array [10,9,2,5, 3,7,101,18], the longest increasing subsequence is [2,3,7,101] and the length is 4.

#### Problem

```Given an array of integers, find the length of the longest increasing subsequence in the array.
Input: [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.```

#### Solution

This solution is in Python. Our A.I. can create solutions in multiple languages.
```# Solution:
# The solution is to use dynamic programming.
# The idea is to create a list of the same length as the input list.
# The list will contain the length of the longest increasing subsequence ending at the index of the input list.
# The list will be filled in a bottom-up manner.
# The base case is when the list is empty.
# The length of the longest increasing subsequence ending at the first index is 1.
# The length of the longest increasing subsequence ending at the second index is 1.
# The length of the longest increasing subsequence ending at the third index is 1.
# The length of the longest increasing subsequence ending at the fourth index is 2.
# The length of the longest increasing subsequence ending at the fifth index is 2.
# The length of the longest increasing subsequence ending at the sixth index is 3.
# The length of the longest increasing subsequence ending at the seventh index is 4.
# The length of the longest increasing subsequence ending at the eighth index is 4.
# The solution is the maximum value in the list.
# The time complexity is O(n^2) and the space complexity is O(n).

def lengthOfLIS(nums):
if not nums:
return 0
dp =  * len(nums)
for i in range(1, len(nums)):
for j in range(i):
if nums[i] > nums[j]:
dp[i] = max(dp[i], dp[j] + 1)
return max(dp)

print(lengthOfLIS([10,9,2,5,3,7,101,18]))```

A.I. Evaluation of the Solution

The candidate's solution is correct and solves the problem. The approach is to use dynamic programming. The idea is to create a list of the same length as the input list. The list will contain the length of the longest increasing subsequence ending at the index of the input list. The list will be filled in a bottom-up manner. The base case is when the list is empty. The length of the longest increasing subsequence ending at the first index is 1. The length of the longest increasing subsequence ending at the second index is 1. The length of the longest increasing subsequence ending at the third index is 1. The length of the longest increasing subsequence ending at the fourth index is 2. The length of the longest increasing subsequence ending at the fifth index is 2. The length of the longest increasing subsequence ending at the sixth index is 3. The length of the longest increasing subsequence ending at the seventh index is 4. The length of the longest increasing subsequence ending at the eighth index is 4. The solution is the maximum value in the list. The time complexity is O(n^2) and the space complexity is O(n).

Evaluated at: 2022-11-19 12:16:07