Longest Increasing Subsequence: A Dynamic Programming Problem

#### Problem Statement:

Given an integer array nums, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Example 1:
Input: nums = [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.

Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4

Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1

#### Recurrence Relation:

Let arr[0..n-1] be the input array and L(i) be the length of the LIS ending at index i such that arr[i] is the last element of the LIS.
``` L(i) = 1 + max( L(j) ) where 0 < j < i and arr[j] < arr[i]; or ```
``` L(i) = 1, if no such j exists. ```

To find the LIS for a given array, we need to return max(L(i)) where 0 < i < n.
The length of the longest increasing subsequence ending at index i, will be 1 greater than the maximum of lengths of all longest increasing subsequences ending at indices j such that j is before i, where arr[j] < arr[i].

#### Java Code:

``````
class Solution {
public int lengthOfLIS(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int len = nums.length;
int[] dp = new int[len];
Arrays.fill(dp, 1);
int max = 1;
for (int i = 1; i < len; i++) {
for (int j = 0; j < i; j++) {
if (nums[i] > nums[j]) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
max = Math.max(max, dp[i]);
}
return max;
}
}
```
```

#### Python Code:

``````
class Solution(object):

# :type nums: List[int]
# :rtype: int
def lengthOfLIS(self, nums):
if nums == None or len(nums) == 0:
return 0

length = len(nums)
dp = [1] * length
maximum = 1

for i in range(length):
for j in range(i):
if nums[i] > nums[j]:
dp[i] = max(dp[i], dp[j] + 1)
maximum = max(maximum, dp[i])

return maximum
```
```

#### Track Path:

Now we will concentrate on printing the longest increasing subsequence. This is how we achieve this: for every element in the array we keep track of the element that comes immediately before it in the longest increasing subsequence. We store this data in parent[] array. Initially all the elements are the parent of its own. The code below will make this process clearer.

#### Use Cases:

Most problems where ``` you are given an array (or list) of items and you'd have to find the largest subset of the items which maintains certain condition ``` could be solved using Longest Increasing Subsequence technique.

#### Template:

Take a look at the following problems and you would be able to have a strong grasp on how to effortlessly use the above template to solve a wide variety of problems of the pattern described in the Use Case section: