How can Dynamic Programming be used to solve the Longest Increasing Subsequence in an Array problem?

Dynamic Programming can be used to solve the Longest Increasing Subsequence (LIS) problem by breaking it down into smaller subproblems and using the solutions of these subproblems to build the solution for the larger problem.

The basic idea behind using Dynamic Programming for the LIS problem is to maintain an array, let's call it "dp", where dp[i] represents the length of the longest increasing subsequence ending at index i of the given array.

To find the LIS, we iterate through each element of the array and for each element, we compare it with all the previous elements. If the current element is greater than the previous element, we update dp[i] as dp[j] + 1, where j is the index of the previous element. We then update the maximum length of the LIS found so far.

Here is the step-by-step approach to solve the LIS problem using Dynamic Programming:

1. Initialize an array dp of the same length as the given array, with all elements set to 1. This is because the minimum length of any subsequence is 1 (the element itself).

2. Iterate through each element of the given array from left to right.

3. For each element, compare it with all the previous elements (from 0 to i-1).

4. If the current element is greater than the previous element, update dp[i] as dp[j] + 1, where j is the index of the previous element.

5. Update the maximum length of the LIS found so far.

6. Finally, return the maximum length of the LIS.

The time complexity of this approach is O(n^2), where n is the length of the given array. This is because we have nested loops to compare each element with all the previous elements.

However, there is also an optimized approach using Binary Search that reduces the time complexity to O(n log n). This approach involves maintaining a separate array to store the LIS elements and using Binary Search to find the correct position to insert the current element in the LIS array.

Overall, Dynamic Programming provides an efficient solution to the Longest Increasing Subsequence problem by breaking it down into smaller subproblems and using the solutions of these subproblems to build the solution for the larger problem.