Dynamic Programming Questions Medium
The time complexity of the Dynamic Programming approach for solving the Longest Increasing Subsequence Size problem is O(n^2), where n is the length of the input sequence.
In this approach, we use a dynamic programming table to store the lengths of the longest increasing subsequences ending at each index of the input sequence. We initialize the table with all values set to 1, as the minimum length of any subsequence is 1.
Then, we iterate through the input sequence from left to right, and for each index i, we compare the value at index i with all the previous indices j (0 <= j < i). If the value at index i is greater than the value at index j, we update the length of the longest increasing subsequence ending at index i as the maximum of its current length and the length of the longest increasing subsequence ending at index j plus 1.
Finally, we find the maximum value in the dynamic programming table, which represents the length of the longest increasing subsequence in the input sequence.
The time complexity of this approach is O(n^2) because we have nested loops. The outer loop iterates through the input sequence of length n, and the inner loop iterates from 0 to i, where i is the current index of the outer loop. Therefore, the total number of iterations is approximately n*(n+1)/2, which is in the order of n^2.
In some cases, the time complexity can be improved to O(n log n) using more efficient algorithms like the Patience Sorting algorithm or Binary Search based approaches.