Longest Increasing Subsequence (Dynamic Programming)
def length_of_lis(nums): """Calculates the length of the longest strictly increasing subsequence. Args: nums: A list of integers. Returns: The length of the longest increasing subsequence. """ if not nums: return 0 n = len(nums) # dp[i] will store the length of the longest increa...
This function computes the length of the longest strictly increasing subsequence (LIS) within a given array of integers. An increasing subsequence is a sequence of elements from the original array that are in strictly in...
The `length_of_lis` function utilizes dynamic programming to solve the Longest Increasing Subsequence problem. It defines `dp[i]` as the length of the longest increasing subsequence ending at index `i`. The base case is that every element itself forms an increasing subsequence of length 1, so `dp` is initialized with all ones. The transition involves iterating through all previous elements `j` for each element `i`. If `nums[i]` is greater than `nums[j]`, it means `nums[i]` can extend the increasing subsequence ending at `j`. Therefore, `dp[i]` is updated to be the maximum of its current value and `dp[j] + 1`. The final answer is the maximum value in the `dp` array. The time complexity is O(n^2) due to the nested loops, and the space complexity is O(n) to store the `dp` array. Edge cases like an empty input array are handled by returning 0.
function length_of_lis(nums): if nums is empty: return 0 n = length of nums dp = array of size n, initialized with 1s for i from 0 to n-1: for j from 0 to i-1: if nums[i] > nums[j]: dp[i] = max(dp[i], dp[j] + 1) return m...