text function lengthOfLIS(nums): n = length(nums) dp = array of size n filled with 1 for i from 1 to n - 1: for j from 0 to i - 1: if nums[j] < nums[i]: dp[i] = max(dp[i], dp[j] + 1) return max(dp) Time complexity: for this approach. with binary search. Space complexity: for the dp array.
##### ###### ##### ### # # ### # # ###### ## ## ## ## ## ## ## # # # # # ## ##### #### ##### # # # # # # # #### ## # ## ## ## ## # # # # # ## ## # ###### ## ### # ### # ######
$ curl repovive.com/roadmaps/maang-interview-prep/dynamic-programming/lis-pseudocode
░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░████████████████████████████████████████████████████████████████████████████████████