#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/dynamic-programming/longest-increasing-subsequence/lis-implementation

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░██████████████████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/dynamic-programming/longest-increasing-subsequence/lis-implementation

Repovive

LeetCode 300 Longest Increasing Subsequence - Implementation - Longest Increasing Subsequence | Dynamic Programming | Repovive

Repovive.

Dynamic Programming21 sections · 918 units60h total

Open in Course

LeetCode 300 Longest Increasing Subsequence - Implementation

The code

Here's the full solution:

function lengthOfLIS(nums)
    n := length of nums
    dp := array of size n, all set to 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)
    maxLen := 0
    for i from 0 to n - 1
        maxLen := max(maxLen, dp[i])
    return maxLen

Time: $O(n^2)$ . Space: $O(n)$ .

For each position $i$ , you check all earlier positions $j$ . If nums[j] < nums[i], you can extend that subsequence. You take the best extension.

This is the classic DP approach. It's correct but slow for large inputs. The binary search version runs in $O(n \log n)$ .