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

$ curl repovive.com/roadmaps/pattern-22/segment-tree/longest-increasing-subsequence-ii-implementation

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

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

$ curl repovive.com/roadmaps/pattern-22/segment-tree/longest-increasing-subsequence-ii-implementation

Repovive

LeetCode 2407 Longest Increasing Subsequence II - Implementation - Segment Tree | Pattern 22: LeetCode Interview Patterns | Repovive

Repovive.

Pattern 22: LeetCode Interview Patterns24 sections · 736 units32h total

Open in Course

LeetCode 2407 Longest Increasing Subsequence II - Implementation

The code

class SegmentTree:
    def query(l, r):  // max in range
    def update(i, val):  // point update

tree = SegmentTree(max(nums) + 1)

for num in nums:
    left = max(0, num - k)
    right = num - 1
    maxLen = tree.query(left, right) if left <= right else 0
    tree.update(num, maxLen + 1)

return tree.query(0, max(nums))

Time: $O(n \log m)$ where $m$ is max value. Space: $O(m)$ for tree.