LC-2407 Hard LeetCode

2407. Longest Increasing Subsequence II

Array Divide and Conquer Dynamic Programming Binary Indexed Tree Segment Tree Queue Monotonic Queue

Read the full problem statement on LeetCode.

Difficulty: hard Acceptance: 25% Topics: Array, Divide and Conquer, Dynamic Programming, Binary Indexed Tree, Segment Tree, Queue, Monotonic Queue

View full problem on LeetCode

Reading material

Arrays & Hashing 12 min Divide & Conquer 11 min 1-D Dynamic Programming 16 min

Reference solution (spoiler · python)

# Time:  O(nlogn)
# Space: O(n)

import bisect


# Range Maximum Query
class SegmentTree(object):
    def __init__(self, N,
                 build_fn=lambda _: 0,
                 query_fn=lambda x, y: y if x is None else x if y is None else max(x, y),
                 update_fn=lambda x: x):
        self.tree = [None]*(2*2**((N-1).bit_length()))
        self.base = len(self.tree)//2
        self.query_fn = query_fn
        self.update_fn = update_fn
        for i in xrange(self.base, self.base+N):
            self.tree[i] = build_fn(i-self.base)
        for i in reversed(xrange(1, self.base)):
            self.tree[i] = query_fn(self.tree[2*i], self.tree[2*i+1])

    def update(self, i, h):
        x = self.base+i
        self.tree[x] = self.update_fn(h)
        while x > 1:
            x //= 2
            self.tree[x] = self.query_fn(self.tree[x*2], self.tree[x*2+1])

    def query(self, L, R):
        if L > R:
            return 0
        L += self.base
        R += self.base
        left = right = None
        while L <= R:
            if L & 1:
                left = self.query_fn(left, self.tree[L])
                L += 1
            if R & 1 == 0:
                right = self.query_fn(self.tree[R], right)
                R -= 1
            L //= 2
            R //= 2
        return self.query_fn(left, right)


# segment tree with coordinate compression
class Solution(object):
    def lengthOfLIS(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        sorted_nums = sorted({x-1 for x in nums})
        num_to_idx = {x:i for i, x in enumerate(sorted_nums)}
        st = SegmentTree(len(num_to_idx))
        for x in nums:
            x -= 1
            st.update(num_to_idx[x], st.query(bisect.bisect_left(sorted_nums, x-k), num_to_idx[x]-1)+1)
        return st.tree[1]  # st.query(0, len(num_to_idx)-1)

Solution from kamyu104/LeetCode-Solutions · MIT