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花花酱 LeetCode 295. Find Median from Data Stream O(logn) + O(1)


Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.


[2,3,4] , the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

Design a data structure that supports the following two operations:

  • void addNum(int num) – Add a integer number from the data stream to the data structure.
  • double findMedian() – Return the median of all elements so far.

For example:



  1. Min/Max heap
  2. Balanced binary search tree

Time Complexity:

add(num): O(logn)

findMedian(): O(logn)



Solution 2:


Related Problems

花花酱 LeetCode 146. LRU Cache O(1)


Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and put.

get(key) – Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
put(key, value) – Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

Follow up:
Could you do both operations in O(1) time complexity?


Use Hashtable for fast mapping and double linked list for fast manipulation.
Time Complexity:
Put O(1)
Get O(1)


花花酱 LeetCode 676. Implement Magic Dictionary


Implement a magic directory with buildDict, and search methods.

For the method buildDict, you’ll be given a list of non-repetitive words to build a dictionary.

For the method search, you’ll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:


  1. You may assume that all the inputs are consist of lowercase letters a-z.
  2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
  3. Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.
Fuzzy match
Time Complexity:
buildDict: O(n*m)
n: numbers of words
m: length of word
search: O(m)
m: length of word
Space Complexity:

Java / Trie

Java / Trie v2


花花酱 LeetCode 300. Longest Increasing Subsequence


Given an unsorted array of integers, find the length of longest increasing subsequence.

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.

Your algorithm should run in O(n2) complexity.

Follow up: Could you improve it to O(n log n) time complexity?


Dynamic Programming

Time Complexity:


Solution 1:

Solution 2: DP + Binary Search / Patience Sort

dp[i] := smallest tailing number of a increasing subsequence of length i + 1.

dp is an increasing array, we can use binary search to find the index to insert/update the array.

ans = len(dp)

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



Related Problems:

花花酱 LeetCode 674. Longest Continuous Increasing Subsequence


Given an unsorted array of integers, find the length of longest continuous increasing subsequence.

Example 1:

Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. Even though [1,3,5,7] is also an increasing subsequence, it’s not a continuous one where 5 and 7 are separated by 4.

Example 2:

Explanation: The longest continuous increasing subsequence is [2], its length is 1.

Note: Length of the array will not exceed 10,000.

Dynamic Programming