liyunrui

class Solution:
    """
    Brute Force: 已當前高度為邊界,去找出所有可能的rectangle
        we traverse the array.
        For each cur_h, we enumerate all valid rectangle and calculater area
        by w * min_h in the subarray.
        T:O(n^2)
        S:O(1)
    
    Can we optimise?

liyunrui

class MedianFinder:
    """
    Binary Search + Insert
    T: O(logn) +O(n) = O(n) for add function O(1) for findMedian.
    S: O(n)
    我們用一個array不斷存放新進來的data, as time goes, it's become inefficient in terms of space(not scalable)
    
    Two heaps: max heap+min_heap+balance
    T: 3*O(logn) for add function O(1) for findMedian.
    S: O(n)

liyunrui

class KthLargest:
    """
    Binary Search+ Insert
    [2,4,5,8]
    k=3
    add 3 -> [2,3,4,5,8]-->return 4
                  -
    add 4 -> [2,3,4,5,5,8]-->return 5
                    -
    [2,3,4,5,5,8,10]

liyunrui

class Solution:
    def balancedStringSplit(self, s: str) -> int:
        balance = 0
        n = len(s)
        ans = 0
        for i in range(n):
            if s[i] == "R":
                balance+=1
            else:
                balance-=1

liyunrui

class Solution:
    """
    what's distribution of input string s:
        ---> s consists of English letters and digits. At most, 26+10 numbers
        
    Heap solution and Sorting takes O(nlogn)
    
    Bucket sort 
    
    freq array=[0,1,2,3,..,max_freq]

liyunrui

class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        return self.bucketSort(nums)
    def bucketSort(self, A):
        """
        1. It's in-place sorting. 
        2. Bucket sort is mainly useful when input is uniformly distributed over a range.(floating number is allowed)
        
        step1: create n buckets
        step2: put element into buckets

liyunrui

class Solution:
    """
    {
    i:2
    love:2
    leetcode:1
    codong:1
    }
    怎麼處理k=1
    heqp可能會丟掉i而不是丟掉love

liyunrui

class Solution:
    """
    Thought Process:
        Sorting
        PQ:
        if k == n:
            we return min elememnt 
        So we use a pq with size k to maintain a top k elements out of n.
        after looping over all elements in the given array. Return the topmost one, the smallest one inside the pq with size k, which is kth largest elememnt
    

liyunrui

"""
Given a bst
           5
          / \
         3   7
        / \  /
       1  4 6
pre-order: 5->  3    ->     7   ==> 5->3->1->4->7->6
              /  \         /
              1   4       6

liyunrui

class Solution:
    def sortArray(self, nums: List[int]) -> List[int]:
        return self.insertionSort(nums)

    def insertionSort(self, A):
        """
        思路：
            i這個指針的元素代表我們需要插入的元素, 經過一輪處理後我們希望i的左邊是partial sorted的.
            所以當i走到最後, 整個array就sort完.
            為了做到這部, 每次都create 一個指針j從i開始遞減到index 0.每一次遞減都跟上一個元素比如果比較大就交換.