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

class Solution:
    def rotate(self, matrix: List[List[int]]) -> None:
        """
        Do not return anything, modify matrix in-place instead.
        """
        m = len(matrix)
        n = len(matrix[0])
        row_b = 0 # 0-> 1
        row_e = m-1 # 2->1
        col_b = 0

liyunrui

class Solution:
    """
    [3,5,2,1,6,4]
     n = 6
     mid = 3
     [1,2,3,5,6,4]
        i     j
     i = mid-1 if n%2==0
     tmp_arr = [3,4,2,6,1,5]
     在把tmp_arr猜回array

liyunrui

class Solution:   
    """
    Divide and Conquer:
    我們可以把我們要求的問題拆成不重複的子問題
        1.左邊subarray的最大連續合
        2.右邊subarray的最大連續合
        3.包含當前元素的的最大連續和
    用一個最小的例子[-2, 1]去展示你的idea
    [-2,1,-3,4,-1,2,1,-5,4]
      0 1. 2 3. 4 5. 6 7 8

liyunrui

class Solution:
    """
    PC: there might be multiple valid answer
         <---i
    s = "babad"
          i
             j
            
    i從後面遍歷到前面, j從i遍歷到後面
         

liyunrui

class UnionFind:
    def __init__(self, n):
        self._parents = [node for node in range(n)]
        self._rank = [1 for _ in range(n)]
        
    def find(self, u):
        while u != self._parents[u]:
            self._parents[u] = self._parents[self._parents[u]]
            u = self._parents[u]
        return u 

liyunrui

class Solution:
    def findMin(self, nums: List[int]) -> int:
        def find_pivot_element(nums, l, r):
            mid = l + (r-l) // 2
            if r-l <= 1:
                return min(nums[l], nums[r])
            
            if nums[l] < nums[r]:
                return nums[l]
            left_min = find_pivot_element(nums, l, mid-1)

liyunrui

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    """
    Solution: using stack -> O(n) in space
    
        we use stack to simulate addition.

liyunrui

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    """
    Naive Solution(如果你不懂binary representation)
        step1: reverse the linked list
        step2: travser the linked list and progressivly build the res