NigelJu

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def largestValues(self, root: TreeNode) -> List[int]:
      

NigelJu

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        oddDummy = ListNode(0)
        oddNode = oddDummy

NigelJu

class Solution:
    def minRemoveToMakeValid(self, s: str) -> str:
        validParenthese = 0
        leftCount = 0
        
        for val in s:
            if val == "(":
                leftCount += 1
            elif val == ")" and leftCount > 0:
                leftCount -= 1

NigelJu

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
                        
        head = ListNode(0)

NigelJu

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
                
                

NigelJu

class Solution:
    def twoCitySchedCost(self, costs: List[List[int]]) -> int:
        res = []
        bSum = 0
        for cost in costs:
            res.append(cost[1] - cost[0])
            bSum += cost[1]
        
        res.sort()
#         print(res)

NigelJu

class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
            return min(text.count("b"), text.count("a"), 
                       text.count("l") //2, text.count("o") // 2, text.count("n"))

NigelJu

class Solution:
    def maxNumberOfBalloons(self, text: str) -> int:
            bCount = 0
            aCount = 0
            lCount = 0
            oCount = 0
            nCount = 0
            
            for ch in text:
                if ch == "b":

NigelJu

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t): return False
        
        for ch in s:
            if s.count(ch) != t.count(ch):
                return False
            
        return True

NigelJu

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        
        sList = list(set(list(s)))
        tList = list(set(list(t)))
        
        if len(sList) != len(tList): return False
        
        for ch in sList:
            # print(sList.count(ch), tList.count(ch))