leetcode
Link: https://leetcode.com/problems/flip-equivalent-binary-trees/ Thought If 2 nodes are equivalent, either their left and right side match, or they are flipped match. So using a recursion(why recursion? It is a Tree!), and try to match each node pairs from root1 and root2. For each
leetcode
link: https://leetcode.com/problems/reveal-cards-in-increasing-order/ Thought So the rule is if you pop anything from front, then you shall move next head to tail. Reversely, if we want to build the original sequence, we could assume if you want to insert anything in head, you
leetcode
link: https://leetcode.com/problems/largest-time-for-given-digits/ Thought Its easy to think use DFS to solve it: it has limit data(4 digits), and DFS will enumerates all possible combinations Code class Solution: def largestTimeFromDigits(self, A): """ :type A: List[int] :rtype: str "
leetcode
Link: Most Stones Removed with Same Row or Column Thought When I saw the question, I realized its a Union Find question - For any union, I can keep only one stone form it. Then it convert to a count how many unique union finds exist question. So, only need to
leetcode
Link: validate stack sequences Thought Its pretty straight forward, just stimulate a stack and push/pop as the sequence shows. Code class Solution(object): def validateStackSequences(self, pushed, popped): """ :type pushed: List[int] :type popped: List[int] :rtype: bool """ stack = [] visited = set() N = len(
leetcode
Link: Bag of Tokens Thought For this question, I find we could do 2 things: spend any cost form the tokens, and get 1 point. Or, spend 1 point, get any power from the tokens. So it turns to be pretty clear that I will firstly sort all tokens by
leetcode
Link Minimum Increment to Make Array Unique Thought Firstly, I try to enumerate all existing number, and I got TLE :(. Then I realized I could sort the origin array and maintain a upper boundary, for any number, if it smaller than upper boundary, means I can only move it to
leetcode
link: Minimum Area Rectangle Thought I plan to use sweep line to solve. Firstly I put a dictionary with x-val as key, all y-val with coordinate x-val as value. Then, once I check x1, x2, if they has intersection, means there could be a rectangle. So I
leetcode
link: Range Sum of BST Thought When first saw this question, I think using tree traverse could solve it and it did works. Code # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def rangeSumBST(self,
leetcode
link Distinct Subsequences II Thought The very first thought here is use DFS to find all distinct subsequences. However it will time out. So my next idea is using dynamic programming to solve it. I'll define a DP array, where DP[i + 1] stands for how many distinct
leetcode
link: Number of Recent Calls Thought Question ask for numbers of calls in 3000. So sliding windows is my choice here - it will move and drop calls earlier than 3000 - which I don't care, and keep tracing incoming calls. Code class RecentCounter: def __init__(self): self.t = collections.
leetcode
link Knight Dialer Thought Since it ask how many distinct numbers could be generated, DP jumps to my mind immediately. Firstly, the relation is, if knight stop at number 6, it has to come from 1, 7, or 0. So I keep a DP array, DP[j][i] stands for