8月刷题/Mock interview接龙活动（缺米刷题的来）

每日三题 day12。。
Longest Valid Parentheses - 想出了DP公式，但是好多边界条件处理出错。。有点像#22 Generate Parentheses，串联两个valid子串会出问题；后来看了答案改成stack以后简化很多，但还是漏了串联的处理
Group Anagrams - 找出字符串对应的feature然后group就行
Pascal's Triangle II - 最开始用二项式定理乘出来经常越界变成负数，改成long可破。但是Leetcode好像出了问题怎么都交不上，害得我又用递归重写了一遍

8/12 打卡 七道题

20200807:
new
3:
4: 0030, 0639
5: 0381

review
3: 0015, 0016, 0259, 0119, 0270, 0275, 0560, 0611, 0680, 0923, 1249
4: 0018, 0120
5: 0212

8/12 今日刷了3道DP，都是同一个类型匹配型DP
1. Longest common subsequence:  构建2D matrix，更新时考虑左，上和左上三个状态。
2. Longest Common Substring：注意substring和subsequence的区别，只需要考虑左上的状态，并返回全局最大。
3. Distinct Subsequences: 在2D状态数组中，只需要考虑左上和上的状态。

求加米

8月12日 打卡第七天

8/12/2020
76. Minimum Window Substring

567. Permutation in String

438. Find All Anagrams in a String

加油！！

8.12 打卡。结束binary search topic。 三道二分两道sql

#1064. Fixed Point
The tricky point is I need to find the smallest index. My first solution failed on this test case: [0,1,3,7,8,9]
(0, 5, 2)
(0, 2, 1)
I was returning mid in the while loop, but actually I should return in the end:
return l if A[l] == l else -1
math proof:

A[i] is distinct and ascending.
A[i] - i is non-descending array.
Binary search the first 0 in the array of A[i] - i.
A[i] < A[i + 1]
A[i] <= A[i + 1] - 1
A[i] - i <= A[i + 1] - i - 1
#1337. The K weakest rows in a matrix
binary search + priority queue

#1533. Find the index of the large integer
Tricky part is to understand that l, r and x,y should have equal number of elements to get valid results from reader.compareSub().

8.12

今日成果！