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Weekeew/Python-practise

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Python-practise

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1-20

Index Name source code Improved Note
1 Two Sum Ok
2 Add Two Numbers Yes move the pointer separately: while l1 != None or l2 != None: if l1 != None: if l2 != None:
3 Longest Substring Without Repeating Characters Yes
4 Median of Two Sorted Arrays Yes Two Cases:""" x is searching region case1: A[k/2] < B[k/2] ----(k/2)xxxx xxx(k/2)----- 
 case2: A[k/2] > B[k/2]
 xxxx(k/2)-----
 ---(k/2)xxxxxx
 """
 ----------> How come is log(n): Although the program kill 1/4 of the total length. But the base case is for each individual list, for each list is 1/2.</td>
5 Longest Palindromic Substring Bad Need improve
6 ZigZag Conversion Yes
7 Reverse Integer Yes
8 String to Integer (atoi) Yes
9 Palindrome Number Yes
10 Regular Expression Matching
11 Container With Most Water Yes
12 Integer to Roman Yes
13 Roman to Integer Yes
14 Longest Common Prefix Yes My approach, 1, compare two shorest string, 2, compare char by char
15 3Sum Yes 1, can use hashmap, too. 2, can choose set, 3, my approach, less space cost
16 3Sum Closest Yes Trick, if find target, return immediately
17 Letter Combinations of a Phone Number Yes IMPORTANT, Bracktracking method, need to redo it again
18 4 Sum Yes 2, 3, 4 sum summary Best approach is O(kn^2), k is the hashvalue average length 哈嘻表, LESSONS Learned, When using API like set.add(). be careful. It is very costly
19 Remove Nth Node From End of List Ok My apporoach: 1, change the node, delete the last one. 2, inorder traverse(n = total - n). Other approach: Remove directly or using list as extra storage. TRY recursive later
20 Valid Parentheses Yes

21-40

Index Name source code Improved Note
21 Merge Two Sorted Lists Yes Note, not best performance need to improve
22 Generate Parentheses Yes poor performance, I used iteration, Can try recursion, I also have it in CC150
23
24 Swap Nodes in Pairs Yes It is not a hard problem, but it is very easy to make mistake. if seperate the reverse pair into two case: 1, the init case and other cases. The inital case is simple and do not need to think about the connections between previous nodes
25
26 Remove Duplicates from Sorted Array Yes The question itself does not state clear, Two points: 1, return length 2, new list with no-duplicate elements and it doesn't matter what you leave beyond the new length.
27 Remove Element Yes Tips, be careful about changing and iterating the list at the same time
28 Implement strStr() Yes
29 Divide Two Integers Yes NNNEED to redo it. Some tricks. 1, two nested while loop, only decrease in the inner loop. 2, the difference between the <<= and <<: 1, << shifts things, but if you don not assign the result to anything, nothing will be record. ex: y = 8. y<<1 -> this gives you 16, but y will still be 8. But for <<=, y = 8. y <<=1 -> y becomes 16
30
31 Next Permutation Yes Detailed alg, 3 major steps, 1, find, 2, swapping, 3 sorting.Trick, do not return, you need an in-place sorting alg
32
33
34 Search for a Range Yes Tips, if I wanna 6, I search for both 5.5 and 6.5, then find both lower and upper boundaries for 6.
35 Search Insert Position Yes Easy, maybe have some other cool way to solve it
36 Valid Sudoku Yes
37
38 Count and Say Yes
39 Combination Sum Yes First, sorted. Then, dfs: index keep the same position for recursion
40 Combination Sum II Yes Performance matters, the recursive call need to prevent the duplicate cases. First, increasing index, Second, the consecutive duplicated integer should be ignored. tracking the current value. GOOD LUCK

41-60

Index Name source code Improved Notes
41
42
43 Multiply Strings Yes May try some other methods, the API is pretty simple. Do not see the point adding char by char
44
45
46 Permutations Yes Permutations: 1, bracktrack question using recursion to solve, faster. 2, the increment normally used in the input, not seperate. the function will auto becoming append and pop. Not hard question overall
47 Permutations II Yes Need to redo by iteration. Tricks, 1, similar to closet sum. avoid duplicate iteration. 2, using a list of bool to track the integer used or not.
48 Rotate Image Yes for i in( 0, len/2) , then 1, first = i, 2, last = len - i - 1, inner for j in (first, last). Finally, offset = last -i. Neet to draw the matrix. It is not hard, but easy to make a tiny mistake
49 Group Anagrams Yes NEED huge improve, the normal way is not efficient. maybe string compression
50 Pow(x, n) Yes The odd or even power is different. Also, the positive and negative power is different
51
52
53 Maximum Subarray Yes Poor perfornamce, the DP method will be better, Need to redo later
54 Spiral Matrix Yes Trick for 2d Matrix, 1, the matrix: [[]]: len(matrix) == 1 2, the matrix: []: len(matrix) == 0. So we need to check both of cases, before we do the pop operation
55 Jump Game Yes First, I disagree this question uses greedy alg, it is not taking the sub-optimal solution for every step. COOL question, track the max every iteration, step increase, max-1. O(n) method can be found
56
57
58 Length of Last Word Yes Nothing hard
59 Spiral Matrix II Yes The odd and even cases need to be careful, the logic is similar to array rotating 90 degree
60 Permutation Sequence Yes EX: [1, 2, 3, 4], we will have 6! permutation first digit is "1". The next iteration: factorial number: f2 = 6!/i, i = 3. => f2 = 2!

61-80

Index Name source code Improved Notes
61 Rotate List Yes The rotation k can be larger than len(list). First, k = k % len(list)
62 Unique Paths Yes Dp program, using the bottom-up approach to save the result for the subproblems. However, the trick is the initialize the matrix at the beginning
63 Unique Paths II Yes Similar method as Question 62. Set 1 to None
64 Minimum Path Sum Yes Need an improvement
65
66 Plus One Ok May be need some other approach, nothing hard
67 Add Binary Ok Need an improvement
68
69 Sqrt(x) Yes
70 Climbing Stairs Yes
71 Simplify Path Yes
72
73 Set Matrix Zeroes Yes
74 Search a 2D Matrix Yes
75 Sort Colors Yes
76
77 Combinations Yes
78 Subsets Yes
79 Word Search Yes
80 Remove Duplicates from Sorted Array II Yes

80-100

Index Name source code Improved Notes
81 Search in Rotated Sorted Array II Yes
82 Remove Duplicates from Sorted List II Yes Trick, dummyhead, return dummyhead.next
83 Remove Duplicates from Sorted List Yes
84
85
86 Partition List Yes
87
88 Merge Sorted Array Yes
89 Gray Code Yes
90 Subsets II Yes
91 Decode Ways Yes
92 Reverse Linked List II Yes
93 Restore IP Addresses Yes
94 Binary Tree Inorder Traversal Yes Iteration also have space complexity is O(1)
95 Unique Binary Search Trees II Yes
96 Unique Binary Search Trees Yes
97
98 Validate Binary Search Tree Yes
99
100 Same Tree Yes

101-120

Index Name source code Improved Notes
101 Symmetric Tree
102 Binary Tree Level Order Traversal
103 Binary Tree Zigzag Level Order Traversal
104 Maximum Depth of Binary Tree
105 Construct Binary Tree from Preorder and Inorder Traversal Preorder to define the root position, inorder to divide right, left subtree
106 Construct Binary Tree from Inorder and Postorder Traversal postorder to define root position, inorder to divide right, left subtree
107 Binary Tree Level Order Traversal II
108 Convert Sorted Array to Binary Search Tree similar to the binary search, find mid and insert into the tree
109 Convert Sorted List to Binary Search Tree inorder walk for building the tree, build the root for each level
110 Balanced Binary Tree initial check for the empty tree
111 Minimum Depth of Binary Tree
112 Path Sum
113 Path Sum II
114 Flatten Binary Tree to Linked List
115
116 Populating Next Right Pointers in Each Node
117 Populating Next Right Pointers in Each Node II
118 Pascal's Triangle
119 Pascal's Triangle II
120 Triangle

121-140

Index Name source code Improved Notes
121 Best Time to Buy and Sell Stock
122 Best Time to Buy and Sell Stock II
123 Best Time to Buy and Sell Stock III
124 Binary Tree Maximum Path Sum
125 Valid Palindrome
126
127 Word Ladder BFS: two diff method, buffer swapping or size counting(better)
128
129 Sum Root to Leaf Numbers
130
131 Palindrome Partitioning
132
133 Clone Graph Hashmap track and store, continue to skip the recursion. can also use the interation
134 Gas Station Pick starting point, if the temporary sum is smaller than 0, it means that previous steps cannot be the starting points
135
136 Single Number Xor method
137
138
139 Word Break
140 Word Break II Using the hashmap to track the previous result

141-160

Index Name source code Improved Notes
141 Linked List Cycle
142 Linked List Cycle II s+k == loop size
143 Reorder List Break by the middle point, the second list is reverse order
144
145
146
147 Insertion Sort List
148 Sort List O(nlog(n)) - > merge sort. First partition(find mid), then merge two sorted list
149 Max Points on a Line
150
151
152
153
154
155 Min Stack
156
157
158
159
160 Intersection of Two Linked Lists Two different approach. Shortcut, if the order is sorted, then the list can keep move on to the next node

221-240

Index Name source code Improved Notes
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236 Lowest Common Ancestor of a Binary Tree 1, Three cases: // 在root为根的二叉树中找A,B的LCA: // 如果找到了就返回这个LCA // 如果只碰到A,就返回A // 如果只碰到B,就返回B // 如果都没有,就返回null
237
238
239
240

61-80

Index Name source code Improved Notes
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
Index Name source code Improved
36 Valid Sudoku Yes
37 Count and Say Ok
146 LRU Cache Ok, need other approach
202 Happy Number Yes

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