Added more hints and tags.

Author: aliasvishnu

README.md

@@ -1,39 +1,60 @@
 # Leetcode
 
-Hints + solutions for ~250 leetcode problems. Mostly medium, some hard. Currently adding more hints as I revisit my solutions. Please help me add more hints if you can.
+Hints for ~250 leetcode problems. Mostly medium, some hard.
+Good practice list for internships/job hunt.
 
-I commit a lot, many a times unnecessarily in this repo, it could be better solution or some optimization, or alternate solution, or variable renaming. I apologize for that as well.
+Solutions <a href = "https://github.com/aliasvishnu/leetcode/tree/master/Solutions" > here </a>
 
-## First hint list
+## Abbreviations
+* BF - Brute Force,
+* DP - Dynamic Programming
+* Array - Problem specific array logic
+* D&C - Divide & Conquer
 
+## First hint list
 
-| # | Hints | Complexity |
-|---| ----- | ---------- |
-| 1 | Use hash map to store all values until current value. | O(n) |
-| 2 | Brute force. | O(n) |
-| 3 | Use a window. Extend if next element is not in the current window. | O(n) |
-| 5 | Build recurrence relation (may help to solve longest common substring problem) | O(n^2) |
-| 7 | Brute force. | O(log(n)) |
-| 9 | Brute force. | O(log(n)) |
-| 10 | Use a 2D matrix. DP[i][j] = whether p[0:i] matches with str[0:j]. | O(mn) |
-| 15 | Solve 2 sum before attempting this problem. | O(n^2) |
-| 16 | Similar to 3 sum, but use a sorted array and make front++, back-- updates to minimize abs(target - sum). | O(n^2) |
-| 17 | Possible approaches are trie, or more easy - building on smaller solution. | O(4^n)|
-| 22 | If less than n open brackets are open, we can add open bracket. But to add a close bracket you need count(')') < count('(') | . |
-| 23 | Use a data structure which can give you minimum element at low cost.| . |
-| 26 | Maintain length of unique elements till current element.| O(n)|
-| 27 | vector.erase() will not work, as vector overwrites the element every time erase is called.| O(n)|
-| 28 | Brute force. | O(mn) |
-| 31 | Find the location closest from the end which has segment ab of the form a < b, and make it ba. | O(n log(n))|
-| 33 | Array increasing, sudden drop and then increasing again. Find out where the middle element is, can it's location help you? | O(log(n))|
-| 36 | Brute force.| O(1)|
-| 42 | Find next greater element index for each index, and prefix sum for each element. Can you build the solution from this? Check out problem 496 for next greater element. | |
-| 43 | Normal multiplication, like in school. | O(n log(d)) |
-| 44 | Solve problem 10 first. Construct a 2D box of size len(pattern)*len(string). DP[i][j] = whether p[0:i] matches with str[0:j] | O(mn) |
-| 46 | Combine str[0] with permutations of str[1:], use recursion. | O(n!) |
-| 47 |
-| 108 | Where does the root element fall during an inorder (sorted) traversal? | |
-| 128 | DP problem, find recurrence relation. | |
+| # | Tags | Hints | Complexity |
+|---| ---- | ----- | ---------- |
+| 1 | Hash | Use hash map to store all values until current value. | O(n) |
+| 2 | BF | Brute force. | O(n) |
+| 3 | Array | Use a window. Extend if next element is not in the current window. | O(n) |
+| 5 | DP | Build recurrence relation (may help to solve longest common substring problem) | O(n^2) |
+| 7 | BF | Brute force. | O(log(n)) |
+| 9 | BF | Brute force. | O(log(n)) |
+| 10 | DP | Use a 2D matrix. DP[i][j] = whether p[0:i] matches with str[0:j]. | O(mn) |
+| 15 | Array | Solve 2 sum before attempting this problem. | O(n^2) |
+| 16 | Array | Similar to 3 sum, but use a sorted array and make front++, back-- updates to minimize abs(target - sum). | O(n^2) |
+| 17 | Recursion | Possible approaches are trie, or more easy - building on smaller solution. | O(4^n)|
+| 22 | Recursion | If less than n open brackets are open, we can add open bracket. But to add a close bracket you need count(')') < count('(') | . |
+| 23 | Heap | Use a data structure which can give you minimum element at low cost.| . |
+| 26 | Array | Maintain length of unique elements till current element.| O(n)|
+| 27 | Array | vector.erase() will not work, as vector overwrites the element every time erase is called.| O(n)|
+| 28 | BF | Brute force. | O(mn) |
+| 31 | Array, Sorting | Find the location closest from the end which has segment ab of the form a < b, and make it ba. | O(n log(n))|
+| 33 | Binary Search | Array increasing, sudden drop and then increasing again. Find out where the middle element is, can it's location help you? | O(log(n))|
+| 36 | BF | Brute force.| O(1)|
+| 42 | Stack + Array | Find next greater element index for each index, and prefix sum for each element. Can you build the solution from this? Check out problem 496 for next greater element. | O(n) |
+| 43 | BF | Normal multiplication, like in school. | O(n log(d)) |
+| 44 | DP | Solve problem 10 first. Construct a 2D box of size len(pattern)*len(string). DP[i][j] = whether p[0:i] matches with str[0:j] | O(mn) |
+| 46 | Recursion | Combine str[0] with permutations of str[1:], use recursion. | O(n!) |
+| 47 | Recursion | Eg, aaaabbcc, only one of each kind {a, b, c} can come at 0th index. Recurse. | O(n!) |
+| 48 | Array | index <i, j> is i positions from left, j positions from top. Find a mapping to 3 other locations, and swap the values. | O(n^2) |
+| 49 | Hash | Hashing | O(n) if O(1) hash |
+| 50 | D&C/Recursion | 2^n = 2^(n/2)*2^(n/2) | O(log(n)) |
+| 51 | Backtracking + Pruning | If a queen is in <i, j>, block that diagonal, row & column.  | O(n!) |
+| 52 | Backtracking | Solution for 51 is more exhaustive than this problem. | O(n!) |
+| 53 | Array | Keep extending window <i, j> to the right until sum > 0, if sum < 0, set i = j+1 | O(n) |
+| 54 | Array/BF | Brute Force | O(n^2)|
+| 56 | Sorting | Sort based on start time. | O(n log(n)) |
+| 57 | Sorting | Insert interval and do problem 56. | O(n log(n)) |
+| 66 | BF | Brute force | O(n) |
+| 67 | BF | Simple array addition | O(log(n)) |
+| 68 | Array | If you can fit in k words in a line, there must by atleast k-1 spaces, so L-(k-1)-sum(char's in words) must be divided mentioned in the question. | O(n) |
+| 69 | Binary Search | start = 0, end = x/2, binary search | O(log(n)) |
+| 70 | DP | DP[i] = DP[i-1] + DP[i-2] | O(n)|
+| 71 | BF | Split by '/', use stack to handle '..' | O(n) |
+| 108 | Inorder traversal | Where does the root element fall during an inorder (sorted) traversal? | |
+| 128 | DP | DP problem, find recurrence relation. | |
 
 ## Second hint list