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Best way to implement friends list into a database? MySQL

So my project has a "friends list" and in the MySQL database I have created a table:
nameA
nameB
Primary Key (nameA, nameB)
This will lead to a lot of entries, but to ensure that my database is normalised I'm not sure how else to achieve this?
My project also uses Redis.. I could store them there.
When a person joins the server, I would then have to search for all of the entries to see if their name is nameA or nameB, and then put those two names together as friends, this may also be inefficient.
Cheers.

The task is quite common. You want to store pairs where A|B has the same meaning as B|A. As a table has columns, one of the two will be stored in the first column and the other in the second, but who to store first and who second and why?
One solution is to always store the lesser ID first and the greater ID second:
userid1 | userid2
--------+--------
1 | 2
2 | 5
2 | 6
4 | 5
This has the advantage that you store each pair only once, as feels natural, but has the disadvantage that you must look up a person in both coumns and find their friend sometimes in the first and sometimes in the second column. That may make queries kind of clumsy.
Another method is to store the pairs redundantly (by using a trigger typically):
userid1 | userid2
--------+--------
1 | 2
2 | 1
2 | 5
2 | 6
4 | 5
5 | 2
5 | 4
6 | 2
Here querying is easier: Look the person up in one column and find their friends in the other. However, it looks kind of weird to have all pairs duplicated. And you rely on a trigger, which some people don't like.
A third method is to store numbered friendships:
friendship | user_id
-----------+--------
1 | 1
1 | 2
2 | 2
2 | 5
3 | 2
3 | 6
4 | 4
4 | 5
This gives both users in the pair equal value. But in order to find friends, you need two passes: find the friendships for a user, find the friends in these friendships. However, the design is very clear and even extensible, i.e. you could have friendships of three four or more users.
No method is really much better than the other.

How to print all possible sequences with intermixed sequence of push and pop operation

Suppose that an intermixed sequence of push and pop operations are performed on a LIFO stack.How to print all possible sequences? I can just judge it's about recursion. For example, if order 1 2 3 is given, output is
1 2 3
1 3 2
2 1 3
2 3 1
3 2 1

Use Google GUAVA's method https://google.github.io/guava/releases/19.0/api/docs/com/google/common/collect/Collections2.html#orderedPermutations(java.lang.Iterable) to get all possible permutations and then for each permutation reverse the order using https://docs.oracle.com/javase/7/docs/api/java/util/Collections.html#reverse(java.util.List)

Interpret sentence and convert into their corresponding format

There are some formats of inputs and their corresponding outputs
1. 7 years 10 months ---> YRS:7 MNHS:10
2. 7 kgs 10 grms ---> KGS:7 GRMS:10
3. 7 kilograms 10 grams ---> KGS:7 GRMS:10
4. 7 thousand 9 hundread ---> 7900
5. seven years ten months --> YRS:7 MNHS:10
6. seven kgs ten grms ---> KGS:7 GRMS:10
7. triple seven double five --> 77755
I wrote separate modules for all by storing informations in **HashMap. And it is working fine.**
Then I need to write one main module in which input is one sentence(utterance), and I need to replace all above substrings into corresponding substring output.
For example,
Input :- Dial number triple eight triple four three nine eight.
Output :- Dial number 888444398.
and many such utterances.
My doubts :-
I used numbers of HashMap for smaller modules to store meaning of keys, just like - triple means 3 times, double means 2 times and all. But this has limitation that if I need to add anything I have to add that entry in HashMap. Suggest some good technique for this.
I am confused in main module, how to extract useful substring given in above examples from given utterances. So suggest some good technique for this also.
Project Lanuguage : Java.

You should look at Illinos Quantifier package:
http://cogcomp.cs.illinois.edu/page/software_view/Quantifier
http://cogcomp.cs.illinois.edu/demo/quantities/results.php

You might want to use some kind of formal grammar parser. Just doing design of a grammar can clear a view of the problem. In the most simple case your grammar could look like:
STRING -> "" | STRING MEASUREMENT | STRING NUMBER | STRING WORD
MEASUREMENT -> NUMBER UNITS
UNITS -> kgs | grms | years | months | ...
NUMBER -> THOUSAND HUNDRED NUMBER_BELOW_HUNDRED | THOUSAND HUNDRED
THOUSAND -> "" | NUMBER_BELOW_HUNDRED thousand
HUNDRED -> "" | NUMBER_BELOW_HUNDRED hundred
NUMBER_BELOW_HUNDRED -> one | two | three | ... | ninety nine | 99 | 98 | ... | 1
WORD -> /* all other */
You can write a parser by yourself (in this case it seems to be pretty easy) or use a ready solution like Bison/Flex.
The usual alternative for your HashMaps are configuration files.

Searching words in array [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How to find list of possible words from a letter matrix [Boggle Solver]
I have a String[][] array such as
h,b,c,d
e,e,g,h
i,l,k,l
m,l,o,p
I need to match an ArrayList against this array to find the words specified in the ArrayList. When searching for word hello, I need to get a positive match and the locations of the letters, for example in this case (0,0), (1,1), (2,1), (3,1) and (3,2).
When going letter by letter and we suppose we are successfully located the first l letter, the program should try to find the next letter (l) in the places next to it. So it should match against e, e, g, k, o, l, m and i meaning all the letters around it: horizontally, vertically and diagonally. The same position can't be found in the word twice, so (0,0), (1,1), (2,1), (2,1) and (3,2) wouldn't be acceptable because the position (2,1) is matched twice. In this case, both will match the word because diagonally location is allowed but it needs to match the another l due to the requirement that a position can not be used more than once.
This case should also be matched
h,b,c,d
e,e,g,h
l,l,k,l
m,o,f,p
If we suppose that we try to search for helllo, it won't match. Either (x1, y1) (x1, y1) or (x1, y1) (x2, y2) (x1, y1) can't be matched.
What I want to know what is the best way to implement this kind of feature. If I have 4x4 String[][] array and 100 000 words in an ArrayList, what is the most efficient and the easiest way to do this?

I think you will probably spend most of your time trying to match words that can't possibly be built by your grid. So, the first thing I would do is try to speed up that step and that should get you most of the way there.
I would re-express the grid as a table of possible moves that you index by the letter. Start by assigning each letter a number (usually A=0, B=1, C=2, ... and so forth). For your example, let's just use the alphabet of the letters you have (in the second grid where the last row reads " m o f p "):
b | c | d | e | f | g | h | k | l | m | o | p
---+---+---+---+---+---+---+---+---+---+----+----
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11
Then you make a 2D boolean array that tells whether you have a certain letter transition available:
| 0 1 2 3 4 5 6 7 8 9 10 11 <- from letter
| b c d e f g h k l m o p
-----+--------------------------------------
0 b | T T T T
1 c | T T T T T
2 d | T T T
3 e | T T T T T T T
4 f | T T T T
5 g | T T T T T T T
6 h | T T T T T T T
7 k | T T T T T T T
8 l | T T T T T T T T T
9 m | T T
10 o | T T T T
11 p | T T T
^
to letter
Now go through your word list and convert the words to transitions (you can pre-compute this):
hello (6, 3, 8, 8, 10):
6 -> 3, 3 -> 8, 8 -> 8, 8 -> 10
Then check if these transitions are allowed by looking them up in your table:
[6][3] : T
[3][8] : T
[8][8] : T
[8][10] : T
If they are all allowed, there's a chance that this word might be found.
For example the word "helmet" can be ruled out on the 4th transition (m to e: helMEt), since that entry in your table is false.
And the word hamster can be ruled out, since the first (h to a) transition is not allowed (doesn't even exist in your table).
Now, for the remaining words that you didn't eliminate, try to actually find them in the grid the way you're doing it now or as suggested in some of the other answers here. This is to avoid false positives that result from jumps between identical letters in your grid. For example the word "help" is allowed by the table, but not by the grid
Let me know when your boggle phone-app is done! ;)

Although I am sure there is a beatiful and efficient answer for this question academically, you can use the same approach, but with a list possibilities. so, for the word 'hello', when you find the letter 'h', next you will add possible 'e' letters and so on. Every possibility will form a path of letters.

I would start by thinking of your grid as a graph, where each grid position is a node and each node connect to its eight neighbors (you shouldn't need to explicitly code it as a graph in code, however). Once you find the potential starting letters, all you need to do is to do a depth first search of the graph from each start position. The key is to remember where you've already searched, so you don't end up making more work for yourself (or worse, get stuck in a cycle).
Depending on the size of character space being used, you might also be able to benefit from building lookup tables. Let's assume English (26 contiguous character codepoints); if you start by building a 26-element List<Point>[] array, you can populate that array once from your grid, and then can quickly get a list of locations to start your search for any word. For example, to get the locations of h I would write arr['h'-'a']
You can even leverage this further if you apply the same strategy and build lookup tables for each edge list in the graph. Instead of having to search all 8 edges for each node, you already know which edges to search (if any).
(One note - if your character space is non-contiguous, you can still do a lookup table, but you'll need to use a HashMap<Character,List<Point>> and map.get('h') instead.)

One approach to investigate is to generate all the possible sequences of letters (strings) from the grid, then check if each word exists in this set of strings, rather than checking each word against the grid. E.g. starting at h in your first grid:
h
hb
he
he // duplicate, but different path
hbc
hbg
hbe
hbe // ditto
heb
hec
heg
...
This is only likely to be faster for very large lists of words because of the overhead of generating the sequences. For small lists of words it would be much faster to test them individually against the grid.
You would either need to store the entire path (including coordinates) or have a separate step to work out the path for the words that match. Which is faster will depend on the hit rate (i.e. what proportion of input words you actually find in the grid).
Depending on what you need to achieve, you could perhaps compare the sequences against a list of dictionary words to eliminate the non-words before beginning the matching.
Update 2 in the linked question there are several working, fast solutions that generate sequences from the grid, deepening recursively to generate longer sequences. However, they test these against a Trie generated from the words list, which enables them to abandon a subtree of sequences early - this prunes the search and improves efficiency greatly. This has a similar effect to the transition filtering suggested by Markus.

Sorting an array while moving duplicates to the end?

This was a question in one my friend's programming class.
Q. How do you sort an array of ints and then arrange them such that all duplicate elements appear at the end of the array?
For example, given the input
{5, 2, 7, 6, 1, 1, 5, 6, 2}
The output would be
{1, 2, 5, 6, 7, 1, 2, 5, 6}
Note that the numbers are sorted and duplicate numbers are after 7, which is the maximum in the array.
This has to be achieved with out using any Java library packages/utils.
I suggested to sort the array first using insertion or bubble sort, and then go over the array, perform something like the following :
for (int i = 0; i < nums.length - 2; i++) {
for (int j = i + 1; j < nums.length; j++) {
//current and next are same, move elements up
//and place the next number at the end.
if (nums[i] == nums[j]) {
int temp = nums[j];
for (int k = j; k < nums.length - 1; k++) {
nums[k] = nums[k + 1];
}
nums[nums.length - 1] = temp;
break;
}
}
}
I tried this myself later (and that is how the code above) - As I try this out, I think this could be achieved by using less code, be more efficiently. And may be I gave a wrong advice.
Any thoughts?

Depending on the parameters of your problem, there are many approaches to solving this.
If you are not allowed to use O(n) external memory, then one option would be to use a standard sorting algorithm to sort the array in-place in O(n log n) time, then to run a second pass over it to move the duplicates to the end (as you've suggested). The code you posted above takes O(n2) time, but I think that this step can be done in O(n log n) time using a slightly more complicated algorithm. The idea works in two steps. In the first step, in O(n log n) time you bring all non-duplicated elements to the front in sorted order and bring all the duplicates to the back in non-sorted order. Once you've done that, you then sort the back half of the array in O(n log n) time using the sorting algorithm from the first step.
I'm not going to go into the code to sort the array. I really love sorting, but there are so many other good resources on how to sort arrays in-place that it's not a good use of my time/space here to go into them. If it helps, here's links to Java implementations of heapsort, quicksort, and smoothsort, all of which runs in O(n log n) time. Heapsort and smoothsort use only O(1) external memory, while quicksort can use O(n) in the worst case (though good implementations can limit this to O(log n) using cute tricks).
The interesting code is the logic to bring all the non-duplicated elements to the front of the range. Intuitively, the code works by storing two pointers - a read pointer and a write pointer. The read pointer points to the next element to read, while the write pointer points to the location where the next unique element should be placed. For example, given this array:
1 1 1 1 2 2 3 4 5 5
We start with the read and write pointers initially pointing at 1:
write v
1 1 1 1 2 2 3 4 5 5
read ^
Next, we skip the read pointer ahead to the next element that isn't 1. This finds 2:
write v
1 1 1 1 2 2 3 4 5 5
read ^
Then, we bump the write pointer to the next location:
write v
1 1 1 1 2 2 3 4 5 5
read ^
Now, we swap the 2 into the spot held by the write pointer:
write v
1 2 1 1 1 2 3 4 5 5
read ^
advance the read pointer to the next value that isn't 2:
write v
1 2 1 1 1 2 3 4 5 5
read ^
then advance the write pointer:
write v
1 2 1 1 1 2 3 4 5 5
read ^
Again, we exchange the values pointed at by 'read' and 'write' and move the write pointer forward, then move the read pointer to the next unique value:
write v
1 2 3 1 1 2 1 4 5 5
read ^
Once more yields
write v
1 2 3 4 1 2 1 1 5 5
read ^
and the final iteration gives
write v
1 2 3 4 5 2 1 1 1 5
read ^
If we now sort from the write pointer to the read pointer, we get
write v
1 2 3 4 5 1 1 1 2 5
read ^
and bingo! We've got the answer we're looking for.
In (untested, sorry...) Java code, this fixup step might look like this:
int read = 0;
int write = 0;
while (read < array.length) {
/* Swap the values pointed at by read and write. */
int temp = array[write];
array[write] = array[read];
array[read] = temp;
/* Advance the read pointer forward to the next unique value. Since we
* moved the unique value to the write location, we compare values
* against array[write] instead of array[read].
*/
while (read < array.length && array[write] == array[read])
++ read;
/* Advance the write pointer. */
++ write;
}
This algorithm runs in O(n) time, which leads to an overall O(n log n) algorithm for the problem. Since the reordering step uses O(1) memory, the overall memory usage would be either O(1) (for something like smoothsort or heapsort) or O(log n) (for something like quicksort).
EDIT: After talking this over with a friend, I think that there is a much more elegant solution to the problem based on a modification of quicksort. Typically, when you run quicksort, you end up partitioning the array into three regions:
+----------------+----------------+----------------+
| values < pivot | values = pivot | values > pivot |
+----------------+----------------+----------------+
The recursion then sorts the first and last regions to put them into sorted order. However, we can modify this for our version of the problem. We'll need as a primitive the rotation algorithm, which takes two adjacent blocks of values in an array and exchanges them in O(n) time. It does not change the relative order of the elements in those blocks. For example, we could use rotation to convert the array
1 2 3 4 5 6 7 8
into
3 4 5 6 7 8 1 2
and can do so in O(n) time.
The modified version of quicksort would work by using the Bentley-McIlroy three-way partition algortihm (described here) to, using O(1) extra space, rearrange the array elements into the configuration shown above. Next, we apply a rotation to reorder the elements so that they look like this:
+----------------+----------------+----------------+
| values < pivot | values > pivot | values = pivot |
+----------------+----------------+----------------+
Next, we perform a swap so that we move exactly one copy of the pivot element into the set of elements at least as large as the pivot. This may have extra copies of the pivot behind. We then recursively apply the sorting algorithm to the < and > ranges. When we do this, the resulting array will look like this:
+---------+-------------+---------+-------------+---------+
| < pivot | dup < pivot | > pivot | dup > pivot | = pivot |
+---------+-------------+---------+-------------+---------+
We then apply two rotations to the range to put it into the final order. First, rotate the duplicate values less than the pivot with the values greater than the pivot. This gives
+---------+---------+-------------+-------------+---------+
| < pivot | > pivot | dup < pivot | dup > pivot | = pivot |
+---------+---------+-------------+-------------+---------+
At this point, this first range is the unique elements in ascending order:
+---------------------+-------------+-------------+---------+
| sorted unique elems | dup < pivot | dup > pivot | = pivot |
+---------------------+-------------+-------------+---------+
Finally, do one last rotation of the duplicate elements greater than the pivot and the elements equal to the pivot to yield this:
+---------------------+-------------+---------+-------------+
| sorted unique elems | dup < pivot | = pivot | dup > pivot |
+---------------------+-------------+---------+-------------+
Notice that these last three blocks are just the sorted duplicate values:
+---------------------+-------------------------------------+
| sorted unique elems | sorted duplicate elements |
+---------------------+-------------------------------------+
and voila! We've got everything in the order we want. Using the same analysis that you'd do for normal quicksort, plus the fact that we're only doing O(n) work at each level (three extra rotations), this works out to O(n log n) in the best case with O(log n) memory usage. It's still O(n2) in the worst case with O(log n) memory, but that happens with extremely low probability.
If you are allowed to use O(n) memory, one option would be to build a balanced binary search tree out of all of the elements that stores key/value pairs, where each key is an element of the array and the value is the number of times it appears. You could then sort the array in your format as follows:
For each element in the array:
If that element already exists in the BST, increment its count.
Otherwise, add a new node to the BST with that element having count 1.
Do an inorder walk of the BST. When encountering a node, output its key.
Do a second inorder walk of the BST. When encountering a node, if it has count greater than one, output n - 1 copies of that node, where n is the number of times it appears.
The runtime of this algorithm is O(n log n), but it would be pretty tricky to code up a BST from scratch. It also requires external space, which I'm not sure you're allowed to do.
However, if you are allowed external space and the arrays you are sorting are small and contain small integers, you could modify the above approach by using a modified counting sort. Just replace the BST with an array large enough for each integer in the original array to be a key. This reduces the runtime to O(n + k), with memory usage O(k), where k is the largest element in the array.
Hope this helps!

a modified merge sort could do the trick: on the last merge pass keep track of the last number you pushed on the front of result array and if the lowest of the next numbers is equal add to the end instead of front

Welcome to the world of Data Structures and Algorithms. You're absolutely right in that you could sort that faster. You could also do it a dozen different ways. PHD's are spent on this stuff :)
Here's a link where you can see an optimized bubble sort
You might also want to check out Big O Notation
Have fun and good luck!

Use quicksort to sort the array. When implementing the sort you can modify it slightly by adding all duplicates to a seperate duplicate array. When done simply append the duplicate array to the end of the sorted array.