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Suppose I have a file similar to the following:
123
123
234
234
123
345
I would like to find how many times '123' was duplicated, how many times '234' was duplicated, etc.
So ideally, the output would be like:
123 3
234 2
345 1
Source: (StackOverflow)
Does anyone know how to count the number of times a value appears in a matrix?
For example, if I have a 1500 x 1 matrix M (vector) which stores the values of weekdays (1 - 7), how could I count how many Sundays (1), Mondays(2), ... , Saturdays(7) are stored in M?
Source: (StackOverflow)
I've got a list of Python objects that I'd like to sort by an attribute of the objects themselves. The list looks like:
>>> ut
[<Tag: 128>, <Tag: 2008>, <Tag: <>, <Tag: actionscript>, <Tag: addresses>, <Tag: aes>, <Tag: ajax> ...]
Each object has a count:
>>> ut[1].count
1L
I need to sort the list by number of counts descending.
I've seen several methods for this, but I'm looking for best practice in Python.
Source: (StackOverflow)
What's the fastest way to count the number of keys/properties of an object? It it possible to do this without iterating over the object? i.e. without doing
var count = 0;
for (k in myobj) if (myobj.hasOwnProperty(k)) count++;
(Firefox did provide a magic __count__ property, but this was removed somewhere around version 4.)
Source: (StackOverflow)
What's the simplest way to count the number of occurrences of a character in a string?
e.g. count the number of times 'a' appears in 'Mary had a little lamb'
Source: (StackOverflow)
This question came to my mind, when I had something like
enum Folders {FA, FB, FC};
and wanted to create an array of containers for each folder:
ContainerClass*m_containers[3];
....
m_containers[FA] = ...; // etc.
(Using maps it's much more elegant to use: std::map<Folders, ContainerClass*> m_containers;)
But to come back to my original question: What if I do not want to hard-code the array size, is there a way to figure out how many items are in Folders? (Without relying on e.g. FC being the last item in the list which would allow something like ContainerClass*m_containers[FC+1] if I'm not mistaken.)
Source: (StackOverflow)
I have a stdClass object created from json_decode that won't return the right number when I run the count($obj) function. The object has 30 properties, but the return on the count() function is say 1.
Any ideas?
Below is an example of one of the objects. (I'm requesting the daily trend information from Twitter). If this object had more than one property, the count($obj) would equal 1.
[trends] => stdClass Object
(
[2009-08-21 11:05] => Array
(
[0] => stdClass Object
(
[query] => "Follow Friday"
[name] => Follow Friday
)
[1] => stdClass Object
(
[query] => "Inglourious Basterds" OR "Inglorious Basterds"
[name] => Inglourious Basterds
)
[2] => stdClass Object
(
[query] => Inglourious
[name] => Inglourious
)
[3] => stdClass Object
(
[query] => #songsincode
[name] => #songsincode
)
[4] => stdClass Object
(
[query] => #shoutout
[name] => #shoutout
)
[5] => stdClass Object
(
[query] => "District 9"
[name] => District 9
)
[6] => stdClass Object
(
[query] => #howmanypeople
[name] => #howmanypeople
)
[7] => stdClass Object
(
[query] => Ashes OR #ashes
[name] => Ashes
)
[8] => stdClass Object
(
[query] => #youtubefail
[name] => #youtubefail
)
[9] => stdClass Object
(
[query] => TGIF
[name] => TGIF
)
[10] => stdClass Object
(
[query] => #wish09
[name] => #wish09
)
[11] => stdClass Object
(
[query] => #watch
[name] => #watch
)
[12] => stdClass Object
(
[query] => Avatar
[name] => Avatar
)
[13] => stdClass Object
(
[query] => Ramadhan
[name] => Ramadhan
)
[14] => stdClass Object
(
[query] => Goodnight
[name] => Goodnight
)
[15] => stdClass Object
(
[query] => iPhone
[name] => iPhone
)
[16] => stdClass Object
(
[query] => #iranelection
[name] => #iranelection
)
[17] => stdClass Object
(
[query] => Apple
[name] => Apple
)
[18] => stdClass Object
(
[query] => "Usain Bolt"
[name] => Usain Bolt
)
[19] => stdClass Object
(
[query] => H1N1
[name] => H1N1
)
)
)
Source: (StackOverflow)
Imagine you sell those metallic digits used to number houses, locker doors, hotel rooms, etc. You need to find how many of each digit to ship when your customer needs to number doors/houses:
- 1 to 100
- 51 to 300
- 1 to 2,000 with zeros to the left
The obvious solution is to do a loop from the first to the last number, convert the counter to a string with or without zeros to the left, extract each digit and use it as an index to increment an array of 10 integers.
I wonder if there is a better way to solve this, without having to loop through the entire integers range.
Solutions in any language or pseudocode are welcome.
Edit:
Answers review
John at CashCommons and Wayne Conrad comment that my current approach is good and fast enough. Let me use a silly analogy: If you were given the task of counting the squares in a chess board in less than 1 minute, you could finish the task by counting the squares one by one, but a better solution is to count the sides and do a multiplication, because you later may be asked to count the tiles in a building.
Alex Reisner points to a very interesting mathematical law that, unfortunately, doesn’t seem to be relevant to this problem.
Andres suggests the same algorithm I’m using, but extracting digits with %10 operations instead of substrings.
John at CashCommons and phord propose pre-calculating the digits required and storing them in a lookup table or, for raw speed, an array. This could be a good solution if we had an absolute, unmovable, set in stone, maximum integer value. I’ve never seen one of those.
High-Performance Mark and strainer computed the needed digits for various ranges. The result for one millon seems to indicate there is a proportion, but the results for other number show different proportions.
strainer found some formulas that may be used to count digit for number which are a power of ten.
Robert Harvey had a very interesting experience posting the question at MathOverflow. One of the math guys wrote a solution using mathematical notation.
Aaronaught developed and tested a solution using mathematics. After posting it he reviewed the formulas originated from Math Overflow and found a flaw in it (point to Stackoverflow :).
noahlavine developed an algorithm and presented it in pseudocode.
A new solution
After reading all the answers, and doing some experiments, I found that for a range of integer from 1 to 10n-1:
- For digits 1 to 9, n*10(n-1) pieces are needed
- For digit 0, if not using leading zeros, n*10n-1 - ((10n-1) / 9) are needed
- For digit 0, if using leading zeros, n*10n-1 - n are needed
The first formula was found by strainer (and probably by others), and I found the other two by trial and error (but they may be included in other answers).
For example, if n = 6, range is 1 to 999,999:
- For digits 1 to 9 we need 6*105 = 600,000 of each one
- For digit 0, without leading zeros, we need 6*105 – (106-1)/9 = 600,000 - 111,111 = 488,889
- For digit 0, with leading zeros, we need 6*105 – 6 = 599,994
These numbers can be checked using High-Performance Mark results.
Using these formulas, I improved the original algorithm. It still loops from the first to the last number in the range of integers, but, if it finds a number which is a power of ten, it uses the formulas to add to the digits count the quantity for a full range of 1 to 9 or 1 to 99 or 1 to 999 etc. Here's the algorithm in pseudocode:
integer First,Last //First and last number in the range
integer Number //Current number in the loop
integer Power //Power is the n in 10^n in the formulas
integer Nines //Nines is the resut of 10^n - 1, 10^5 - 1 = 99999
integer Prefix //First digits in a number. For 14,200, prefix is 142
array 0..9 Digits //Will hold the count for all the digits
FOR Number = First TO Last
CALL TallyDigitsForOneNumber WITH Number,1 //Tally the count of each digit
//in the number, increment by 1
//Start of optimization. Comments are for Number = 1,000 and Last = 8,000.
Power = Zeros at the end of number //For 1,000, Power = 3
IF Power > 0 //The number ends in 0 00 000 etc
Nines = 10^Power-1 //Nines = 10^3 - 1 = 1000 - 1 = 999
IF Number+Nines <= Last //If 1,000+999 < 8,000, add a full set
Digits[0-9] += Power*10^(Power-1) //Add 3*10^(3-1) = 300 to digits 0 to 9
Digits[0] -= -Power //Adjust digit 0 (leading zeros formula)
Prefix = First digits of Number //For 1000, prefix is 1
CALL TallyDigitsForOneNumber WITH Prefix,Nines //Tally the count of each
//digit in prefix,
//increment by 999
Number += Nines //Increment the loop counter 999 cycles
ENDIF
ENDIF
//End of optimization
ENDFOR
SUBROUTINE TallyDigitsForOneNumber PARAMS Number,Count
REPEAT
Digits [ Number % 10 ] += Count
Number = Number / 10
UNTIL Number = 0
For example, for range 786 to 3,021, the counter will be incremented:
- By 1 from 786 to 790 (5 cycles)
- By 9 from 790 to 799 (1 cycle)
- By 1 from 799 to 800
- By 99 from 800 to 899
- By 1 from 899 to 900
- By 99 from 900 to 999
- By 1 from 999 to 1000
- By 999 from 1000 to 1999
- By 1 from 1999 to 2000
- By 999 from 2000 to 2999
- By 1 from 2999 to 3000
- By 1 from 3000 to 3010 (10 cycles)
- By 9 from 3010 to 3019 (1 cycle)
- By 1 from 3019 to 3021 (2 cycles)
Total: 28 cycles
Without optimization: 2,235 cycles
Note that this algorithm solves the problem without leading zeros. To use it with leading zeros, I used a hack:
If range 700 to 1,000 with leading zeros is needed, use the algorithm for 10,700 to 11,000 and then substract 1,000 - 700 = 300 from the count of digit 1.
Benchmark and Source code
I tested the original approach, the same approach using %10 and the new solution for some large ranges, with these results:
Original 104.78 seconds
With %10 83.66
With Powers of Ten 0.07
A screenshot of the benchmark application:
If you would like to see the full source code or run the benchmark, use these links:
Accepted answer
noahlavine solution may be correct, but l just couldn’t follow the pseudo code, I think there are some details missing or not completely explained.
Aaronaught solution seems to be correct, but the code is just too complex for my taste.
I accepted strainer’s answer, because his line of thought guided me to develop this new solution.
Source: (StackOverflow)
When searching for number of occurrences of a string in a file, I generally use:
grep pattern file | wc -l
However, this only finds one occurrence per line, because of the way grep works. How can I search for the number of times a string appears in a file, regardless of whether they are on the same or different lines?
Also, what if I'm searching for a regex pattern, not a simple string? How can I count those, or, even better, print each match on a new line?
Source: (StackOverflow)
I have a collection of boxes with the properties weight, volume and owner.
I want to use LINQ to get a summarized list (by owner) of the box information
e.g.
**Owner, Boxes, Total Weight, Total Volume**
Jim, 5, 1430.00, 3.65
George, 2, 37.50, 1.22
Can someone show me how to do this with Lambda expressions?
Source: (StackOverflow)
I have a list of words in a dictionary with the value = the repetition of the keyword but I only want a list of distinct words so i wanted to count the number of keywords. Is there a to count the number of keywords or is there another way I should look for distinct words?
Source: (StackOverflow)
I have two tables one is for news and other one is for comments and i want to get the count for the comments whose status has been set as approved.
SELECT ccc_news . * ,
count(if(ccc_news_comments.id = 'approved', ccc_news_comments.id, 0)) AS comments
FROM ccc_news
LEFT JOIN ccc_news_comments ON ccc_news_comments.news_id = ccc_news.news_id
WHERE `ccc_news`.`category` = 'news_layer2'
AND `ccc_news`.`status` = 'Active'
GROUP BY ccc_news.news_id
ORDER BY ccc_news.set_order ASC
LIMIT 20
But the problem with this query is that minimum value that is fetched for the comments column is 1 whether there is any comment existent corresponding to that news or not.
Any help would be highly appreciable.
Source: (StackOverflow)
In Javascript, I'm trying to take an initial array of number values and count the elements inside it. Ideally, the result would be two new arrays, the first specifying each unique element, and the second containing the number of times each element occurs. However, I'm open to suggestions on the format of the output.
For example, if the initial array was:
5, 5, 5, 2, 2, 2, 2, 2, 9, 4
Then two new arrays would be created. The first would contain the name of each unique element:
5, 2, 9, 4
The second would contain the number of times that element occurred in the initial array:
3, 5, 1, 1
Because the number 5 occurs three times in the initial array, the number 2 occurs five times and 9 and 4 both appear once.
I've searched a lot for a solution, but nothing seems to work, and everything I've tried myself has wound up being ridiculously complex. Any help would be appreciated!
Thanks :)
Source: (StackOverflow)
Let's say I have
v = rep(c(1,2, 2, 2), 25)
Now, I want to count the number of times each unique value appears. unique(v) returns what the unique values are, but not how many they are.
> unique(v)
[1] 1 2
I want something that gives me
length(v[v==1])
[1] 25
length(v[v==2])
[1] 75
but as a more general one-liner :) Something close (but not quite) like this:
#<doesn't work right> length(v[v==unique(v)])
Source: (StackOverflow)