Given an integer n such that (1<=n<=10^18)
We need to calculate f(1)+f(2)+f(3)+f(4)+....+f(n).
f(x) is given as :-
Say, x = 1112222333,
then f(x)=1002000300.
Whenever we see a contiguous subsequence of same numbers, we replace it with the first number and zeroes all behind it.
Formally, f(x) = Sum over all (first element of the contiguous subsequence * 10^i ), where i is the index of first element from left of a particular contiguous subsequence.
f(x)=1*10^9 + 2*10^6 + 3*10^2 = 1002000300.
In, x=1112222333,
Element at index '9':-1
and so on...
We follow zero based indexing :-)
For, x=1234.
Element at index-'0':-4,element at index -'1':3,element at index '2':-2,element at index 3:-1
How to calculate f(1)+f(2)+f(3)+....+f(n)?
I want to generate an algorithm which calculates this sum efficiently.
There is nothing to calculate.
Multiplying each position in the array od numbers will yeild thebsame number.
So all you want to do is end up with 0s on a repeated number
IE lets populate some static values in an array in psuedo code
$As[1]='0'
$As[2]='00'
$As[3]='000'
...etc
$As[18]='000000000000000000'```
these are the "results" of 10^index
Given a value n of `1234`
```1&000 + 2&00 +3 & 0 + 4```
Results in `1234`
So, if you are putting this on a chip, then probably your most efficient method is to do a bitwise XOR between each register and the next up the line as a single operation
Then you will have 0s in all the spots you care about, and just retrive the values in the registers with a 1
In code, I think it would be most efficient to do the following
```$n = arbitrary value 11223334
$x=$n*10
$zeros=($x-$n)/10```
Okay yeah we can just do bit shifting to get a value like 100200300400 etc.
To approach this problem, it could help to begin with one digit numbers and see what sum you get.
I mean like this:
Let's say, we define , then we have:
F(1)= 45 # =10*9/2 by Euler's sum formula
F(2)= F(1)*9 + F(1)*100 # F(1)*9 is the part that comes from the last digit
# because for each of the 10 possible digits in the
# first position, we have 9 digits in the last
# because both can't be equal and so one out of ten
# becomse zero. F(1)*100 comes from the leading digit
# which is multiplied by 100 (10 because we add the
# second digit and another factor of 10 because we
# get the digit ten times in that position)
If you now continue with this scheme, for k>=1 in general you get
F(k+1)= F(k)*100+10^(k-1)*45*9
The rest is probably straightforward.
Can you tell me, which Hackerrank task this is? I guess one of the Project Euler tasks right?
We have the related and well known problems:
1) PARTITION (decision problem):
Given a set S of n natural numbers. Is it possible to find a subset T of S such that the sum of the numbers of T is equal to the sum of the numbers of T\S?
2) PARTITION (general problem):
Given a set S of n natural numbers. Assuming the answer of the decision problem 1) to this set is 'yes' then find such a subset.
Simple question: How can we solve 2) in polynomial time if we have an algorithm that solves 1) in polynomial time?
Solution: Suppose we want to partition a set S of n natural numbers with the sum equal to a number b and we have a blackbox algorithm that solves the decision problem in polynomial time.
1: If the answer of the partition problem of S is no, then return.
2: Pick an element x of S.
3: If x is equal to b/2 return S\x (partition found).
4. Merge x with another element y of S (set x=x+y and set S=S\y) which is not processed yet.
5: Solve the decision problem for S.
6: If the answer is no then revert step 4, mark y as processed.
7: Go back to step 3.
Each time we repeat step 2 we have to solve a decision problem in polynomial time. Since we only have to repeat step 2 at most n-1 times, the overall time complexity is also polynomial.
I am new to objective C and trying to understand arc4random().
There are so many conflicting explanations on the web. Please clear my confusion, which of the following is correct:
// 1.
arc4random() % (toNumber - fromNumber) + fromNumber;
OR
//2.
arc4random() % ((toNumber - fromNumber) + 1) + fromNumber;
//toNumber-fromNumbers are any range of numbers like random # between 7-90.
This code will get you a random number between 7 and 90.
NSUInteger random = 7 + arc4random_uniform(90 - 7);
Use arc4random_uniform to avoid modulo bias.
Adam's answer is correct. However, just to clarify the difference between the two, the second one raises the possible range by one to make the range inclusive. The important thing to remember is that modulo is remainder division, so while there are toNumber possible outcomes, one of them is zero (if the result of arc4random() is a multiple of toNumber) and toNumber itself can not be the remainder.
// 1.
arc4random() % (10 - 5) + 5;
This results in a range of 0 + 5 to 4 + 5, which is 5 to 9.
//2.
arc4random() % ((10 - 5) + 1) + 5;
This results in a range of 0 + 5 to (4 + 1) + 5, which is 5 to 10.
Neither is correct or incorrect if you wish to use modulo. One is exclusive of the upper range while the other is inclusive of the upper range. However, if you think about how remainder division works and think of the pool of numbers returned by any PRNG in terms of cycles the length of your total range, then you'll realize that if the range does not divide evenly into the maximum range of the pool you'll get biased results. For instance, if arc4random() returned a result from 1 to 5 (it doesn't, obviously) and you wanted a number from 0 to 2, and you used arc4random() % 3, these are the possible results.
1 % 3 = 1
2 % 3 = 2
3 % 3 = 0
4 % 3 = 1
5 % 3 = 2
Note that there are two ones and two twos, but only one zero. This is because our range of 3 does not evenly divide into the PRNG's range of 5. The result is that (humorously enough) PRNG range % desired range numbers at the end of the cycle need to be culled because they are "biased"–the numbers themselves aren't really biased, but it's easier to cull from the end. Failing to do this results in the lower numbers of the range becoming more likely to appear.
We can cull the numbers by calculating the upper range of the numbers we can generate, modulo it with the desired range and then pull those numbers off of the end. By "pull those numbers off of the end" I really mean "loop infinitely until we get a number that isn't one of the end numbers".
Some would say that's bad practice; you could theoretically loop forever. In practice, however, the expected number of retries is always less than one since the modulo bias is never more than half the pool (usually much less than that) of the PRNG's numbers. I once wrote a wrapper for rand using this technique.
You can see an example of this in the source for OpenBSD, where arc4random_uniform calls arc4random in a loop until a number is determined to be clean.
One of my friends was asked this question recently:
You have to count how many binary strings are possible of length "K".
Constraint: Every 0 has a 1 in its immediate left.
This question can be reworded:
How many binary sequences of length K are posible if there are no two consecutive 0s, but the first element should be 1 (else the constrains fails). Let us forget about the first element (we can do it bcause it is always fixed).
Then we got a very famous task that sounds like this: "What is the number of binary sequences of length K-1 that have no consecutive 0's." The explanation can be found, for example, here
Then the answer will be F(K+1) where F(K) is the K`th fibonacci number starting from (1 1 2 ...).
∑ From n=0 to ⌊K/2⌋ of (K-n)Cn; n is the number of zeros in the string
The idea is to group every 0 with a 1 and find the number of combinations of the string, for n zeros there will be n ones grouped to them so the string becomes (k-n) elements long. There can be no more than of K/2 zeros as there would not have enough ones to be to the immediate left of each zero.
E.g. 111111[10][10]1[10] for K = 13, n = 3
I know the modulus (%) operator calculates the remainder of a division. How can I identify a situation where I would need to use the modulus operator?
I know I can use the modulus operator to see whether a number is even or odd and prime or composite, but that's about it. I don't often think in terms of remainders. I'm sure the modulus operator is useful, and I would like to learn to take advantage of it.
I just have problems identifying where the modulus operator is applicable. In various programming situations, it is difficult for me to see a problem and realize "Hey! The remainder of division would work here!".
Imagine that you have an elapsed time in seconds and you want to convert this to hours, minutes, and seconds:
h = s / 3600;
m = (s / 60) % 60;
s = s % 60;
0 % 3 = 0;
1 % 3 = 1;
2 % 3 = 2;
3 % 3 = 0;
Did you see what it did? At the last step it went back to zero. This could be used in situations like:
To check if N is divisible by M (for example, odd or even)
or
N is a multiple of M.
To put a cap of a particular value. In this case 3.
To get the last M digits of a number -> N % (10^M).
I use it for progress bars and the like that mark progress through a big loop. The progress is only reported every nth time through the loop, or when count%n == 0.
I've used it when restricting a number to a certain multiple:
temp = x - (x % 10); //Restrict x to being a multiple of 10
Wrapping values (like a clock).
Provide finite fields to symmetric key algorithms.
Bitwise operations.
And so on.
One use case I saw recently was when you need to reverse a number. So that 123456 becomes 654321 for example.
int number = 123456;
int reversed = 0;
while ( number > 0 ) {
# The modulus here retrieves the last digit in the specified number
# In the first iteration of this loop it's going to be 6, then 5, ...
# We are multiplying reversed by 10 first, to move the number one decimal place to the left.
# For example, if we are at the second iteration of this loop,
# reversed gonna be 6, so 6 * 10 + 12345 % 10 => 60 + 5
reversed = reversed * 10 + number % 10;
number = number / 10;
}
Example. You have message of X bytes, but in your protocol maximum size is Y and Y < X. Try to write small app that splits message into packets and you will run into mod :)
There are many instances where it is useful.
If you need to restrict a number to be within a certain range you can use mod. For example, to generate a random number between 0 and 99 you might say:
num = MyRandFunction() % 100;
Any time you have division and want to express the remainder other than in decimal, the mod operator is appropriate. Things that come to mind are generally when you want to do something human-readable with the remainder. Listing how many items you could put into buckets and saying "5 left over" is good.
Also, if you're ever in a situation where you may be accruing rounding errors, modulo division is good. If you're dividing by 3 quite often, for example, you don't want to be passing .33333 around as the remainder. Passing the remainder and divisor (i.e. the fraction) is appropriate.
As #jweyrich says, wrapping values. I've found mod very handy when I have a finite list and I want to iterate over it in a loop - like a fixed list of colors for some UI elements, like chart series, where I want all the series to be different, to the extent possible, but when I've run out of colors, just to start over at the beginning. This can also be used with, say, patterns, so that the second time red comes around, it's dashed; the third time, dotted, etc. - but mod is just used to get red, green, blue, red, green, blue, forever.
Calculation of prime numbers
The modulo can be useful to convert and split total minutes to "hours and minutes":
hours = minutes / 60
minutes_left = minutes % 60
In the hours bit we need to strip the decimal portion and that will depend on the language you are using.
We can then rearrange the output accordingly.
Converting linear data structure to matrix structure:
where a is index of linear data, and b is number of items per row:
row = a/b
column = a mod b
Note above is simplified logic: a must be offset -1 before dividing & the result must be normalized +1.
Example: (3 rows of 4)
1 2 3 4
5 6 7 8
9 10 11 12
(7 - 1)/4 + 1 = 2
7 is in row 2
(7 - 1) mod 4 + 1 = 3
7 is in column 3
Another common use of modulus: hashing a number by place. Suppose you wanted to store year & month in a six digit number 195810. month = 195810 mod 100 all digits 3rd from right are divisible by 100 so the remainder is the 2 rightmost digits in this case the month is 10. To extract the year 195810 / 100 yields 1958.
Modulus is also very useful if for some crazy reason you need to do integer division and get a decimal out, and you can't convert the integer into a number that supports decimal division, or if you need to return a fraction instead of a decimal.
I'll be using % as the modulus operator
For example
2/4 = 0
where doing this
2/4 = 0 and 2 % 4 = 2
So you can be really crazy and let's say that you want to allow the user to input a numerator and a divisor, and then show them the result as a whole number, and then a fractional number.
whole Number = numerator/divisor
fractionNumerator = numerator % divisor
fractionDenominator = divisor
Another case where modulus division is useful is if you are increasing or decreasing a number and you want to contain the number to a certain range of number, but when you get to the top or bottom you don't want to just stop. You want to loop up to the bottom or top of the list respectively.
Imagine a function where you are looping through an array.
Function increase Or Decrease(variable As Integer) As Void
n = (n + variable) % (listString.maxIndex + 1)
Print listString[n]
End Function
The reason that it is n = (n + variable) % (listString.maxIndex + 1) is to allow for the max index to be accounted.
Those are just a few of the things that I have had to use modulus for in my programming of not just desktop applications, but in robotics and simulation environments.
Computing the greatest common divisor
Determining if a number is a palindrome
Determining if a number consists of only ...
Determining how many ... a number consists of...
My favorite use is for iteration.
Say you have a counter you are incrementing and want to then grab from a known list a corresponding items, but you only have n items to choose from and you want to repeat a cycle.
var indexFromB = (counter-1)%n+1;
Results (counter=indexFromB) given n=3:
`1=1`
`2=2`
`3=3`
`4=1`
`5=2`
`6=3`
...
Best use of modulus operator I have seen so for is to check if the Array we have is a rotated version of original array.
A = [1,2,3,4,5,6]
B = [5,6,1,2,3,4]
Now how to check if B is rotated version of A ?
Step 1: If A's length is not same as B's length then for sure its not a rotated version.
Step 2: Check the index of first element of A in B. Here first element of A is 1. And its index in B is 2(assuming your programming language has zero based index).
lets store that index in variable "Key"
Step 3: Now how to check that if B is rotated version of A how ??
This is where modulus function rocks :
for (int i = 0; i< A.length; i++)
{
// here modulus function would check the proper order. Key here is 2 which we recieved from Step 2
int j = [Key+i]%A.length;
if (A[i] != B[j])
{
return false;
}
}
return true;
It's an easy way to tell if a number is even or odd. Just do # mod 2, if it is 0 it is even, 1 it is odd.
Often, in a loop, you want to do something every k'th iteration, where k is 0 < k < n, assuming 0 is the start index and n is the length of the loop.
So, you'd do something like:
int k = 5;
int n = 50;
for(int i = 0;i < n;++i)
{
if(i % k == 0) // true at 0, 5, 10, 15..
{
// do something
}
}
Or, you want to keep something whitin a certain bound. Remember, when you take an arbitrary number mod something, it must produce a value between 0 and that number - 1.