Suppose we have a boolean array of length X. The only rule is, TRUE must not occur twice in adjacent places. Especially the array with only false values is allowed. E.g. this is forbidden: [1,1,0,0,0] and these is allowed: [1,0,0,0,0], [0,0,0,0,0], [1,0,1,0,1] etc. How can I use dynamic programming to determine how many different valid arrays of length X there are?
Let Ti be the number of arrays of length i that meet your criterion and end in 1, and let Fi be the number of arrays of length i that meet your criterion and do not end in 1.
Then:
T0 = 0
F0 = 1
Ti+1 = Fi. (Each array of length i+1 that meets your criterion and ends in 1 consists of an array of length i that meets your criterion and does not end in 1, plus an extra 1 at the end.)
Fi+1 = Fi + Ti. (Each array of length i+1 that meets your criterion and does not end in 1 consists of an array of length i that meets your criterion, plus an extra 0 at the end.)
You want FX + TX.
So you can just write a loop that calculates Fi and Ti for each i from 0 to X, and then return FX + TX.
(This isn't even dynamic programming, per se, because you don't need to store partial values; Fi+1 and Ti+1 depend only on Fi and Ti. So this is O(X) time and O(1) space.)
I think you can calculate the number without using DP. Since you know the total number of arrays for length N, that is `2^N'.
Now you need to deduct those bad arrays that do not qualify if they have adjacent 1's. For an array of length N, there are these cases
1. the array has no 1's, only one case, and it is a valid array
2. the array has one 1's, all cases are valid
3. the array has two 1's, there are N - 1 cases which are not valid
4. the array has three 1's, there are (N-1) * (N-2) / 2 cases which are not valid
...
The dp solution would have two state parameters. One is the position of the array and the other is the previous position’s value. If the previous position’s value is 1 then you can only choose 0. If the previous position’s value is 0 then you can choose either 1 or 0. Hope this helps.
You don't really need dynamic programming.
For array length X the number of valid arrays is Fib(X+1), where Fib is the array of fibonacci numbers.
X=1: valid arrays: 2
X=2: valid arrays: 3
X=3: valid arrays: 5
X=4: valid arrays: 8
and so on...
Demonstration:
Let's assume we are looking for the arrays for X and we know the number of valid arrays for X-1. We can freely add a zero to the end of each of these X-1 length arrays, so that's F(X-1) so far. We can also add a '1' to the end of each X-1 arrays which ends with 0. But how many of those arrays are? Well, it's exactly F(X-2) because we could generate the zero ending X-1 length arrays exactly the same way: by adding a zero at the end of each X-2 length arrays. So F(X) = F(X-1) + F(X-2)
And that's exactly the definition of the Fibonacci array.
All we have to do is manually calculate the first two element to determinate if it's exactly the Fibonacci array or is it shifted.
You can even find a formula to calculate the Nth element of the fibonacci array, so it can be solved in O(1).
Related
If I had four arrays with the same size, how do I determine the number of ways to choose one element from each array such that the four elements have the same sum?
For example, there are 81 ways to choose choose an element from each array with a sum of 4.
A. 1 1 1
B. 1 1 1
C. 1 1 1
D. 1 1 1
I am not sure how to do this without some sort of brute forcing.
The idea
Number of ways to get sum k from 1 array = number of occurences of k in the array.
Number of ways to get sum k from 2 arrays = sum(count(k-x) for each element x in array 2), where count(y) is number of ways of getting sum y from the first array. If you memoize the results from point #1, getting count(y) can be done in O(1).
Number of ways to get sum k from 3 arrays = sum(count(k-x) for each element x in array 3), where count(y) is number of ways of getting sum y from the first two arrays. Again, if you memoize the results from point #2, getting count(y) can be done in O(1).
You should get the concept by now; number of ways to get sum k from n arrays = sum(count(k-x) for each element x in the nth array), where count(y) is number of ways of getting sum y from the first n-1 arrays.
Dynamic programming
You need to build a memoization table which basically tells you the number of ways of getting sum x from the first y arrays. If you have this table for the first n-1 arrays, you can calculate all sums including the n-th array efficiently and update the table.
I found this question on topcoder:
Your friend Lucas gave you a sequence S of positive integers.
For a while, you two played a simple game with S: Lucas would pick a number, and you had to select some elements of S such that the sum of all numbers you selected is the number chosen by Lucas. For example, if S={2,1,2,7} and Lucas chose the number 11, you would answer that 2+2+7 = 11.
Lucas now wants to trick you by choosing a number X such that there will be no valid answer. For example, if S={2,1,2,7}, it is not possible to select elements of S that sum up to 6.
You are given the int[] S. Find the smallest positive integer X that cannot be obtained as the sum of some (possibly all) elements of S.
Constraints: - S will contain between 1 and 20 elements, inclusive. - Each element of S will be between 1 and 100,000, inclusive.
But in the editorial solution it has been written:
How about finding the smallest impossible sum? Well, we can try the following naive algorithm: First try with x = 1, if this is not a valid sum (found using the methods in the previous section), then we can return x, else we increment x and try again, and again until we find the smallest number that is not a valid sum.
Let's find an upper bound for the number of iterations, the number of values of x we will need to try before we find a result. First of all, the maximum sum possible in this problem is 100000 * 20 (All numbers are the maximum 100000), this means that 100000 * 20 + 1 will not be an impossible value. We can be certain to need at most 2000001 steps.
How good is this upper bound? If we had 100000 in each of the 20 numbers, 1 wouldn't be a possible sum. So we actually need one iteration in that case. If we want 1 to be a possible sum, we should have 1 in the initial elements. Then we need a 2 (Else we would only need 2 iterations), then a 4 (3 can be found by adding 1+2), then 8 (Numbers from 5 to 7 can be found by adding some of the first 3 powers of two), then 16, 32, .... It turns out that with the powers of 2, we can easily make inputs that require many iterations. With the first 17 powers of two, we can cover up to the first 262143 integer numbers. That should be a good estimation for the largest number. (We cannot use 2^18 in the input, smaller than 100000).
Up to 262143 times, we need to query if a number x is in the set of possible sums. We can just use a boolean array here. It appears that even O(log(n)) data structures should be fast enough, however.
I did understand the first paragraph. But after that they have explained something about "How good is this upper bound?...". I couldnt understand that paragraph. How did they deduce to the fact that we need to query 262143 times if a number x is in the set of possible sums?
I am a newbie at dynamic programming and so it would be great if somebody could explain this to me.
Thank you.
The idea is as follows:
If the input sequence contains the first k powers of two: 2^0, 2^1, ... 2^(k-1), then the sum can be any integer between 0 and (2^k) - 1. Since the greatest power of two that can appear in the sequence is 2^17, the greatest sum that you can build from 18 numbers is 2^18 - 1=262,143. If a power of two would be missing, there would be a smaller sum that was not possible to achieve.
However, the statement is missing that there may be 2 more numbers in the sequence (at most 20). From these two numbers, you can repeat the same process. Hence, the maximum number to check is actually (2^18) - 1 + (2^2) - 1.
You may wonder why we use powers of two and not any other powers. The reason is the binary selection that we perform on the numbers in the input sequence. Either we add a number to the sum or we don't. So, if we represent this selection for number ni as a selection variable si (either 0 or 1), then the possible sum is:
s = s0 * n0 + s1 * n1 + s2 * n2 + ...
Now, if we choose the ni to be powers of two ni = 2^i, then:
s = s0 * 2^0 + s1 * 2^1 + s2 * 2^2 + ...
= sum si * 2^i
This is equivalent to the binary representations of numbers (see Positional Notation). By definition, different choices for the selection variables will produce different sums. Hence, the number of possible sums is maximal by choosing powers of two in the input sequence.
I want to know whether the task explained below is even theoretically possible, and if so how I could do it.
You are given a space of N elements (i.e. all numbers between 0 and N-1.) Let's look at the space of all permutations on that space, and call it S. The ith member of S, which can be marked S[i], is the permutation with the lexicographic number i.
For example, if N is 3, then S is this list of permutations:
S[0]: 0, 1, 2
S[1]: 0, 2, 1
S[2]: 1, 0, 2
S[3]: 1, 2, 0
S[4]: 2, 0, 1
S[5]: 2, 1, 0
(Of course, when looking at a big N, this space becomes very large, N! to be exact.)
Now, I already know how to get the permutation by its index number i, and I already know how to do the reverse (get the lexicographic number of a given permutation.) But I want something better.
Some permutations can be huge by themselves. For example, if you're looking at N=10^20. (The size of S would be (10^20)! which I believe is the biggest number I ever mentioned in a Stack Overflow question :)
If you're looking at just a random permutation on that space, it would be so big that you wouldn't be able to store the whole thing on your harddrive, let alone calculate each one of the items by lexicographic number. What I want is to be able to do item access on that permutation, and also get the index of each item. That is, given N and i to specify a permutation, have one function that takes an index number and find the number that resides in that index, and another function that takes a number and finds in which index it resides. I want to do that in O(1), so I don't need to store or iterate over each member in the permutation.
Crazy, you say? Impossible? That may be. But consider this: A block cipher, like AES, is essentially a permutation, and it almost accomplishes the tasks I outlined above. AES has a block size of 16 bytes, meaning that N is 256^16 which is around 10^38. (The size of S, not that it matters, is a staggering (256^16)!, or around 10^85070591730234615865843651857942052838, which beats my recent record for "biggest number mentioned on Stack Overflow" :)
Each AES encryption key specifies a single permutation on N=256^16. That permutation couldn't be stored whole on your computer, because it has more members than there are atoms in the solar system. But, it allows you item access. By encrypting data using AES, you're looking at the data block by block, and for each block (member of range(N)) you output the encrypted block, which the member of range(N) that is in the index number of the original block in the permutation. And when you're decrypting, you're doing the reverse (Finding the index number of a block.) I believe this is done in O(1), I'm not sure but in any case it's very fast.
The problem with using AES or any other block cipher is that it limits you to very specific N, and it probably only captures a tiny fraction of the possible permutations, while I want to be able to use any N I like, and do item access on any permutation S[i] that I like.
Is it possible to get O(1) item access on a permutation, given size N and permutation number i? If so, how?
(If I'm lucky enough to get code answers here, I'd appreciate if they'll be in Python.)
UPDATE:
Some people pointed out the sad fact that the permutation number itself would be so huge, that just reading the number would make the task non-feasible. Then, I'd like to revise my question: Given access to the factoradic representation of a permutation's lexicographic number, is it possible to get any item in the permutation in O(as small as possible)?
The secret to doing this is to "count in base factorial".
In the same way that 134 = 1*10^2+3*10 + 4, 134 = 5! + 2 * 3! + 2! => 10210 in factorial notation (include 1!, exclude 0!). If you want to represent N!, you will then need N^2 base ten digits. (For each factorial digit N, the maximum number it can hold is N). Up to a bit of confusion about what you call 0, this factorial representation is exactly the lexicographic number of a permutation.
You can use this insight to solve Euler Problem 24 by hand. So I will do that here, and you will see how to solve your problem. We want the millionth permutation of 0-9. In factorial representation we take 1000000 => 26625122. Now to convert that to the permutation, I take my digits 0,1,2,3,4,5,6,7,8,9, and The first number is 2, which is the third (it could be 0), so I select 2 as the first digit, then I have a new list 0,1,3,4,5,6,7,8,9 and I take the seventh number which is 8 etc, and I get 2783915604.
However, this assumes that you start your lexicographic ordering at 0, if you actually start it at one, you have to subtract 1 from it, which gives 2783915460. Which is indeed the millionth permutation of the numbers 0-9.
You can obviously reverse this procedure, and hence convert backwards and forwards easily between the lexiographic number and the permutation that it represents.
I am not entirely clear what it is that you want to do here, but understanding the above procedure should help. For example, its clear that the lexiographic number represents an ordering which could be used as the key in a hashtable. And you can order numbers by comparing digits left to right so once you have inserted a number you never have to work outs it factorial.
Your question is a bit moot, because your input size for an arbitrary permutation index has size log(N!) (assuming you want to represent all possible permutations) which is Theta(N log N), so if N is really large then just reading the input of the permutation index would take too long, certainly much longer than O(1). It may be possible to store the permutation index in such a way that if you already had it stored, then you could access elements in O(1) time. But probably any such method would be equivalent to just storing the permutation in contiguous memory (which also has Theta(N log N) size), and if you store the permutation directly in memory then the question becomes trivial assuming you can do O(1) memory access. (However you still need to account for the size of the bit encoding of the element, which is O(log N)).
In the spirit of your encryption analogy, perhaps you should specify a small SUBSET of permutations according to some property, and ask if O(1) or O(log N) element access is possible for that small subset.
Edit:
I misunderstood the question, but it was not in waste. My algorithms let me understand: the factoradic representation of a permutation's lexicographic number is almost the same as the permutation itself. In fact the first digit of the factoradic representation is the same as the first element of the corresponding permutation (assuming your space consists of numbers from 0 to N-1). Knowing this there is not really a point in storing the index rather than the permutation itself . To see how to convert the lexicographic number into a permutation, read below.
See also this wikipedia link about Lehmer code.
Original post:
In the S space there are N elements that can fill the first slot, meaning that there are (N-1)! elements that start with 0. So i/(N-1)! is the first element (lets call it 'a'). The subset of S that starts with 0 consists of (N-1)! elements. These are the possible permutations of the set N{a}. Now you can get the second element: its the i(%((N-1)!)/(N-2)!). Repeat the process and you got the permutation.
Reverse is just as simple. Start with i=0. Get the 2nd last element of the permutation. Make a set of the last two elements, and find the element's position in it (its either the 0th element or the 1st), lets call this position j. Then i+=j*2!. Repeat the process (you can start with the last element too, but it will always be the 0th element of the possibilities).
Java-ish pesudo code:
find_by_index(List N, int i){
String str = "";
for(int l = N.length-1; i >= 0; i--){
int pos = i/fact(l);
str += N.get(pos);
N.remove(pos);
i %= fact(l);
}
return str;
}
find_index(String str){
OrderedList N;
int i = 0;
for(int l = str.length-1; l >= 0; l--){
String item = str.charAt(l);
int pos = N.add(item);
i += pos*fact(str.length-l)
}
return i;
}
find_by_index should run in O(n) assuming that N is pre ordered, while find_index is O(n*log(n)) (where n is the size of the N space)
After some research in Wikipedia, I desgined this algorithm:
def getPick(fact_num_list):
"""fact_num_list should be a list with the factorial number representation,
getPick will return a tuple"""
result = [] #Desired pick
#This will hold all the numbers pickable; not actually a set, but a list
#instead
inputset = range(len(fact_num_list))
for fnl in fact_num_list:
result.append(inputset[fnl])
del inputset[fnl] #Make sure we can't pick the number again
return tuple(result)
Obviously, this won't reach O(1) due the factor we need to "pick" every number. Due we do a for loop and thus, assuming all operations are O(1), getPick will run in O(n).
If we need to convert from base 10 to factorial base, this is an aux function:
import math
def base10_baseFactorial(number):
"""Converts a base10 number into a factorial base number. Output is a list
for better handle of units over 36! (after using all 0-9 and A-Z)"""
loop = 1
#Make sure n! <= number
while math.factorial(loop) <= number:
loop += 1
result = []
if not math.factorial(loop) == number:
loop -= 1 #Prevent dividing over a smaller number than denominator
while loop > 0:
denominator = math.factorial(loop)
number, rem = divmod(number, denominator)
result.append(rem)
loop -= 1
result.append(0) #Don't forget to divide to 0! as well!
return result
Again, this will run in O(n) due to the whiles.
Summing all, the best time we can find is O(n).
PS: I'm not a native English speaker, so spelling and phrasing errors may appear. Apologies in advance, and let me know if you can't get around something.
All correct algorithms for accessing the kth item of a permutation stored in factoradic form must read the first k digits. This is because, regardless of the values of the other digits among the first k, it makes a difference whether an unread digit is a 0 or takes on its maximum value. That this is the case can be seen by tracing the canonical correct decoding program in two parallel executions.
For example, if we want to decode the third digit of the permutation 1?0, then for 100, that digit is 0, and for 110, that digit is 2.
I tried to find a solution to this but couldn't get much out of my head.
We are given two unsorted integer arrays A and B. We have to check whether array B is a permutation of A. How can this be done.? Even XORing the numbers wont work as there can be several counterexamples which have same XOR value bt are not permutation of each other.
A solution needs to be O(n) time and with space O(1)
Any help is welcome!!
Thanks.
The question is theoretical but you can do it in O(n) time and o(1) space. Allocate an array of 232 counters and set them all to zero. This is O(1) step because the array has constant size. Then iterate through the two arrays. For array A, increment the counters corresponding to the integers read. For array B, decrement them. If you run into a negative counter value during iteration of array B, stop --- the arrays are not permutations of each others. Otherwise at the end (assuming A and B have the same size, a prerequisite) the counter array is all zero and the two arrays are permutations of each other.
This is O(1) space and O(n) time solution. However it is not practical, but would easily pass as a solution to the interview question. At least it should.
More obscure solutions
Using a nondeterministic model of computation, checking that the two arrays are not permutations of each others can be done in O(1) space, O(n) time by guessing an element that has differing count on the two arrays, and then counting the instances of that element on both of the arrays.
In randomized model of computation, construct a random commutative hash function and calculate the hash values for the two arrays. If the hash values differ, the arrays are not permutations of each others. Otherwise they might be. Repeat many times to bring the probability of error below desired threshold. Also on O(1) space O(n) time approach, but randomized.
In parallel computation model, let 'n' be the size of the input array. Allocate 'n' threads. Every thread i = 1 .. n reads the ith number from the first array; let that be x. Then the same thread counts the number of occurrences of x in the first array, and then check for the same count on the second array. Every single thread uses O(1) space and O(n) time.
Interpret an integer array [ a1, ..., an ] as polynomial xa1 + xa2 + ... + xan where x is a free variable and the check numerically for the equivalence of the two polynomials obtained. Use floating point arithmetics for O(1) space and O(n) time operation. Not an exact method because of rounding errors and because numerical checking for equivalence is probabilistic. Alternatively, interpret the polynomial over integers modulo a prime number, and perform the same probabilistic check.
If we are allowed to freely access a large list of primes, you can solve this problem by leveraging properties of prime factorization.
For both arrays, calculate the product of Prime[i] for each integer i, where Prime[i] is the ith prime number. The value of the products of the arrays are equal iff they are permutations of one another.
Prime factorization helps here for two reasons.
Multiplication is transitive, and so the ordering of the operands to calculate the product is irrelevant. (Some alluded to the fact that if the arrays were sorted, this problem would be trivial. By multiplying, we are implicitly sorting.)
Prime numbers multiply losslessly. If we are given a number and told it is the product of only prime numbers, we can calculate exactly which prime numbers were fed into it and exactly how many.
Example:
a = 1,1,3,4
b = 4,1,3,1
Product of ith primes in a = 2 * 2 * 5 * 7 = 140
Product of ith primes in b = 7 * 2 * 5 * 2 = 140
That said, we probably aren't allowed access to a list of primes, but this seems a good solution otherwise, so I thought I'd post it.
I apologize for posting this as an answer as it should really be a comment on antti.huima's answer, but I don't have the reputation yet to comment.
The size of the counter array seems to be O(log(n)) as it is dependent on the number of instances of a given value in the input array.
For example, let the input array A be all 1's with a length of (2^32) + 1. This will require a counter of size 33 bits to encode (which, in practice, would double the size of the array, but let's stay with theory). Double the size of A (still all 1 values) and you need 65 bits for each counter, and so on.
This is a very nit-picky argument, but these interview questions tend to be very nit-picky.
If we need not sort this in-place, then the following approach might work:
Create a HashMap, Key as array element, Value as number of occurances. (To handle multiple occurrences of the same number)
Traverse array A.
Insert the array elements in the HashMap.
Next, traverse array B.
Search every element of B in the HashMap. If the corresponding value is 1, delete the entry. Else, decrement the value by 1.
If we are able to process entire array B and the HashMap is empty at that time, Success. else Failure.
HashMap will use constant space and you will traverse each array only once.
Not sure if this is what you are looking for. Let me know if I have missed any constraint about space/time.
You're given two constraints: Computational O(n), where n means the total length of both A and B and memory O(1).
If two series A, B are permutations of each other, then theres also a series C resulting from permutation of either A or B. So the problem is permuting both A and B into series C_A and C_B and compare them.
One such permutation would be sorting. There are several sorting algorithms which work in place, so you can sort A and B in place. Now in a best case scenario Smooth Sort sorts with O(n) computational and O(1) memory complexity, in the worst case with O(n log n) / O(1).
The per element comparision then happens at O(n), but since in O notation O(2*n) = O(n), using a Smooth Sort and comparison will give you a O(n) / O(1) check if two series are permutations of each other. However in the worst case it will be O(n log n)/O(1)
The solution needs to be O(n) time and with space O(1).
This leaves out sorting and the space O(1) requirement is a hint that you probably should make a hash of the strings and compare them.
If you have access to a prime number list do as cheeken's solution.
Note: If the interviewer says you don't have access to a prime number list. Then generate the prime numbers and store them. This is O(1) because the Alphabet length is a constant.
Else here's my alternative idea. I will define the Alphabet as = {a,b,c,d,e} for simplicity.
The values for the letters are defined as:
a, b, c, d, e
1, 2, 4, 8, 16
note: if the interviewer says this is not allowed, then make a lookup table for the Alphabet, this takes O(1) space because the size of the Alphabet is a constant
Define a function which can find the distinct letters in a string.
// set bit value of char c in variable i and return result
distinct(char c, int i) : int
E.g. distinct('a', 0) returns 1
E.g. distinct('a', 1) returns 1
E.g. distinct('b', 1) returns 3
Thus if you iterate the string "aab" the distinct function should give 3 as the result
Define a function which can calculate the sum of the letters in a string.
// return sum of c and i
sum(char c, int i) : int
E.g. sum('a', 0) returns 1
E.g. sum('a', 1) returns 2
E.g. sum('b', 2) returns 4
Thus if you iterate the string "aab" the sum function should give 4 as the result
Define a function which can calculate the length of the letters in a string.
// return length of string s
length(string s) : int
E.g. length("aab") returns 3
Running the methods on two strings and comparing the results takes O(n) running time. Storing the hash values takes O(1) in space.
e.g.
distinct of "aab" => 3
distinct of "aba" => 3
sum of "aab => 4
sum of "aba => 4
length of "aab => 3
length of "aba => 3
Since all the values are equal for both strings, they must be a permutation of each other.
EDIT: The solutions is not correct with the given alphabet values as pointed out in the comments.
You can convert one of the two arrays into an in-place hashtable. This will not be exactly O(N), but it will come close, in non-pathological cases.
Just use [number % N] as it's desired index or in the chain that starts there. If any element has to be replaced, it can be placed at the index where the offending element started. Rinse , wash, repeat.
UPDATE:
This is a similar (N=M) hash table It did use chaining, but it could be downgraded to open addressing.
I'd use a randomized algorithm that has a low chance of error.
The key is to use a universal hash function.
def hash(array, hash_fn):
cur = 0
for item in array:
cur ^= hash_item(item)
return cur
def are_perm(a1, a2):
hash_fn = pick_random_universal_hash_func()
return hash_fn(a1, hash_fn) == hash_fn(a2, hash_fn)
If the arrays are permutations, it will always be right. If they are different, the algorithm might incorrectly say that they are the same, but it will do so with very low probability. Further, you can get an exponential decrease in chance for error with a linear amount of work by asking many are_perm() questions on the same input, if it ever says no, then they are definitely not permutations of each other.
I just find a counterexample. So, the assumption below is incorrect.
I can not prove it, but I think this may be possible true.
Since all elements of the arrays are integers, suppose each array has 2 elements,
and we have
a1 + a2 = s
a1 * a2 = m
b1 + b2 = s
b1 * b2 = m
then {a1, a2} == {b1, b2}
if this is true, it's true for arrays have n-elements.
So we compare the sum and product of each array, if they equal, one is the permutation
of the other.
i have one problem
for example we have array
int a[]=new int[]{a1,a2,a3,a4,..........an}:
task is fill the same array by elements which are not in the array
for example
a={1,3,4,5,6,7} should be filled by any numbers {2,8,9,12,13,90}or others but not by elements which are in array this must not be{1,12,13,14,110} because 1 is in the array a
thanks
Interesting problem.
If the array is of signed integers, I believe it is possible in O(n) time and O(1) space, with no overflows, assuming the length is small enough to permit such a thing to happen.
The basic idea is as follows:
We have n numbers. Now on dividing those numbers by n+1, we get n remainders. So atleast one of the remainders in {0,1,2, ..., n} must be missing (say r). We fill the array with numbers whose remainders are r.
First, we add a multiple of n+1 to all negative numbers to make them positive.
Next we walk the array and find the remainder of each number with n+1. If remainder is r, we set a[r] to be -a[r] if a[r] was positive. (If we encounter negative numbers when walking, we use the negated version when taking remainder).
We also have an extra int for remainder = n.
At the end, we walk the array again to see if there are any positive numbers (there will be one, or the extra int for remainder = n will be unset).
Once we have the remainder, it is easy to generate n numbers with that remainder. Of course we could always generate just one number and fill it with that, as the problem never said anything about unique numbers.
If the array was of unsigned integers, we could probably still do this with better book-keeping.
For instance we could try using the first n/logn integers as our bitarray to denote which remainders have been seen and use some extra O(1) integers to hold the numbers temporarily.
For eg, you do tmp = a[0], find remainder and set the appropriate bit of a[0] (after setting it to zero first). tmp = a[1], set bit etc. We will never overwrite a number before we need it to find its remainder.
Just get the highest and lowest number in the array, create a new array with elements from your lower bound value to n.
Obtaining the highest and lowest number can be done in the same loop.
Assuming 12,4,3,5,7,8,89, it'll detect 3 as the lowest, 89 as the highest value. It then creates a new array and fills it with 3..89; then discard the old array.