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I have been attempting to solve the following problem:
You are given an array of n+1 integers where all the elements lies in [1,n]. You are also given that one of the elements is duplicated a certain number of times, whilst the others are distinct. Develop an algorithm to find both the duplicated number and the number of times it is duplicated.
Here is my solution where I let k = number of duplications:
struct LatticePoint{ // to hold duplicate and k
int a;
int b;
LatticePoint(int a_, int b_) : a(a_), b(b_) {}
}
LatticePoint findDuplicateAndK(const std::vector<int>& A){
int n = A.size() - 1;
std::vector<int> Numbers (n);
for(int i = 0; i < n + 1; ++i){
++Numbers[A[i] - 1]; // A[i] in range [1,n] so no out-of-access
}
int i = 0;
while(i < n){
if(Numbers[i] > 1) {
int duplicate = i + 1;
int k = Numbers[i] - 1;
LatticePoint result{duplicate, k};
return LatticePoint;
}
So, the basic idea is this: we go along the array and each time we see the number A[i] we increment the value of Numbers[A[i]]. Since only the duplicate appears more than once, the index of the entry of Numbers with value greater than 1 must be the duplicate number with the value of the entry the number of duplications - 1. This algorithm of O(n) in time complexity and O(n) in space.
I was wondering if someone had a solution that is better in time and/or space? (or indeed if there are any errors in my solution...)
You can reduce the scratch space to n bits instead of n ints, provided you either have or are willing to write a bitset with run-time specified size (see boost::dynamic_bitset).
You don't need to collect duplicate counts until you know which element is duplicated, and then you only need to keep that count. So all you need to track is whether you have previously seen the value (hence, n bits). Once you find the duplicated value, set count to 2 and run through the rest of the vector, incrementing count each time you hit an instance of the value. (You initialise count to 2, since by the time you get there, you will have seen exactly two of them.)
That's still O(n) space, but the constant factor is a lot smaller.
The idea of your code works.
But, thanks to the n+1 elements, we can achieve other tradeoffs of time and space.
If we have some number of buckets we're dividing numbers between, putting n+1 numbers in means that some bucket has to wind up with more than expected. This is a variant on the well-known pigeonhole principle.
So we use 2 buckets, one for the range 1..floor(n/2) and one for floor(n/2)+1..n. After one pass through the array, we know which half the answer is in. We then divide that half into halves, make another pass, and so on. This leads to a binary search which will get the answer with O(1) data, and with ceil(log_2(n)) passes, each taking time O(n). Therefore we get the answer in time O(n log(n)).
Now we don't need to use 2 buckets. If we used 3, we'd take ceil(log_3(n)) passes. So as we increased the fixed number of buckets, we take more space and save time. Are there other tradeoffs?
Well you showed how to do it in 1 pass with n buckets. How many buckets do you need to do it in 2 passes? The answer turns out to be at least sqrt(n) bucekts. And 3 passes is possible with the cube root. And so on.
So you get a whole family of tradeoffs where the more buckets you have, the more space you need, but the fewer passes. And your solution is merely at the extreme end, taking the most spaces and the least time.
Here's a cheekier algorithm, which requires only constant space but rearranges the input vector. (It only reorders; all the original elements are still present at the end.)
It's still O(n) time, although that might not be completely obvious.
The idea is to try to rearrange the array so that A[i] is i, until we find the duplicate. The duplicate will show up when we try to put an element at the right index and it turns out that that index already holds that element. With that, we've found the duplicate; we have a value we want to move to A[j] but the same value is already at A[j]. We then scan through the rest of the array, incrementing the count every time we find another instance.
#include <utility>
#include <vector>
std::pair<int, int> count_dup(std::vector<int> A) {
/* Try to put each element in its "home" position (that is,
* where the value is the same as the index). Since the
* values start at 1, A[0] isn't home to anyone, so we start
* the loop at 1.
*/
int n = A.size();
for (int i = 1; i < n; ++i) {
while (A[i] != i) {
int j = A[i];
if (A[j] == j) {
/* j is the duplicate. Now we need to count them.
* We have one at i. There's one at j, too, but we only
* need to add it if we're not going to run into it in
* the scan. And there might be one at position 0. After that,
* we just scan through the rest of the array.
*/
int count = 1;
if (A[0] == j) ++count;
if (j < i) ++count;
for (++i; i < n; ++i) {
if (A[i] == j) ++count;
}
return std::make_pair(j, count);
}
/* This swap can only happen once per element. */
std::swap(A[i], A[j]);
}
}
/* If we get here, every element from 1 to n is at home.
* So the duplicate must be A[0], and the duplicate count
* must be 2.
*/
return std::make_pair(A[0], 2);
}
A parallel solution with O(1) complexity is possible.
Introduce an array of atomic booleans and two atomic integers called duplicate and count. First set count to 1. Then access the array in parallel at the index positions of the numbers and perform a test-and-set operation on the boolean. If a boolean is set already, assign the number to duplicate and increment count.
This solution may not always perform better than the suggested sequential alternatives. Certainly not if all numbers are duplicates. Still, it has constant complexity in theory. Or maybe linear complexity in the number of duplicates. I am not quite sure. However, it should perform well when using many cores and especially if the test-and-set and increment operations are lock-free.
Is it possible to create arrays based of their index as in
int x = 4;
int y = 5;
int someNr = 123;
int foo[x][y] = someNr;
dynamically/on the run, without creating foo[0...3][0...4]?
If not, is there a data structure that allow me to do something similar to this in C?
No.
As written your code make no sense at all. You need foo to be declared somewhere and then you can index into it with foo[x][y] = someNr;. But you cant just make foo spring into existence which is what it looks like you are trying to do.
Either create foo with correct sizes (only you can say what they are) int foo[16][16]; for example or use a different data structure.
In C++ you could do a map<pair<int, int>, int>
Variable Length Arrays
Even if x and y were replaced by constants, you could not initialize the array using the notation shown. You'd need to use:
int fixed[3][4] = { someNr };
or similar (extra braces, perhaps; more values perhaps). You can, however, declare/define variable length arrays (VLA), but you cannot initialize them at all. So, you could write:
int x = 4;
int y = 5;
int someNr = 123;
int foo[x][y];
for (int i = 0; i < x; i++)
{
for (int j = 0; j < y; j++)
foo[i][j] = someNr + i * (x + 1) + j;
}
Obviously, you can't use x and y as indexes without writing (or reading) outside the bounds of the array. The onus is on you to ensure that there is enough space on the stack for the values chosen as the limits on the arrays (it won't be a problem at 3x4; it might be at 300x400 though, and will be at 3000x4000). You can also use dynamic allocation of VLAs to handle bigger matrices.
VLA support is mandatory in C99, optional in C11 and C18, and non-existent in strict C90.
Sparse arrays
If what you want is 'sparse array support', there is no built-in facility in C that will assist you. You have to devise (or find) code that will handle that for you. It can certainly be done; Fortran programmers used to have to do it quite often in the bad old days when megabytes of memory were a luxury and MIPS meant millions of instruction per second and people were happy when their computer could do double-digit MIPS (and the Fortran 90 standard was still years in the future).
You'll need to devise a structure and a set of functions to handle the sparse array. You will probably need to decide whether you have values in every row, or whether you only record the data in some rows. You'll need a function to assign a value to a cell, and another to retrieve the value from a cell. You'll need to think what the value is when there is no explicit entry. (The thinking probably isn't hard. The default value is usually zero, but an infinity or a NaN (not a number) might be appropriate, depending on context.) You'd also need a function to allocate the base structure (would you specify the maximum sizes?) and another to release it.
Most efficient way to create a dynamic index of an array is to create an empty array of the same data type that the array to index is holding.
Let's imagine we are using integers in sake of simplicity. You can then stretch the concept to any other data type.
The ideal index depth will depend on the length of the data to index and will be somewhere close to the length of the data.
Let's say you have 1 million 64 bit integers in the array to index.
First of all you should order the data and eliminate duplicates. That's something easy to achieve by using qsort() (the quick sort C built in function) and some remove duplicate function such as
uint64_t remove_dupes(char *unord_arr, char *ord_arr, uint64_t arr_size)
{
uint64_t i, j=0;
for (i=1;i<arr_size;i++)
{
if ( strcmp(unord_arr[i], unord_arr[i-1]) != 0 ){
strcpy(ord_arr[j],unord_arr[i-1]);
j++;
}
if ( i == arr_size-1 ){
strcpy(ord_arr[j],unord_arr[i]);
j++;
}
}
return j;
}
Adapt the code above to your needs, you should free() the unordered array when the function finishes ordering it to the ordered array. The function above is very fast, it will return zero entries when the array to order contains one element, but that's probably something you can live with.
Once the data is ordered and unique, create an index with a length close to that of the data. It does not need to be of an exact length, although pledging to powers of 10 will make everything easier, in case of integers.
uint64_t* idx = calloc(pow(10, indexdepth), sizeof(uint64_t));
This will create an empty index array.
Then populate the index. Traverse your array to index just once and every time you detect a change in the number of significant figures (same as index depth) to the left add the position where that new number was detected.
If you choose an indexdepth of 2 you will have 10² = 100 possible values in your index, typically going from 0 to 99.
When you detect that some number starts by 10 (103456), you add an entry to the index, let's say that 103456 was detected at position 733, your index entry would be:
index[10] = 733;
Next entry begining by 11 should be added in the next index slot, let's say that first number beginning by 11 is found at position 2023
index[11] = 2023;
And so on.
When you later need to find some number in your original array storing 1 million entries, you don't have to iterate the whole array, you just need to check where in your index the first number starting by the first two significant digits is stored. Entry index[10] tells you where the first number starting by 10 is stored. You can then iterate forward until you find your match.
In my example I employed a small index, thus the average number of iterations that you will need to perform will be 1000000/100 = 10000
If you enlarge your index to somewhere close the length of the data the number of iterations will tend to 1, making any search blazing fast.
What I like to do is to create some simple algorithm that tells me what's the ideal depth of the index after knowing the type and length of the data to index.
Please, note that in the example that I have posed, 64 bit numbers are indexed by their first index depth significant figures, thus 10 and 100001 will be stored in the same index segment. That's not a problem on its own, nonetheless each master has his small book of secrets. Treating numbers as a fixed length hexadecimal string can help keeping a strict numerical order.
You don't have to change the base though, you could consider 10 to be 0000010 to keep it in the 00 index segment and keep base 10 numbers ordered, using different numerical bases is nonetheless trivial in C, which is of great help for this task.
As you make your index depth become larger, the amount of entries per index segment will be reduced
Please, do note that programming, especially lower level like C consists in comprehending the tradeof between CPU cycles and memory use in great part.
Creating the proposed index is a way to reduce the number of CPU cycles required to locate a value at the cost of using more memory as the index becomes larger. This is nonetheless the way to go nowadays, as masive amounts of memory are cheap.
As SSDs' speed become closer to that of RAM, using files to store indexes is to be taken on account. Nevertheless modern OSs tend to load in RAM as much as they can, thus using files would end up in something similar from a performance point of view.
I am struggling to decide between two optimisations for building a numerical solver for the poisson equation.
Essentially, I have a two dimensional array, of which I require n doubles in the first row, n/2 in the second n/4 in the third and so on...
Now my difficulty is deciding whether or not to use a contiguous 2d array grid[m][n], which for a large n would have many unused zeroes but would probably reduce the chance of a cache miss. The other, and more memory efficient method, would be to dynamically allocate an array of pointers to arrays of decreasing size. This is considerably more efficient in terms of memory storage but would it potentially hinder performance?
I don't think I clearly understand the trade-offs in this situation. Could anybody help?
For reference, I made a nice plot of the memory requirements in each case:
There is no hard and fast answer to this one. If your algorithm needs more memory than you expect to be given then you need to find one which is possibly slower but fits within your constraints.
Beyond that, the only option is to implement both and then compare their performance. If saving memory results in a 10% slowdown is that acceptable for your use? If the version using more memory is 50% faster but only runs on the biggest computers will it be used? These are the questions that we have to grapple with in Computer Science. But you can only look at them once you have numbers. Otherwise you are just guessing and a fair amount of the time our intuition when it comes to optimizations are not correct.
Build a custom array that will follow the rules you have set.
The implementation will use a simple 1d contiguous array. You will need a function that will return the start of array given the row. Something like this:
int* Get( int* array , int n , int row ) //might contain logical errors
{
int pos = 0 ;
while( row-- )
{
pos += n ;
n /= 2 ;
}
return array + pos ;
}
Where n is the same n you described and is rounded down on every iteration.
You will have to call this function only once per entire row.
This function will never take more that O(log n) time, but if you want you can replace it with a single expression: http://en.wikipedia.org/wiki/Geometric_series#Formula
You could use a single array and just calculate your offset yourself
size_t get_offset(int n, int row, int column) {
size_t offset = column;
while (row--) {
offset += n;
n << 1;
}
return offset;
}
double * array = calloc(sizeof(double), get_offset(n, 64, 0));
access via
array[get_offset(column, row)]
I have an array of doubles which is the result of the FFT applied on an array, that contains the audio data of a Wav audio file in which i have added a 1000Hz tone.
I obtained this array thought the DREALFT defined in "Numerical Recipes".(I must use it).
(The original array has a length that is power of two.)
Mine array has this structure:
array[0] = first real valued component of the complex transform
array[1] = last real valued component of the complex transform
array[2] = real part of the second element
array[3] = imaginary part of the second element
etc......
Now, i know that this array represent the frequency domain.
I want to determine and kill the 1000Hz frequency.
I have tried this formula for finding the index of the array which should contain the 1000Hz frequency:
index = 1000. * NElements /44100;
Also, since I assume that this index refers to an array with real values only, i have determined the correct(?) position in my array, that contains imaginary values too:
int correctIndex=2;
for(k=0;k<index;k++){
correctIndex+=2;
}
(I know that surely there is a way easier but it is the first that came to mind)
Then, i find this value: 16275892957.123705, which i suppose to be the real part of the 1000Hz frequency.(Sorry if this is an imprecise affermation but at the moment I do not care to know more about it)
So i have tried to suppress it:
array[index]=-copy[index]*0.1f;
I don't know exactly why i used this formula but is the only one that gives some results, in fact the 1000hz tone appears to decrease slightly.
This is the part of the code in question:
double *copy = malloc( nCampioni * sizeof(double));
int nSamples;
/*...Fill copy with audio data...*/
/*...Apply ZERO PADDING and reach the length of 8388608 samples,
or rather 8388608 double values...*/
/*Apply the FFT (Sure this works)*/
drealft(copy - 1, nSamples, 1);
/*I determine the REAL(?) array index*/
i= 1000. * nSamples /44100;
/*I determine MINE(?) array index*/
int j=2;
for(k=0;k<i;k++){
j+=2;
}
/*I reduce the array value, AND some other values aroud it as an attempt*/
for(i=-12;i<12;i+=2){
copy[j-i]=-copy[i-j]*0.1f;
printf("%d\n",j-i);
}
/*Apply the inverse FFT*/
drealft(copy - 1, nSamples, -1);
/*...Write the audio data on the file...*/
NOTE: for simplicity I omitted the part where I get an array of double from an array of int16_t
How can i determine and totally kill the 1000Hz frequency?
Thank you!
As Oli Charlesworth writes, because your target frequency is not exactly one of the FFT bins (your index, TargetFrequency * NumberOfElements / SamplingRate, is not exactly an integer), the energy of the target frequency will be spread across all bins. For a start, you can eliminate some of the frequency by zeroing the bin closest to the target frequency. This will of course affect other frequencies somewhat too, since it is slightly off target. To better suppress the target frequency, you will need to consider a more sophisticated filter.
However, for educational purposes: To suppress the frequency corresponding to a bin, simply set that bin to zero. You must set both the real and the imaginary components of the bin to zero, which you can do with:
copy[index*2 + 0] = 0;
copy[index*2 + 1] = 1;
Some notes about this:
You had this code to calculate the position in the array:
int correctIndex = 2;
for (k = 0; k < index; k++) {
correctIndex += 2;
}
That is equivalent to:
correctIndex = 2*(index+1);
I believe you want 2*index, not 2*(index+1). So you were likely reducing the wrong bin.
At one point in your question, you wrote array[index] = -copy[index]*0.1f;. I do not know what array is. You appeared to be working in place in copy. I also do not know why you multiplied by 1/10. If you want to eliminate a frequency, just set it to zero. Multiplying it by 1/10 only reduces it to 10% of its original magnitude.
I understand that you must pass copy-1 to drealft because the Numerical Recipes code uses one-based indexing. However, the C standard does not support the way you are doing it. The behavior of the expression copy-1 is not defined by the standard. It will work in most C implementations. However, to write supported portable code, you should do this instead:
// Allocate one extra element.
double *memory = malloc((nCampioni+1) * sizeof *memory);
// Make a pointer that is convenient for your work.
double *copy = memory+1;
…
// Pass the necessary base address to drealft.
drealft(memory, nSamples, 1);
// Suppress a frequency.
copy[index*2 + 0] = 0;
copy[index*2 + 1] = 0;
…
// Free the memory.
free(memory);
One experiment I suggest you consider is to initialize an array with just a sine wave at the desired frequency:
for (i = 0; i < nSamples; ++i)
copy[i] = sin(TwoPi * Frequency / SampleRate * i);
(TwoPi is of course 2*3.1415926535897932384626433.) Then apply drealft and look at the results. You will see that much of the energy is at a peak in the closest bin to the target frequency, but much of it has also spread to other bins. Clearly, zeroing a single bin and performing the inverse FFT cannot eliminate all of the frequency. Also, you should see that the peak is in the same bin you calculated for index. If it is not, something is wrong.
[Description] Given two integer arrays with the same length. Design an algorithm which can judge whether they're the same. The definition of "same" is that, if these two arrays were in sorted order, the elements in corresponding position should be the same.
[Example]
<1 2 3 4> = <3 1 2 4>
<1 2 3 4> != <3 4 1 1>
[Limitation] The algorithm should require constant extra space, and O(n) running time.
(Probably too complex for an interview question.)
(You can use O(N) time to check the min, max, sum, sumsq, etc. are equal first.)
Use no-extra-space radix sort to sort the two arrays in-place. O(N) time complexity, O(1) space.
Then compare them using the usual algorithm. O(N) time complexity, O(1) space.
(Provided (max − min) of the arrays is of O(Nk) with a finite k.)
You can try a probabilistic approach - convert the arrays into a number in some huge base B and mod by some prime P, for example sum B^a_i for all i mod some big-ish P. If they both come out to the same number, try again for as many primes as you want. If it's false at any attempts, then they are not correct. If they pass enough challenges, then they are equal, with high probability.
There's a trivial proof for B > N, P > biggest number. So there must be a challenge that cannot be met. This is actually the deterministic approach, though the complexity analysis might be more difficult, depending on how people view the complexity in terms of the size of the input (as opposed to just the number of elements).
I claim that: Unless the range of input is specified, then it is IMPOSSIBLE to solve in onstant extra space, and O(n) running time.
I will be happy to be proven wrong, so that I can learn something new.
Insert all elements from the first array into a hashtable
Try to insert all elements from the second array into the same hashtable - for each insert to element should already be there
Ok, this is not with constant extra space, but the best I could come up at the moment:-). Are there any other constraints imposed on the question, like for example to biggest integer that may be included in the array?
A few answers are basically correct, even though they don't look like it. The hash table approach (for one example) has an upper limit based on the range of the type involved rather than the number of elements in the arrays. At least by by most definitions, that makes the (upper limit on) the space a constant, although the constant may be quite large.
In theory, you could change that from an upper limit to a true constant amount of space. Just for example, if you were working in C or C++, and it was an array of char, you could use something like:
size_t counts[UCHAR_MAX];
Since UCHAR_MAX is a constant, the amount of space used by the array is also a constant.
Edit: I'd note for the record that a bound on the ranges/sizes of items involved is implicit in nearly all descriptions of algorithmic complexity. Just for example, we all "know" that Quicksort is an O(N log N) algorithm. That's only true, however, if we assume that comparing and swapping the items being sorted takes constant time, which can only be true if we bound the range. If the range of items involved is large enough that we can no longer treat a comparison or a swap as taking constant time, then its complexity would become something like O(N log N log R), were R is the range, so log R approximates the number of bits necessary to represent an item.
Is this a trick question? If the authors assumed integers to be within a given range (2^32 etc.) then "extra constant space" might simply be an array of size 2^32 in which you count the occurrences in both lists.
If the integers are unranged, it cannot be done.
You could add each element into a hashmap<Integer, Integer>, with the following rules: Array A is the adder, array B is the remover. When inserting from Array A, if the key does not exist, insert it with a value of 1. If the key exists, increment the value (keep a count). When removing, if the key exists and is greater than 1, reduce it by 1. If the key exists and is 1, remove the element.
Run through array A followed by array B using the rules above. If at any time during the removal phase array B does not find an element, you can immediately return false. If after both the adder and remover are finished the hashmap is empty, the arrays are equivalent.
Edit: The size of the hashtable will be equal to the number of distinct values in the array does this fit the definition of constant space?
I imagine the solution will require some sort of transformation that is both associative and commutative and guarantees a unique result for a unique set of inputs. However I'm not sure if that even exists.
public static boolean match(int[] array1, int[] array2) {
int x, y = 0;
for(x = 0; x < array1.length; x++) {
y = x;
while(array1[x] != array2[y]) {
if (y + 1 == array1.length)
return false;
y++;
}
int swap = array2[x];
array2[x] = array2[y];
array2[y] = swap;
}
return true;
}
For each array, Use Counting sort technique to build the count of number of elements less than or equal to a particular element . Then compare the two built auxillary arrays at every index, if they r equal arrays r equal else they r not . COunting sort requires O(n) and array comparison at every index is again O(n) so totally its O(n) and the space required is equal to the size of two arrays . Here is a link to counting sort http://en.wikipedia.org/wiki/Counting_sort.
given int are in the range -n..+n a simple way to check for equity may be the following (pseudo code):
// a & b are the array
accumulator = 0
arraysize = size(a)
for(i=0 ; i < arraysize; ++i) {
accumulator = accumulator + a[i] - b[i]
if abs(accumulator) > ((arraysize - i) * n) { return FALSE }
}
return (accumulator == 0)
accumulator must be able to store integer with range = +- arraysize * n
How 'bout this - XOR all the numbers in both the arrays. If the result is 0, you got a match.