What do these terminologies mean in C++?
1. off the end value
2. half open range - [begin, off_the_end)
I came across them while reading about for loops.
A half-open range is one which includes the first element, but excludes the last one.
The range [1,5) is half-open, and consists of the values 1, 2, 3 and 4.
"off the end" or "past the end" refers to the element just after the end of a sequence, and is special in that iterators are allowed to point to it (but you may not look at the actual value, because it doesn't exist)
For example, in the following code:
char arr[] = {'a', 'b', 'c', 'd'};
char* first = arr
char* last = arr + 4;
first now points to the first element of the array, while last points one past the end of the array. We are allowed to point one past the end of the array (but not two past), but we're not allowed to try to access the element at that position:
// legal, because first points to a member of the array
char firstChar = *first;
// illegal because last points *past* the end of the array
char lastChar = *last;
Our two pointers, first and last together define a range, of all the elements between them.
If it is a half open range, then it contains the element pointed to by first, and all the elements in between, but not the element pointed to by last (which is good, because it doesn't actually point to a valid element)
In C++, all the standard library algorithms operate on such half open ranges. For example, if I want to copy the entire array to some other location dest, I do this:
std::copy(first, last, dest)
A simple for-loop typically follows a similar pattern:
for (int i = 0; i < 4; ++i) {
// do something with arr[i]
}
This loop goes from 0 to 4, but it excludes the end value, so the range of indices covered is half-open, specifically [0, 4)
These aren't C++ specific terms, they are general maths terms.
[] and () denote whether the range is inclusive/exclusive of the endpoint:
[ includes the endpoint
( excludes the endpoint
[] = 'Closed', includes both endpoints
() = 'Open', excludes both endpoints
[) and (] are both 'half-open', and include only one endpoint
Most C++ for-loops cover a half-open range (you include the first element: e.g for int i=0;, but exclude the final element: i < foo, not i ≤ foo)
As explained on other answers, half-open range is also a mathematical term and usage of this term in programming context, it is implied that the starting point is included and end point is excluded.
What does it actually mean in the context of programming in C/C++ ? Let's say, you are going to print the elements of an integer array. Speaking for the C language, because that you have no any run-time knowledge for the size of the array, you have two choice. Either you have to provide the size of the array and thus, the function signature will be as below;
void printArray(int * array, int size);
or you have to use the half-open range, which means, you have to provide both begin and end pointer (and function is going to process including the begin, excluding the end) additional to the array itself. And the function signature will be as below;
void printArray(int * array, int * begin, int * end);
To illustrate, here is an example for providing the size of the array;
#include <stdio.h>
void printArray(int * array, int size)
{
printf("Array: ");
for(int i = 0; i < size; i++)
printf("%2d ", array[i]);
printf("\n");
}
int main()
{
int array[5] = { 1, 2, 3, 4, 5 };
printArray(array, 5);
return 0;
}
In the example above, we have passed two parameters to the printArray function as it is obvious on the function signature, the pointer to the first element of the array (or the array itself), and the size of the array.
However, as I have written above, we can also use the half-opening range in the function signature which can be seen as below;
#include <stdio.h>
void printArray(int * array, int * begin, int * end)
{
printf("Array: ");
for(int * index = begin; index != end; index++)
printf("%2d ", *index);
printf("\n");
}
int main()
{
int array[5] = { 1, 2, 3, 4, 5 };
printArray(array, array, array+5);
return 0;
}
Both of the code will produce the same output as can be seen below;
Array: 1 2 3 4 5
As you can see, the printArray function prints the function for the range [begin, end). The index which is actually is a pointer to the elements of the integer array, starts from begin, and it includes the begin and the for-loop ends up when index equals to the end pointer, excluding to process the end. This i called half-open range.
Half-open range is the C++ convention.
Related
I have been trying a sorting method in which I subtract each number stored in an array by other elements in the same array. Then, I saw a pattern that the number of differences which come to be negative, is the rank or position of element in the Sorted one. But, things go wrong when I am using repeated entries.
My basic method is :
Take every element of the SampleArray.
subtract it from every element of the SampleArray
check if the difference comes to be negative.
if it is then, increase a variable called counter.
And use this counter as the position of element in sorted array.
For example: lets take (5,2,6,4)
first take 5, subtract it from each of the numbers which will give results (0,-3,1,-1), so counter will become 2, which will be the index of 5 in the sorted Array. And repeat it for each of the elements.
for 5, counter will be 2.
for 2, counter will be 0.
for 6, counter will be 3.
for 4, counter will be 1.
And hence the sorted Array will be {2,4,5,6}.
First, see the code :
#include <iostream>
using namespace std;
void sorting(int myArray[], int sizeofArray);
int main()
{
int checkArray[] = {5,4,2,20,12,13,8,6,10,15,0}; //my sample Arry
int sized;
sized=sizeof checkArray/sizeof(int);//to know the size
cout << sized << endl;
sorting(checkArray, sized);
}
void sorting(int myArray[], int sizeofArray)
{
int tempArray[sizeofArray];
for (int i=0; i<sizeofArray; i++)
{
int counter=0;
for (int j=0;j<sizeofArray; j++ )
{
int checkNum = myArray[j]-myArray[i];
if (checkNum<0)
counter++; //to know the numbers of negatives
else
counter+=0;
}
tempArray[counter]=myArray[i];
}
for (int x=0;x<sizeofArray; x++)
{
cout << tempArray[x] << " " ;
}
}
Now, if we run this program with entries with no repetitions then, it sorts out the array, But if we use repeated entries like
int checkArray[] = {8,2,4,4,6}
the tempArray gets its first element as 2 as counter will be zero.
the tempArray gets its second element as 4 as counter will be 1.
but, the tempArray can't get its third one as counter will be still 1, and thus prints some randomNo in place of this. (here the things go wrong).
Can you please suggest a method to solve this?
This is an odd way of writing insertion sort, https://en.wikipedia.org/wiki/Insertion_sort
I would assume you can change your condition to:
if (checkNum<0 || (checkNum==0 && j<i))
But I would suggest using a proper sorting routine instead
The idea is to separate duplicates by saying that if the values are the same we sort according to their order in the sequence; as if the sequence was a pair of the value and the sequence number (0, 1, 2, 3, 4, 5, ...).
The issue here is that for any 2 equally sized numbers the nested loop will return the same counter value. Thus for such a counter value tempArray[counter + 1] will never be initialized.
The way to solve this would be to maintain a vector<bool> denoting what each position had been written and write to the next valid position if that is the case.
But supporting a second vector is just going to make your O(n2) code slower. Consider using sort instead:
sort(begin(checkArray), end(checkArray))
I am learning C++ and one of my practice exercises is to use pointers to set all the elements in an array to 0. I have no idea how to do this by incrementing the pointer to the next position in the array since my IDE log said that comparison between int and * is forbidden. I only need a small snippet as an example to help me better understand where i'm going wrong. The array I have created is of type int and has a single dimension with 5 elements consisting of 1,2,3,4 and 5.
int array[5] = {1, 2, 3, 4, 5};
for(int *i = &array[0], *end = &array[5]; i != end; i++)
*i = 0;
The code creates a pointer to the start &array[0] and a pointer to one position past the end &array[5]
Then it steps the pointer through the array, setting each element to zero.
A more advanced concept that is very similar is iterators.
You could use std::fill, http://en.cppreference.com/w/cpp/algorithm/fill, as follows.
const size_t dataSize = 10;
int data[dataSize];
std::fill(data, data + dataSize, 0);
I have two integer arrays
int A[] = {2, 4, 3, 5, 6, 7};
int B[] = {9, 2, 7, 6};
And i have to get intersection of these array.
i.e. output will be - 2,6,7
I am thinking to sove it by saving array A in a data strcture and then i want to compare all the element till size A or B and then i will get intersection.
Now i have a problem i need to first store the element of Array A in a container.
shall i follow like -
int size = sizeof(A)/sizeof(int);
To get the size but by doing this i will get size after that i want to access all the elemts too and store in a container.
Here i the code which i am using to find Intersection ->
#include"iostream"
using namespace std;
int A[] = {2, 4, 3, 5, 6, 7};
int B[] = {9, 2, 7, 6};
int main()
{
int sizeA = sizeof(A)/sizeof(int);
int sizeB = sizeof(B)/sizeof(int);
int big = (sizeA > sizeB) ? sizeA : sizeB;
int small = (sizeA > sizeB) ? sizeB : sizeA;
for (int i = 0; i <big ;++i)
{
for (int j = 0; j <small ; ++j)
{
if(A[i] == B[j])
{
cout<<"Element is -->"<<A[i]<<endl;
}
}
}
return 0;
}
Just use a hash table:
#include <unordered_set> // needs C++11 or TR1
// ...
unordered_set<int> setOfA(A, A + sizeA);
Then you can just check for every element in B, whether it's also in A:
for (int i = 0; i < sizeB; ++i) {
if (setOfA.find(B[i]) != setOfA.end()) {
cout << B[i] << endl;
}
}
Runtime is expected O(sizeA + sizeB).
You can sort the two arrays
sort(A, A+sizeA);
sort(B, B+sizeB);
and use a merge-like algorithm to find their intersection:
#include <vector>
...
std::vector<int> intersection;
int idA=0, idB=0;
while(idA < sizeA && idB < sizeB) {
if (A[idA] < B[idB]) idA ++;
else if (B[idB] < A[idA]) idB ++;
else { // => A[idA] = B[idB], we have a common element
intersection.push_back(A[idA]);
idA ++;
idB ++;
}
}
The time complexity of this part of the code is linear. However, due to the sorting of the arrays, the overall complexity becomes O(n * log n), where n = max(sizeA, sizeB).
The additional memory required for this algorithm is optimal (equal to the size of the intersection).
saving array A in a data strcture
Arrays are data structures; there's no need to save A into one.
i want to compare all the element till size A or B and then i will get intersection
This is extremely vague but isn't likely to yield the intersection; notice that you must examine every element in both A and B but "till size A or B" will ignore elements.
What approach i should follow to get size of an unkown size array and store it in a container??
It isn't possible to deal with arrays of unknown size in C unless they have some end-of-array sentinel that allows counting the number of elements (as is the case with NUL-terminated character arrays, commonly referred to in C as "strings"). However, the sizes of your arrays are known because their compile-time sizes are known. You can calculate the number of elements in such arrays with a macro:
#define ARRAY_ELEMENT_COUNT(a) (sizeof(a)/sizeof *(a))
...
int *ptr = new sizeof(A);
[Your question was originally tagged [C], and my comments below refer to that]
This isn't valid C -- new is a C++ keyword.
If you wanted to make copies of your arrays, you could simply do it with, e.g.,
int Acopy[ARRAY_ELEMENT_COUNT(A)];
memcpy(Acopy, A, sizeof A);
or, if for some reason you want to put the copy on the heap,
int* pa = malloc(sizeof A);
if (!pa) /* handle out-of-memory */
memcpy(pa, A, sizeof A);
/* After you're done using pa: */
free(pa);
[In C++ you would used new and delete]
However, there's no need to make copies of your arrays in order to find the intersection, unless you need to sort them (see below) but also need to preserve the original order.
There are a few ways to find the intersection of two arrays. If the values fall within the range of 0-63, you can use two unsigned longs and set the bits corresponding to the values in each array, then use & (bitwise "and") to find the intersection. If the values aren't in that range but the difference between the largest and smallest is < 64, you can use the same method but subtract the smallest value from each value to get the bit number. If the range is not that small but the number of distinct values is <= 64, you can maintain a lookup table (array, binary tree, hash table, etc.) that maps the values to bit numbers and a 64-element array that maps bit numbers back to values.
If your arrays may contain more than 64 distinct values, there are two effective approaches:
1) Sort each array and then compare them element by element to find the common values -- this algorithm resembles a merge sort.
2) Insert the elements of one array into a fast lookup table (hash table, balanced binary tree, etc.), and then look up each element of the other array in the lookup table.
Sort both arrays (e.g., qsort()) and then walk through both arrays one element at a time.
Where there is a match, add it to a third array, which is sized to match the larger of the two input arrays (your result array can be no larger than the largest of the two arrays). Use a negative or other "dummy" value as your terminator.
When walking through input arrays, where one value in the first array is larger than the other, move the index of the second array, and vice versa.
When you're done walking through both arrays, your third array has your answer, up to the terminator value.
We have a sorted array and we would like to increase the value of one index by only 1 unit (array[i]++), such that the resulting array is still sorted. Is this possible in O(1)?
It is fine to use any data structure possible in STL and C++.
In a more specific case, if the array is initialised by all 0 values, and it is always incrementally constructed only by increasing a value of an index by one, is there an O(1) solution?
I haven't worked this out completely, but I think the general idea might help for integers at least. At the cost of more memory, you can maintain a separate data-structure that maintains the ending index of a run of repeated values (since you want to swap your incremented value with the ending index of the repeated value). This is because it's with repeated values that you run into the worst case O(n) runtime: let's say you have [0, 0, 0, 0] and you increment the value at location 0. Then it is O(n) to find out the last location (3).
But let's say that you maintain the data-structure I mentioned (a map would works because it has O(1) lookup). In that case you would have something like this:
0 -> 3
So you have a run of 0 values that end at location 3. When you increment a value, let's say at location i, you check to see if the new value is greater than the value at i + 1. If it is not, you are fine. But if it is, you look to see if there is an entry for this value in the secondary data-structure. If there isn't, you can simply swap. If there is an entry, you look up the ending-index and then swap with the value at that location. You then make any changes you need to the secondary data-structure to reflect the new state of the array.
A more thorough example:
[0, 2, 3, 3, 3, 4, 4, 5, 5, 5, 7]
The secondary data-structure is:
3 -> 4
4 -> 6
5 -> 9
Let's say you increment the value at location 2. So you have incremented 3, to 4. The array now looks like this:
[0, 2, 4, 3, 3, 4, 4, 5, 5, 5, 7]
You look at the next element, which is 3. You then look up the entry for that element in the secondary data-structure. The entry is 4, which means that there is a run of 3's that end at 4. This means that you can swap the value from the current location with the value at index 4:
[0, 2, 3, 3, 4, 4, 4, 5, 5, 5, 7]
Now you will also need to update the secondary data-structure. Specifically, there the run of 3's ends one index early, so you need to decrement that value:
3 -> 3
4 -> 6
5 -> 9
Another check you will need to do is to see if the value is repeated anymore. You can check that by looking at the i - 1th and the i + 1th locations to see if they are the same as the value in question. If neither are equal, then you can remove the entry for this value from the map.
Again, this is just a general idea. I will have to code it out to see if it works out the way I thought about it.
Please feel free to poke holes.
UPDATE
I have an implementation of this algorithm here in JavaScript. I used JavaScript just so I could do it quickly. Also, because I coded it up pretty quickly it can probably be cleaned up. I do have comments though. I'm not doing anything esoteric either, so this should be easily portable to C++.
There are essentially two parts to the algorithm: the incrementing and swapping (if necessary), and book-keeping done on the map that keeps track of our ending indices for runs of repeated values.
The code contains a testing harness that starts with an array of zeroes and increments random locations. At the end of every iteration, there is a test to ensure that the array is sorted.
var array = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0];
var endingIndices = {0: 9};
var increments = 10000;
for(var i = 0; i < increments; i++) {
var index = Math.floor(Math.random() * array.length);
var oldValue = array[index];
var newValue = ++array[index];
if(index == (array.length - 1)) {
//Incremented element is the last element.
//We don't need to swap, but we need to see if we modified a run (if one exists)
if(endingIndices[oldValue]) {
endingIndices[oldValue]--;
}
} else if(index >= 0) {
//Incremented element is not the last element; it is in the middle of
//the array, possibly even the first element
var nextIndexValue = array[index + 1];
if(newValue === nextIndexValue) {
//If the new value is the same as the next value, we don't need to swap anything. But
//we are doing some book-keeping later with the endingIndices map. That code requires
//the ending index (i.e., where we moved the incremented value to). Since we didn't
//move it anywhere, the endingIndex is simply the index of the incremented element.
endingIndex = index;
} else if(newValue > nextIndexValue) {
//If the new value is greater than the next value, we will have to swap it
var swapIndex = -1;
if(!endingIndices[nextIndexValue]) {
//If the next value doesn't have a run, then location we have to swap with
//is just the next index
swapIndex = index + 1;
} else {
//If the next value has a run, we get the swap index from the map
swapIndex = endingIndices[nextIndexValue];
}
array[index] = nextIndexValue;
array[swapIndex] = newValue;
endingIndex = swapIndex;
} else {
//If the next value is already greater, there is nothing we need to swap but we do
//need to do some book-keeping with the endingIndices map later, because it is
//possible that we modified a run (the value might be the same as the value that
//came before it). Since we don't have anything to swap, the endingIndex is
//effectively the index that we are incrementing.
endingIndex = index;
}
//Moving the new value to its new position may have created a new run, so we need to
//check for that. This will only happen if the new position is not at the end of
//the array, and the new value does not have an entry in the map, and the value
//at the position after the new position is the same as the new value
if(endingIndex < (array.length - 1) &&
!endingIndices[newValue] &&
array[endingIndex + 1] == newValue) {
endingIndices[newValue] = endingIndex + 1;
}
//We also need to check to see if the old value had an entry in the
//map because now that run has been shortened by one.
if(endingIndices[oldValue]) {
var newEndingIndex = --endingIndices[oldValue];
if(newEndingIndex == 0 ||
(newEndingIndex > 0 && array[newEndingIndex - 1] != oldValue)) {
//In this case we check to see if the old value only has one entry, in
//which case there is no run of values and so we will need to remove
//its entry from the map. This happens when the new ending-index for this
//value is the first location (0) or if the location before the new
//ending-index doesn't contain the old value.
delete endingIndices[oldValue];
}
}
}
//Make sure that the array is sorted
for(var j = 0; j < array.length - 1; j++) {
if(array[j] > array[j + 1]) {
throw "Array not sorted; Value at location " + j + "(" + array[j] + ") is greater than value at location " + (j + 1) + "(" + array[j + 1] + ")";
}
}
}
In a more specific case, if the array is initialised by all 0 values, and it is always incrementally constructed only by increasing a value of an index by one, is there an O(1) solution?
No. Given an array of all 0's: [0, 0, 0, 0, 0]. If you increment the first value, giving [1, 0, 0, 0, 0], then you will have to make 4 swaps to ensure that it remains sorted.
Given a sorted array with no duplicates, then the answer is yes. But after the first operation (i.e. the first time you increment), then you could potentially have duplicates. The more increments you do, the higher the likelihood is that you'll have duplicates, and the more likely it'll take O(n) to keep that array sorted.
If all you have is the array, it's impossible to guarantee less than O(n) time per increment. If what you're looking for is a data structure that supports sorted order and lookup by index, then you probably want an order stastic tree.
If the values are small, counting sort will work. Represent the array [0,0,0,0] as {4}. Incrementing any zero gives {3,1} : 3 zeroes and a one. In general, to increment any value x, deduct one from the count of x and increment the count of {x+1}. The space efficiency is O(N), though, where N is the highest value.
It depends on how many items can have the same value. If more items can have the same value, then it is not possible to have O(1) with ordinary arrays.
Let's do an example: suppose array[5] = 21, and you want to do array[5]++:
Increment the item:
array[5]++
(which is O(1) because it is an array).
So, now array[5] = 22.
Check the next item (i.e., array[6]):
If array[6] == 21, then you have to keep checking new items (i.e., array[7] and so on) until you find a value higher than 21. At that point you can swap the values. This search is not O(1) because potentially you have to scan the whole array.
Instead, if items cannot have the same value, then you have:
Increment the item:
array[5]++
(which is O(1) because it is an array).
So, now array[5] = 22.
The next item cannot be 21 (because two items cannot have the same value), so it must have a value > 21 and the array is already sorted.
So you take sorted array and hashtable. You go over array to figure out 'flat' areas - where elements are of the same value. For every flat area you have to figure out three things 1) where it starts (index of first element) 2) what is it's value 3) what is the value of next element (the next bigger). Then put this tuple into the hashtable, where the key will be element value. This is prerequisite and it's complexity doesn't really matter.
Then when you increase some element (index i) you look up a table for index of next bigger element (call it j), and swap i with i - 1. Then 1) add new entry to hashtable 2) update existing entry for it's previous value.
With perfect hashtable (or limited range of possible values) it will be almost O(1). The downside: it will not be stable.
Here is some code:
#include <iostream>
#include <unordered_map>
#include <vector>
struct Range {
int start, value, next;
};
void print_ht(std::unordered_map<int, Range>& ht)
{
for (auto i = ht.begin(); i != ht.end(); i++) {
Range& r = (*i).second;
std::cout << '(' << r.start << ", "<< r.value << ", "<< r.next << ") ";
}
std::cout << std::endl;
}
void increment_el(int i, std::vector<int>& array, std::unordered_map<int, Range>& ht)
{
int val = array[i];
array[i]++;
//Pick next bigger element
Range& r = ht[val];
//Do the swapping, so last element of that range will be first
std::swap(array[i], array[ht[r.next].start - 1]);
//Update hashtable
ht[r.next].start--;
}
int main(int argc, const char * argv[])
{
std::vector<int> array = {1, 1, 1, 2, 2, 3};
std::unordered_map<int, Range> ht;
int start = 0;
int value = array[0];
//Build indexing hashtable
for (int i = 0; i <= array.size(); i++) {
int cur_value = i < array.size() ? array[i] : -1;
if (cur_value > value || i == array.size()) {
ht[value] = {start, value, cur_value};
start = i;
value = cur_value;
}
}
print_ht(ht);
//Now let's increment first element
increment_el(0, array, ht);
print_ht(ht);
increment_el(3, array, ht);
print_ht(ht);
for (auto i = array.begin(); i != array.end(); i++)
std::cout << *i << " ";
return 0;
}
Yes and no.
Yes if the list contains only unique integers, as that means you only need to check the next value. No in any other situation. If the values are not unique, incrementing the first of N duplicate values means that it must move N positions. If the values are floating-point, you may have thousands of values between x and x+1
It's important to be very clear about the requirements; the simplest way is to express the problem as an ADT (Abstract Datatype), listing the required operations and complexities.
Here's what I think you are looking for: a datatype which provides the following operations:
Construct(n): Create a new object of size n all of whose values are 0.
Value(i): Return the value at index i.
Increment(i): Increment the value at index i.
Least(): Return the index of the element with least value (or one such element if there are several).
Next(i): Return the index of the next element after element i in a sorted traversal starting at Least(), such that the traversal will return every element.
Aside from the Constructor, we want every one of the above operations to have complexity O(1). We also want the object to occupy O(n) space.
The implementation uses a list of buckets; each bucket has a value and a list of elements. Each element has an index, a pointer to the bucket it is part of. Finally, we have an array of pointers to elements. (In C++, I'd probably use iterators rather than pointers; in another language, I'd probably use intrusive lists.) The invariants are that no bucket is ever empty, and the value of the buckets are strictly monotonically increasing.
We start with a single bucket with value 0 which has a list of n elements.
Value(i) is implemented by returning the value of the bucket of the element referenced by the iterator at element i of the array. Least() is the index of the first element in the first bucket. Next(i) is the index of the next element after the one referenced by the iterator at element i, unless that iterator is already pointing at the end of the the list in which case it is the first element in the next bucket, unless the element's bucket is the last bucket, in which case we're at the end of the element list.
The only interface of interest is Increment(i), which is as follows:
If element i is the only element in its bucket (i.e. there is no next element in the bucket list, and element i is the first element in the bucket list):
Increment the value of the associated bucket.
If the next bucket has the same value, append the next bucket's element list to this bucket's element list (this is O(1), regardless of the list's size, because it is just a pointer swap), and then delete the next bucket.
If element i is not the only element in its bucket, then:
Remove it from its bucket list.
If the next bucket has the next sequential value, then push element i onto the next bucket's list.
Otherwise, the next bucket's value is larger, then create a new bucket with the next sequential value and only element i and insert it between this bucket and the next one.
just iterate along the array from the modified element until you find the correct place, then swap. Average case complexity is O(N) where N is the average number of duplicates. Worst case is O(n) where n is the length of the array. As long as N isn't large and doesn't scale badly with n, you're fine and can probably pretend it's O(1) for practical purposes.
If duplicates are the norm and/or scale strongly with n, then there are better solutions, see other responses.
I think that it is possible without using a hashtable. I have an implementation here:
#include <cstdio>
#include <vector>
#include <cassert>
// This code is a solution for http://stackoverflow.com/questions/19957753/maintain-a-sorted-array-in-o1
//
// """We have a sorted array and we would like to increase the value of one index by only 1 unit
// (array[i]++), such that the resulting array is still sorted. Is this possible in O(1)?"""
// The obvious implementation, which has O(n) worst case increment.
class LinearIncrementor
{
public:
LinearIncrementor(int numElems);
int valueAt(int index) const;
void incrementAt(int index);
private:
std::vector<int> m_values;
};
// Free list to store runs of same values
class RunList
{
public:
struct Run
{
int m_end; // end index of run, inclusive, or next object in free list
int m_value; // value at this run
};
RunList();
int allocateRun(int endIndex, int value);
void freeRun(int index);
Run& runAt(int index);
const Run& runAt(int index) const;
private:
std::vector<Run> m_runs;
int m_firstFree;
};
// More optimal implementation, which increments in O(1) time
class ConstantIncrementor
{
public:
ConstantIncrementor(int numElems);
int valueAt(int index) const;
void incrementAt(int index);
private:
std::vector<int> m_runIndices;
RunList m_runs;
};
LinearIncrementor::LinearIncrementor(int numElems)
: m_values(numElems, 0)
{
}
int LinearIncrementor::valueAt(int index) const
{
return m_values[index];
}
void LinearIncrementor::incrementAt(int index)
{
const int n = static_cast<int>(m_values.size());
const int value = m_values[index];
while (index+1 < n && value == m_values[index+1])
++index;
++m_values[index];
}
RunList::RunList() : m_firstFree(-1)
{
}
int RunList::allocateRun(int endIndex, int value)
{
int runIndex = -1;
if (m_firstFree == -1)
{
runIndex = static_cast<int>(m_runs.size());
m_runs.resize(runIndex + 1);
}
else
{
runIndex = m_firstFree;
m_firstFree = m_runs[runIndex].m_end;
}
Run& run = m_runs[runIndex];
run.m_end = endIndex;
run.m_value = value;
return runIndex;
}
void RunList::freeRun(int index)
{
m_runs[index].m_end = m_firstFree;
m_firstFree = index;
}
RunList::Run& RunList::runAt(int index)
{
return m_runs[index];
}
const RunList::Run& RunList::runAt(int index) const
{
return m_runs[index];
}
ConstantIncrementor::ConstantIncrementor(int numElems) : m_runIndices(numElems, 0)
{
const int runIndex = m_runs.allocateRun(numElems-1, 0);
assert(runIndex == 0);
}
int ConstantIncrementor::valueAt(int index) const
{
return m_runs.runAt(m_runIndices[index]).m_value;
}
void ConstantIncrementor::incrementAt(int index)
{
const int numElems = static_cast<int>(m_runIndices.size());
const int curRunIndex = m_runIndices[index];
RunList::Run& curRun = m_runs.runAt(curRunIndex);
index = curRun.m_end;
const bool freeCurRun = index == 0 || m_runIndices[index-1] != curRunIndex;
RunList::Run* runToMerge = NULL;
int runToMergeIndex = -1;
if (curRun.m_end+1 < numElems)
{
const int nextRunIndex = m_runIndices[curRun.m_end+1];
RunList::Run& nextRun = m_runs.runAt(nextRunIndex);
if (curRun.m_value+1 == nextRun.m_value)
{
runToMerge = &nextRun;
runToMergeIndex = nextRunIndex;
}
}
if (freeCurRun && !runToMerge) // then free and allocate at the same time
{
++curRun.m_value;
}
else
{
if (freeCurRun)
{
m_runs.freeRun(curRunIndex);
}
else
{
--curRun.m_end;
}
if (runToMerge)
{
m_runIndices[index] = runToMergeIndex;
}
else
{
m_runIndices[index] = m_runs.allocateRun(index, curRun.m_value+1);
}
}
}
int main(int argc, char* argv[])
{
const int numElems = 100;
const int numInc = 1000000;
LinearIncrementor linearInc(numElems);
ConstantIncrementor constInc(numElems);
srand(1);
for (int i = 0; i < numInc; ++i)
{
const int index = rand() % numElems;
linearInc.incrementAt(index);
constInc.incrementAt(index);
for (int j = 0; j < numElems; ++j)
{
if (linearInc.valueAt(j) != constInc.valueAt(j))
{
printf("Error: differing values at increment step %d, value at index %d\n", i, j);
}
}
}
return 0;
}
As a complement to the other answers: if you can only have the array, then you cannot indeed guarantee the operation will be constant-time; but because the array is sorted, you can find the end of a run of identical numbers in log n operations, not in n operations. This is simply a binary search.
If we expect most runs of numbers to be short, we should use galloping search, which is a variant where we first find the bounds by looking at positions +1, +2, +4, +8, +16, etc. and then doing binary search inside. You would get a time that is often constant (and extremely fast if the item is unique) but can grow up to log n. Unless for some reason long runs of identical numbers remain common even after many updates, this might outperform any solution that requires keeping additional data.
I got this question at an interview and at the end was told there was a more efficient way to do this but have still not been able to figure it out. You are passing into a function an array of integers and an integer for size of array. In the array you have a lot of numbers, some that repeat for example 1,7,4,8,2,6,8,3,7,9,10. You want to take that array and return an array where all the repeated numbers are put at the end of the array so the above array would turn into 1,7,4,8,2,6,3,9,10,8,7. The numbers I used are not important and I could not use a buffer array. I was going to use a BST, but the order of the numbers must be maintained(except for the duplicate numbers). I could not figure out how to use a hash table so I ended up using a double for loop(n^2 horrible I know). How would I do this more efficiently using c++. Not looking for code, just an idea of how to do it better.
In what follows:
arr is the input array;
seen is a hash set of numbers already encountered;
l is the index where the next unique element will be placed;
r is the index of the next element to be considered.
Since you're not looking for code, here is a pseudo-code solution (which happens to be valid Python):
arr = [1,7,4,8,2,6,8,3,7,9,10]
seen = set()
l = 0
r = 0
while True:
# advance `r` to the next not-yet-seen number
while r < len(arr) and arr[r] in seen:
r += 1
if r == len(arr): break
# add the number to the set
seen.add(arr[r])
# swap arr[l] with arr[r]
arr[l], arr[r] = arr[r], arr[l]
# advance `l`
l += 1
print arr
On your test case, this produces
[1, 7, 4, 8, 2, 6, 3, 9, 10, 8, 7]
I would use an additional map, where the key is the integer value from the array and the value is an integer set to 0 in the beginning. Now I would go through the array and increase the values in the map if the key is already in the map.
In the end I would go again through the array. When the integer from the array has a value of one in the map, I would not change anything. When it has a value of 2 or more in the map I would swap the integer from the array with the last one.
This should result in a runtime of O(n*log(n))
The way I would do this would be to create an array twice the size of the original and create a set of integers.
Then Loop through the original array, add each element to the set, if it already exists add it to the 2nd half of the new array, else add it to the first half of the new array.
In the end you would get an array that looks like: (using your example)
1,7,4,8,2,6,3,9,10,-,-,8,7,-,-,-,-,-,-,-,-,-
Then I would loop through the original array again and make each spot equal to the next non-null position (or 0'd or whatever you decided)
That would make the original array turn into your solution...
This ends up being O(n) which is about as efficient as I can think of
Edit: since you can not use another array, when you find a value that is already in the
set you can move every value after it forward one and set the last value equal to the
number you just checked, this would in effect do the same thing but with a lot more operations.
I have been out of touch for a while, but I'd probably start out with something like this and see how it scales with larger input. I know you didn't ask for code but in some cases it's easier to understand than an explanation.
Edit: Sorry I missed the requirement that you cannot use a buffer array.
// returns new vector with dupes a the end
std::vector<int> move_dupes_to_end(std::vector<int> input)
{
std::set<int> counter;
std::vector<int> result;
std::vector<int> repeats;
for (std::vector<int>::iterator i = input.begin(); i < input.end(); i++)
{
if (counter.find(*i) == counter.end())
result.push_back(*i);
else
repeats.push_back(*i);
counter.insert(*i);
}
result.insert(result.end(), repeats.begin(), repeats.end());
return result;
}
#include <algorithm>
T * array = [your array];
size_t size = [array size];
// Complexity:
sort( array, array + size ); // n * log(n) and could be threaded
// (if merge sort)
T * last = unique( array, array + size ); // n, but the elements after the last
// unique element are not defined
Check sort and unique.
void remove_dup(int* data, int count) {
int* L=data; //place to put next unique number
int* R=data+count; //place to place next repeat number
std::unordered_set<int> found(count); //keep track of what's been seen
for(int* cur=data; cur<R; ++cur) { //until we reach repeats
if(found.insert(*cur).second == false) { //if we've seen it
std::swap(*cur,*--R); //put at the beginning of the repeats
} else //or else
std::swap(*cur,*L++); //put it next in the unique list
}
std::reverse(R, data+count); //reverse the repeats to be in origional order
}
http://ideone.com/3choA
Not that I would turn in code this poorly commented. Also note that unordered_set probably uses it's own array internally, bigger than data. (This has been rewritten based on aix's answer, to be much faster)
If you know the bounds on what the integer values are, B, and the size of the integer array, SZ, then you can do something like the following:
Create an array of booleans seen_before with B elements, initialized to 0.
Create a result array result of integers with SZ elements.
Create two integers, one for front_pos = 0, one for back_pos = SZ - 1.
Iterate across the original list:
Set an integer variable val to the value of the current element
If seen_before[val] is set to 1, put the number at result[back_pos] then decrement back_pos
If seen_before[val] is not set to 1, put the number at result[front_pos] then increment front_pos and set seen_before[val] to 1.
Once you finish iterating across the main list, all the unique numbers will be at the front of the list while the duplicate numbers will be at the back. Fun part is that the entire process is done in one pass. Note that this only works if you know the bounds of the values appearing in the original array.
Edit: It was pointed out that there's no bounds on the integers used, so instead of initializing seen_before as an array with B elements, initialize it as a map<int, bool>, then continue as usual. That should get you n*log(n) performance.
This can be done by iterating the array & marking index of the first change.
later on swaping that mark index value with next unique value
& then incrementing that mark index for next swap
Java Implementation:
public static void solve() {
Integer[] arr = new Integer[] { 1, 7, 4, 8, 2, 6, 8, 3, 7, 9, 10 };
final HashSet<Integer> seen = new HashSet<Integer>();
int l = -1;
for (int i = 0; i < arr.length; i++) {
if (seen.contains(arr[i])) {
if (l == -1) {
l = i;
}
continue;
}
if (l > -1) {
final int temp = arr[i];
arr[i] = arr[l];
arr[l] = temp;
l++;
}
seen.add(arr[i]);
}
}
output is 1 7 4 8 2 6 3 9 10 8 7
It's ugly, but it meets the requirements of moving the duplicates to the end in place (no buffer array)
// warning, some light C++11
void dup2end(int* arr, size_t cnt)
{
std::set<int> k;
auto end = arr + cnt-1;
auto max = arr + cnt;
auto curr = arr;
while(curr < max)
{
auto res = k.insert(*curr);
// first time encountered
if(res.second)
{
++curr;
}
else
{
// duplicate:
std::swap(*curr, *end);
--end;
--max;
}
}
}
void move_duplicates_to_end(vector<int> &A) {
if(A.empty()) return;
int i = 0, tail = A.size()-1;
while(i <= tail) {
bool is_first = true; // check of current number is first-shown
for(int k=0; k<i; k++) { // always compare with numbers before A[i]
if(A[k] == A[i]) {
is_first = false;
break;
}
}
if(is_first == true) i++;
else {
int tmp = A[i]; // swap with tail
A[i] = A[tail];
A[tail] = tmp;
tail--;
}
}
If the input array is {1,7,4,8,2,6,8,3,7,9,10}, then the output is {1,7,4,8,2,6,10,3,9,7,8}. Comparing with your answer {1,7,4,8,2,6,3,9,10,8,7}, the first half is the same, while the right half is different, because I swap all duplicates with the tail of the array. As you mentioned, the order of the duplicates can be arbitrary.