Working on below problem as an algorithm puzzle. Referred a few similar solutions (and post one of them below), tried and they worked. The question is, for the line "swap(num[i], num[k]);", how do we ensure we could always swap to a number which never tried before (e.g. suppose we swap 1 with 2 in current iteration of the for loop, then it is possible later we swap 2 back with 1 in next iterations of the same for loop of the same level/layer of recursive call)? I have the confusion since we pass num by reference, and it is very possible later (lower level/layer) recursive calls modify content of num, which cause numbers we already evaluated swap back. However, I tried and it works for all of my test cases. Wondering if below solution is 100% correct, or happened to pass my test cases? :)
Here are detailed problem statement and code I am debugging,
Given a collection of numbers that might contain duplicates, return all possible unique permutations.
For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1]
class Solution {
public:
void recursion(vector<int> num, int i, int j, vector<vector<int> > &res) {
if (i == j-1) {
res.push_back(num);
return;
}
for (int k = i; k < j; k++) {
if (i != k && num[i] == num[k]) continue;
swap(num[i], num[k]);
recursion(num, i+1, j, res);
}
}
vector<vector<int> > permuteUnique(vector<int> &num) {
sort(num.begin(), num.end());
vector<vector<int> >res;
recursion(num, 0, num.size(), res);
return res;
}
};
thanks in advance,
Lin
As #notmyfriend said in the comments, num is actually copied each function call.
So now it boils down to:
Of all array values, select one to be the first one and place it there.
That in a loop for each value one time, and then recursively:
Of all values after the first one, select one to be the first and place it there...
...and so on, combined with a check to filter out swaps where nothing changes, ie. filter out duplicates.
If num were a real reference, it won't work anymore (at least not without additional steps).
Eg. 1 1 2 is an easy conterexample, it would give the results:
112, 121, 211, 112, 121
ie. there are duplicates despite the check (and probably there
are examples where some permutations are not generated at all, too).
About the comment:
Per default, every normal function parameter in C++ is copied
(normal = without explicit reference symbol '&' etc.).
Maybe you're thinking of C-style arrays: Essentially, what is passed there is a pointer (to the first value). The pointer is copied, but both original and copied pointer point to the same memory location.
While the purpose of std::vector is (too) to contain an array, the vector itself is a single class object (which contains a pointer to the values somewhere). A class can define itself how it should be copied (with a copy constructor).
Technically, the vector class could implement copying as pointer copying, then it would have the same effect as passing the whole vector as reference; but the C++ creators wanted to keep the copy semantics, ie. that copying a container class should make a real copy with all values duplicated.
For non-copying, there are references already...
Below you can find a solution written in Java. Sorry for not providing a solution in C++, I'm not using it for a long time. But the syntax would be similar.
Solution is using Backtracking (https://en.wikipedia.org/wiki/Backtracking)
Also I'm using hashset to check uniqueness, may be there is a solution which does not use any hashset type data structure, becase my solution is using extra memory in order to provide unique solutions.
Sample input and output;
input : [1, 1, 2]
output : [1, 1, 2]
[1, 2, 1]
[2, 1, 1]
And the solution is;
public class permutation {
public static void main(String[] args) {
permutation p = new permutation();
p.permute(new int[] { 1, 1, 2 });
}
HashSet<String> set = new HashSet<String>();
private void permute(int[] arr) {
set.clear();
this.permute(arr, 0, arr.length - 1);
}
private void permute(int[] arr, int l, int r) {
if (l == r) {
String key = Arrays.toString(arr);
if (set.contains(key))
return;
set.add(key);
System.out.println(key);
} else {
for (int i = l; i <= r; i++) {
swap(arr, l, i);
permute(arr, l + 1, r);
swap(arr, i, l);
}
}
}
private void swap(int[] arr, int l, int r) {
int tmp = arr[l];
arr[l] = arr[r];
arr[r] = tmp;
}
}
Related
c++ newbie here. So for an assignment I have to rotate all the elements in a vector to the left one. So, for instance, the elements {1,2,3} should rotate to {2,3,1}.
I'm researching how to do it, and I saw the rotate() function, but I don't think that will work given my code. And then I saw a for loop that could do it, but I'm not sure how to translate that into a return statement. (i tried to adjust it and failed)
This is what I have so far, but it is very wrong (i haven't gotten a single result that hasn't ended in an error yet)
Edit: The vector size I have to deal with is just three, so it doesn't need to account for any sized vector
#include <vector>
using namespace std;
vector<int> rotate(const vector<int>& v)
{
// PUT CODE BELOW THIS LINE. DON'T CHANGE ANYTHING ABOVE.
vector<int> result;
int size = 3;
for (auto i = 0; i < size - 1; ++i)
{
v.at(i) = v.at(i + 1);
result.at(i) = v.at(i);
}
return result;
// PUT CODE ABOVE THIS LINE. DON'T CHANGE ANYTHING BELOW.
}
All my teacher does it upload textbook pages that explain what certain parts of code are supposed to do but the textbook pages offer NO help in trying to figure out how to actually apply this stuff.
So could someone please give me a few pointers?
Since you know exactly how many elements you have, and it's the smallest number that makes sense to rotate, you don't need to do anything fancy - just place the items in the order that you need, and return the result:
vector<int> rotate3(const vector<int>& x) {
return vector<int> { x[1], x[2], x[0] };
}
Note that if your collection always has three elements, you could use std::array instead:
std::array<int,3>
First, just pay attention that you have passed v as const reference (const vector<int>&) so you are forbbiden to modify the state of v in v.at(i) = v.at(i + 1);
Although Sergey has already answered a straight forward solution, you could correct your code like this:
#include <vector>
using namespace std;
vector<int> left_rotate(const vector<int>& v)
{
vector<int> result;
int size = v.size(); // this way you are able to rotate vectors of any size
for (auto i = 1; i < size; ++i)
result.push_back(v.at(i));
// adding first element of v at end of result
result.push_back(v.front());
return result;
}
Use Sergey's answer. This answer deals with why what the asker attempted did not work. They're damn close, so it's worth going though it, explaining the problems, and showing how to fix it.
In
v.at(i) = v.at(i + 1);
v is constant. You can't write to it. The naïve solution (which won't work) is to cut out the middle-man and write directly to the result vector because it is NOT const
result.at(i) = v.at(i + 1);
This doesn't work because
vector<int> result;
defines an empty vector. There is no at(i) to write to, so at throws an exception that terminates the program.
As an aside, the [] operator does not check bounds like at does and will not throw an exception. This can lead you to thinking the program worked when instead it was writing to memory the vector did not own. This would probably crash the program, but it doesn't have to1.
The quick fix here is to ensure usable storage with
vector<int> result(v.size());
The resulting code
vector<int> rotate(const vector<int>& v)
{
// PUT CODE BELOW THIS LINE. DON'T CHANGE ANYTHING ABOVE.
vector<int> result(v.size()); // change here to size the vector
int size = 3;
for (auto i = 0; i < size - 1; ++i)
{
result.at(i) = v.at(i + 1); // change here to directly assign to result
}
return result;
// PUT CODE ABOVE THIS LINE. DON'T CHANGE ANYTHING BELOW.
}
almost works. But when we run it on {1, 2, 3} result holds {2, 3, 0} at the end. We lost the 1. That's because v.at(i + 1) never touches the first element of v. We could increase the number of for loop iterations and use the modulo operator
vector<int> rotate(const vector<int>& v)
{
// PUT CODE BELOW THIS LINE. DON'T CHANGE ANYTHING ABOVE.
vector<int> result(v.size());
int size = 3;
for (auto i = 0; i < size; ++i) // change here to iterate size times
{
result.at(i) = v.at((i + 1) % size); // change here to make i + 1 wrap
}
return result;
// PUT CODE ABOVE THIS LINE. DON'T CHANGE ANYTHING BELOW.
}
and now the output is {2, 3, 1}. But it's just as easy, and probably a bit faster, to just do what we were doing and tack on the missing element after the loop.
vector<int> rotate(const vector<int>& v)
{
// PUT CODE BELOW THIS LINE. DON'T CHANGE ANYTHING ABOVE.
vector<int> result(v.size());
int size = 3;
for (auto i = 0; i < size - 1; ++i)
{
result.at(i) = v.at(i + 1);
}
result.at(size - 1) = v.at(0); // change here to store first element
return result;
// PUT CODE ABOVE THIS LINE. DON'T CHANGE ANYTHING BELOW.
}
Taking this a step further, the size of three is an unnecessary limitation for this function that I would get rid of and since we're guaranteeing that we never go out of bounds in our for loop, we don't need the extra testing in at
vector<int> rotate(const vector<int>& v)
{
// PUT CODE BELOW THIS LINE. DON'T CHANGE ANYTHING ABOVE.
if (v.empty()) // nothing to rotate.
{
return vector<int>{}; // return empty result
}
vector<int> result(v.size());
for (size_t i = 0; i < v.size() - 1; ++i) // Explicitly using size_t because
// 0 is an int, and v.size() is an
// unsigned integer of implementation-
// defined size but cannot be larger
// than size_t
// note v.size() - 1 is only safe because
// we made sure v is not empty above
// otherwise 0 - 1 in unsigned math
// Becomes a very, very large positive
// number
{
result[i] = v[i + 1];
}
result.back() = v.front(); // using direct calls to front and back because it's
// a little easier on my mind than math and []
return result;
// PUT CODE ABOVE THIS LINE. DON'T CHANGE ANYTHING BELOW.
}
We can go further still and use iterators and range-based for loops, but I think this is enough for now. Besides at the end of the day, you throw the function out completely and use std::rotate from the <algorithm> library.
1This is called Undefined Behaviour (UB), and one of the most fearsome things about UB is anything could happen including giving you the expected result. We put up with UB because it makes for very fast, versatile programs. Validity checks are not made where you don't need them (along with where you did) unless the compiler and library writers decide to make those checks and give guaranteed behaviour like an error message and crash. Microsoft, for example, does exactly this in the vector implementation in the implementation used when you make a debug build. The release version of Microsoft's vector make no checks and assumes you wrote the code correctly and would prefer the executable to be as fast as possible.
I saw the rotate() function, but I don't think that will work given my code.
Yes it will work.
When learning there is gain in "reinventing the wheel" (e.g. implementing rotate yourself) and there is also gain in learning how to use the existing pieces (e.g. use standard library algorithm functions).
Here is how you would use std::rotate from the standard library:
std::vector<int> rotate_1(const std::vector<int>& v)
{
std::vector<int> result = v;
std::rotate(result.begin(), result.begin() + 1, result.end());
return result;
}
So I was asked to write a function that changes array's values in a way that:
All of the values that are the smallest aren't changed
if, let's assume, the smallest number is 2 and there is no 3's and 4's then all 5's are changed for 3's etc.
for example, for an array = [2, 5, 7, 5] we would get [2, 3, 4, 3], which generalizes to getting a minimal value of an array which remains unchanged, and every other minimum (not including the first one) is changed depending on which minimum it is. On our example - 5 is the first minimum (besides 2), so it is 2 (first minimum) + 1 = 3, 7 is 2nd smallest after 2, so it is 2+2(as it is 2nd smallest).
I've come up with something like this:
int fillGaps(int arr[], size_t sz){
int min = *min_element(arr, arr+sz);
int w = 1;
for (int i = 0; i<sz; i++){
if (arr[i] == min) {continue;}
else{
int mini = *min_element(arr+i, arr+sz);
for (int j = 0; j<sz; j++){
if (arr[j] == mini){arr[j] = min+w;}
}
w++;}
}
return arr[sz-1];
}
However it works fine only for the 0th and 1st value, it doesnt affect any further items. Could anyone please help me with that?
I don't quite follow the logic of your function, so can't quite comment on that.
Here's how I interpret what needs to be done. Note that my example implementation is written to be as understandable as possible. There might be ways to make it faster.
Note that I'm also using an std::vector, to make things more readable and C++-like. You really shouldn't be passing raw pointers and sizes, that's super error prone. At the very least bundle them in a struct.
#include <algorithm>
#include <set>
#include <unordered_map>
#include <vector>
int fillGaps (std::vector<int> & data) {
// Make sure we don't have to worry about edge cases in the code below.
if (data.empty()) { return 0; }
/* The minimum number of times we need to loop over the data is two.
* First to check which values are in there, which lets us decide
* what each original value should be replaced with. Second to do the
* actual replacing.
*
* So let's trade some memory for speed and start by creating a lookup table.
* Each entry will map an existing value to its new value. Let's use the
* "define lambda and immediately invoke it" to make the scope of variables
* used to calculate all this as small as possible.
*/
auto const valueMapping = [&data] {
// Use an std::set so we get all unique values in sorted order.
std::set<int> values;
for (int e : data) { values.insert(e); }
std::unordered_map<int, int> result;
result.reserve(values.size());
// Map minimum value to itself, and increase replacement value by one for
// each subsequent value present in the data vector.
int replacement = *values.begin();
for (auto e : values) { result.emplace(e, replacement++); }
return result;
}();
// Now the actual algorithm is trivial: loop over the data and replace each
// element with its replacement value.
for (auto & e : data) { e = valueMapping.at(e); }
return data.back();
}
I'm looking for efficient memory allocation when dealing with recursive function. As far as I understand, variables I use in the function will remain allocated in memory until recursion is finished. Is there a way to avoid this as I believe this causes slow run of my code below where state variable is copied every time the function is called (correct me if I'm wrong as I'm new to C++).
#include <fstream>
#include <vector>
using namespace std;
int N = 30;
double MIN_COST = 1000000;
vector<int> MIN_CUT = {};
void minCut(vector<int> state, int index, int nodeValue) {
double currentCost;
if (index >= 0) {
currentCost = getCurrentCost(state); // some magic evaluating state cost
state.push_back(nodeValue);
if (currentCost >= MIN_COST) { // kill branch if incomplete solution is already worse than best achieved solution
return;
}
}
if (index == N - 1) { // check if leaf node
if (currentCost < MIN_COST) {
MIN_COST = currentCost;
MIN_CUT = state;
}
return;
}
minCut(state, index + 1, 1); // left subtree - adding 1 to vector
minCut(state, index + 1, 0); // right subtree - adding 0 to vector
return;
}
int main() {
vector<int> state = {};
minCut(state, -1, NULL);
cout << MIN_COST << "\n";
return 0;
}
Your algorithm is effectively building a tree of paths, but you're using a vector to hold the nodes for each path.
A
/ \
/ \
B C
/ \ / \
D E F G
This is the tree you're traversing.
But you're creating new vectors at every node, which contain the whole path up to that node. So as you're visiting node G, in your stack you have 3 vectors:
vector { A, C, G }
vector { A, C }
vector { A }
It should be clear how this is less efficient as you have noticed, but maybe seeing it this way hints at the correct efficient implementation.
The call stack itself holds the path to the root node. The stack when visiting G would be something like
minCut < visiting G >
minCut < visiting C >
minCut < visiting A >
In order to efficiently exploit this fact, make minCut pass the minimum amount of information. In this case we're talking about something linked-list like.
You have then two options that jump out:
Use vector, but:
Pass it by reference.
And you must then maintain it across calls, pushing and popping nodes to keep synchronized with the actual state.
Use an actual linked list. It should be easy to construct the vector by traversing pointers-to-parent-nodes.
Yes, there is a more efficient way to pass state through each function call. This is called passing by reference and can be achieved like so:
void minCut(vector<int>& state, int index, int nodeValue) { ...
This will result in the original state being referenced instead of copied each time the function is called.
For this to work correctly in the code you posted you will have to make some modifications, this is just the general concept.
I have an array of values e.g. 1, 4, 7, 2.
I also have another array of values and I want to add its values to this first array, but only when they all are different from all values that are already in this array. How can I check it? I've tried many types of loops, but I always ended with an iteration problem.
Could you please tell me how to solve this problem? I code in c++.
int array1[7] = {2,3,7,1,0};
int val1 = rand() % 10;
int val2 = rand() % 10;
int array2[2] = {val1, val2};
and I am trying to put every value from array2 into array1. I tried loop
for (int x:array2)
{
while((val1 && val2) == x)
{
val1 = rand() % 10;
val2 = rand() % 10;
}
}
and many more, but still cannot figure it out. I have this problem because I may have various number of elements for array2. So it makes this "&&" solution infinite.
It is just a sample to show it more clearly, my code has much more lines.
Okay, you have a few problems here. If I understand the problem, here's what you want:
A. You have array1 already populated with several values but with space at the end.
1. How do you identify the number of entries in the array already versus the extras?
B. You have a second array you made from two random values. No problem.
You want to append the values from B to A.
2. If initial length of A plus initial length of B is greater than total space allocated for A, you have a new problem.
Now, other people will tell you to use the standard template library, but if you're having problems at this level, you should know how to do this yourself without the extra help from a confusing library. So this is one solution.
class MyArray {
public:
int * data;
int count;
int allocated;
MyArray() : data(nullptr), count(0), allocated(0) {}
~MyArray() { if (data != nullptr) free(data); }
// Appends value to the list, making more space if necessary
void add(int value) {
if (count >= allocated) {
// Not enough space, so make some.
allocated += 10;
data = (data == nullptr) malloc(allocated * sizeof(int))
: realloc)data, allocated * sizeof(int));
}
data[count++] = value;
}
// Adds value only if not already present.
void addUnique(int value) {
if (indexOf(value) < 0) {
add(value);
}
}
// Returns the index of the value, if found, else -1
int indexOf(int value) {
for (int index = 0; index < count; ++index) {
if (data[index] == value) {
return index;
}
}
return -1;
}
}
This class provides you a dynamic array of integers. It's REALLY basic, but it teaches you the basics. It helps you understand about allocation / reallocating space using old-style C-style malloc/realloc/free. It's the sort of code I was writing back in the 80s.
Now, your main code:
MyArray array;
array.add(2);
array.add(3);
array.add(7);
// etc. Yes, you could write a better initializer, but this is easy to understand
MyArray newValues;
newValues.add(rand() % 10);
newValues.add(rand() % 10);
for (int index = 0; index < newValues.count; ++index) {
array.addUnique(newValues.data[index]);
}
Done.
The key part of this is the addUnique function, which simply checks first whether the value you're adding already is in the array. If not, it appends the value to the array and keeps track of the new count.
Ultimately, when using integer arrays like this instead of the fancier classes available in C++, you HAVE TO keep track of the size of the array yourself. There is no magic .length method on int[]. You can use some magic value that indicates the end of the list, if you want. Or you can do what I did and keep two values, one that holds the current length and one that holds the amount of space you've allocated.
With programming, there are always multiple ways to do this.
Now, this is a lot of code. Using standard libraries, you can reduce all of this to about 4 or 5 lines of code. But you're not ready for that, and you need to understand what's going on under the hood. Don't use the fancy libraries until you can do it manually. That's my belief.
I've developed a method called "rotate" to my stack object class. What I did was that if the stack contains elements: {0,2,3,4,5,6,7} I would needed to rotate the elements forwards and backwards.
Where if i need to rotate forwards by 2 elements, then we would have, {3,4,5,6,7,0,2} in the array. And if I need to rotate backwards, or -3 elements, then, looking at the original array it would be, {5,6,7,0,2,3,4}
So the method that I have developed works fine. Its just terribly ineffecient IMO. I was wondering if I could wrap the array around by using the mod operator? Or if their is useless code hangin' around that I havent realized yet, and so on.
I guess my question is, How can i simplify this method? e.g. using less code. :-)
void stack::rotate(int r)
{
int i = 0;
while ( r > 0 ) // rotate postively.
{
front.n = items[top+1].n;
for ( int j = 0; j < bottom; j++ )
{
items[j] = items[j+1];
}
items[count-1].n = front.n;
r--;
}
while ( r < 0 ) // rotate negatively.
{
if ( i == top+1 )
{
front.n = items[top+1].n;
items[top+1].n = items[count-1].n; // switch last with first
}
back.n = items[++i].n; // second element is the new back
items[i].n = front.n;
if ( i == bottom )
{
items[count-1].n = front.n; // last is first
i = 0;
r++;
continue;
}
else
{
front.n = items[++i].n;
items[i].n = back.n;
if ( i == bottom )
{
i = 0;
r++;
continue;
}
}
}
}
Instead of moving all the items in your stack, you could change the definition of 'beginning'. Have an index that represents the first item in the stack, 0 at the start, which you add to and subtract from using modular arithmetic whenever you want to rotate your stack.
Note that if you take this approach you shouldn't give users of your class access to the underlying array (not that you really should anyway...).
Well, as this is an abstraction around an array, you can store the "zero" index as a member of the abstraction, and index into the array based on this abstract notion of the first element. Roughly...
class WrappedArray
{
int length;
int first;
T *array;
T get(int index)
{
return array[(first + index) % length];
}
int rotateForwards()
{
first++;
if (first == length)
first = 0;
}
}
You've gotten a couple of reasonable answers, already, but perhaps one more won't hurt. My first reaction would be to make your stack a wrapper around an std::deque, in which case moving an element from one end to the other is cheap (O(1)).
What you are after here is a circular list.
If you insist on storing items in an array just use top offset and size for access. This approach makes inserting elements after you reached allocated size expensive though (re-allocation, copying). This can be solved by using doubly-linked list (ala std::list) and an iterator, but arbitrary access into the stack will be O(n).
The function rotate below is based on reminders (do you mean this under the 'mod' operation?)
It is also quite efficient.
// Helper function.
// Finds GCD.
// See http://en.wikipedia.org/wiki/Euclidean_algorithm#Implementations
int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
// Number of assignments of elements in algo is
// equal to (items.size() + gcd(items.size(),r)).
void rotate(std::vector<int>& items, int r) {
int size = (int)items.size();
if (size <= 1) return; // nothing to do
r = (r % size + size) % size; // fits r into [0..size)
int num_cycles = gcd(size, r);
for (int first_index = 0; first_index < num_cycles; ++first_index) {
int mem = items[first_index]; // assignment of items elements
int index = (first_index + r) % size, index_prev = first_index;
while (index != first_index) {
items[index_prev] = items[index]; // assignment of items elements
index_prev = index;
index = (index + r) % size;
};
items[index_prev] = mem; // assignment of items elements
}
}
Of course if it is appropriate for you to change data structure as described in other answers, you can obtain more efficient solution.
And now, the usual "it's already in Boost" answer: There is a Boost.CircularBuffer
If for some reason you'd prefer to perform actual physical rotation of array elements, you might find several alternative solutions in "Programming Pearls" by Jon Bentley (Column 2, 2.3 The Power of Primitives). Actually a Web search for Rotating Algorithms 'Programming Pearls' will tell you everything. The literal approach you are using now has very little practical value.
If you'd prefer to try to solve it yourself, it might help to try looking at the problem differently. You see, "rotating an array" is really the same thing as "swapping two unequal parts of an array". Thinking about this problem in the latter terms might lead you to new solutions :)
For example,
Reversal Approach. Reverse the order of the elements in the entire array. Then reverse the two parts independently. You are done.
For example, let's say we want to rotate abcdefg right by 2
abcdefg -> reverse the whole -> gfedcba -> reverse the two parts -> fgabcde
P.S. Slides for that chapter of "Programming Pearls". Note that in Bentley's experiments the above algorithm proves to be quite efficient (among the three tested).
I don't understand what the variables front and back mean, and why you need .n. Anyway, this is the shortest code I know to rotate the elements of an array, which can also be found in Bentley's book.
#include <algorithm>
std::reverse(array , array + r );
std::reverse(array + r, array + size);
std::reverse(array , array + size);