I am working on C++.. am in a need to swap two blocks of elements in an array..
Say, {1,2,3,4,5,6} is my input array.. block {4,5} should be moved to beginning and the output array should be like {4,5,1,2,3,6}.. all i have is the start index and end index of the block {4,5}.. for doing this i am using a temp array, copying the blocks individually to temp array and moving it back to the original array, which is tedious
but i am sure there will be better methods to do this using memcpy or memmove.. any ideas?
There is a standard algorithm designed specifically for this task called std::rotate():
#include <algorithm>
#include <cstdio>
int main()
{
int inputArray[] = {1, 2, 3, 4, 5, 6};
::printf("Before: ");
for(int i = 0; i < 6; ++i)
{
::printf("%d ", inputArray[i]);
}
::printf("\n");
int startIndex = 3; // refers to the number 4 in inputArray
int endIndex = 5; // refers one-past the number 5 in inputArray
std::rotate(inputArray, inputArray+startIndex, inputArray+endIndex);
::printf("After: ");
for(int i = 0; i < 6; ++i)
{
::printf("%d ", inputArray[i]);
}
::printf("\n");
}
Expected output:
Before: 1 2 3 4 5 6
After: 4 5 1 2 3 6
std::rotate() performs the rotation in-place via std::swap(), so there's no temporary array involved.
Bentley's "Programming Pearls" describes three algorithms for solving this problem. You can find slides for this specific problem here
http://www.cs.bell-labs.com/cm/cs/pearls/s02b.pdf
For example, the simplest algorithms would be the Reversal one. Just reverse the blocks that you need to swap, and then reverse the entire array. Done.
P.S. In your example "the entire array" would stand for the 1,2,3,4,5 subsequence (6 is not included), since these are the blocks that you need to swap.
Reverse the blocks:
3, 2, 1, 5, 4
Reverse the whole thing
4, 5, 1, 2, 3
Related
Introduction
Good day,
I am looking for a grouping algorithm that can do the following:
Let's suppose I have an array of sorted numbers (without any multiple occurences). For example, {0, 2, 5, 6, 7, 10}.
I want to make groups from that array, such that:
I minimize the number of groups,
Each groups needs to contain numbers that are linked with at most n - 1 "bonds" (for example, n = 3, 0 and 2 are neighbours but not 0 and 3).
EDIT
In other words, when I say neighbours, I should speak about integer distance. For example, the distance of 0 to 2 i 2 (and vice versa). The distance of 0 to 3 is 3. You could think of the problem like a set of 1D points, and one needs to find the minimal number of centers, which center contains points that are distant to it of n/2. I hope it is more clear like that.
The example has multiple groups possible but the best along conditions 1 and 2 (n = 3) is {{0, 2}, {5, 6, 7}, {10}}. {{0}, {2, 5}, {6, 7}, {10}} has one group more than the best solution. The ideal solution would happen if all sorted numbers are continuous:
nb_groups* = ceil(v.size() / n);
In addition, there might be multiple solution depending on the algorithm.
What I tried
For now, what I do is:
Compute the array of the distances between the neighbouring elemens,
Check neighbouring conditions with rests from the beginning of the vector to the end (see the code below).
It seems to work (to me), but I was wondering two things:
Does it really work for any cases (have maybe not tested all cases?)?
If so, could I optimize my implementation in a way (better than in.size() - 1 iteration ans with less memory consumption)?
Code
I was considering a function that takes the vector to group, and the max distance. This function would return the indices of the first element of the group.
#include <iostream>
#include <vector>
std::vector<int> groupe(const std::vector<int>& at, const int& n);
int main() {
// Example of input vector
std::vector<int> in = {0, 2, 5, 6, 7, 10, 11, 22, 30, 50, 51};
// Try to group with neighbouring distance of 3
std::vector<int> res = groupe(in, 3);
// Printing the result
for(const int& a : res) {
std::cout << a << " ";
}
}
std::vector<int> groupe(const std::vector<int>& at, const int& n) {
std::vector<int> out;
// Reste keeps tracks of a bigger neighbouring distance (in case we can look for another element to be in the group)
int reste(0);
size_t s = at.size() - 1;
for(int i = 0; i < s; i++) {
// Computing the distance between element i and i + 1
int d = at[i + 1] - at[i];
if(d >= n) {
if(reste == 0) {
out.push_back(i);
}
reste = 0;
} else {
if(reste == 0) {
out.push_back(i);
}
reste += d;
if(reste >= n) {
reste = 0;
}
}
}
if(reste == 0 || reste >= n) {
out.push_back(s);
}
return out;
}
OUTPUT
0 2 5 7 8 9
Note
If the original vector was not sorted, I guess we could have sorted it first and then achieved this step (or maybe there is another algorithm mor efficient?).
I thank you in advance for your time and help.
I want to compare all element of vector x to all elements of vector y and if I find a element greater in vector y than being compared to, I have to take that particular element of vector y and compare to all elements of vector z and if it is true return true else if i don't find a greater element in first iteration i,e when elements of vector x are compared to vector y i have to break the loop and return false.
I tried to iterate through all the elements of stackarmies but I don't know how to take the first element of vector one and compare with all the elements of vector, since all the vectors are merged into the last vector.
vector<int> stack;
int noofstack, noofoperations, stackno, OperationType;
// Taking the input number of stacks
cin >> noofstack;
vector<vector<int>> stackarmies;
for (int i = 0; i < noofstack; i++)
{
int stacksize;
//Since vectors are dynamic and we don't need to declare the size but as per the problem statement I've added it/
cin >> stacksize;
for (int k = 0; k < stacksize; k++)
{
//Taking the input of all the vectors one by one and then adding all the vectors into one vector
int armyheight;
cin>>armyheight;
stack.push_back(armyheight);
}
stackarmies.push_back(stack);
Test cases
Input 1
2
3 3 5 4
3 1 1 2
Resulting stackarmies: { {3, 5, 4}, {3, 5, 4, 1, 1, 2} }
Desired output: False
We will take first element of vector 1 : 3 and compare with all
elements of vector 2 , in vector 2 no element is greater than 3.
Input 2
2
3 1 0 4
3 2 1 3
Resulting stackarmies: { {1, 0, 4}, {1, 0, 4, 2, 1, 3} }
Desired output: True
We will take first element of vector 1 : 1 and compare with all
elements of vector 2, in vector 2, the first element is greater than 1,
so true
Input 3
2
3 1 9 0
2 0 11
Resulting stackarmies: { {1, 9, 0}, {1, 9, 0, 0, 11} }
Desired output: True
We will take first element of vector 1 : 1 and compare with all
elements of vector 2, in vector 2, the last element is greater than 1,
so true
Input 4
3
3 0 8 0
3 4 0 11
3 0 9 0
Resulting stackarmies: { {0, 8, 0}, {0, 8, 0, 4, 0, 11} , {0, 8, 0, 4, 0, 11, 0, 9, 0} }
Desired output: True
We will take the second element of vector 1: 8 and compare with
all elements of vector 2 , 11 is greater than 8 so we will compare 11 of
vector 2 with vector , since there are no values greater than 11, so it's
false
I don't know how to take the first element of vector one and compare with all the elements of vector, since all the vectors are merged into the last vector.
You're getting ahead of yourself. Why do you want all the vectors merged into the last vector? Answer: you don't; that's just what happened. Why did all the vectors merge into the last vector? Answer: because you have a bug in your code that reads the data. Fix that bug instead of spending ten times as much effort trying to handle the malformed data.
That whole spiel about what you intend to do next is nothing more than a distraction that wastes the time of the people from whom you are asking help. Ask for help with the real problem (the loading bug) instead of driving people away with a confusing question that assumes bad data is good.
There are several ways to fix the bug. I think the most helpful approach is one that would have avoided the bug in the first place. You try to do too much in a single function. Divide and conquer; when you have a non-trivial sub-step, create a function to handle it. Good programming practices lead to fewer bugs.
Specifically, reading the heights of the fighters in a stack is non-trivial. Delegate that to a helper and reduce the body of your outer for loop to a single line.
for (int i = 0; i < noofstack; i++)
{
//* This is non-trivial, so use a helper function.
stackarmies.push_back(read_fighter_heights());
}
This helper function is responsible for reading a line of data, generating a stack (a vector<int>) from it, and returning that stack. That covers most of the body of your loop, leaving only the simple task of pushing the returned stack onto your vector of stacks.
Creating this helper function from your existing code is fairly simple. Mostly, just move the body of the loop into an appropriate function definition. In addition, you should notice that stack is needed (only) in this function, so also move that variable's declaration into the new function's definition.
vector<int> read_fighter_heights()
{
vector<int> stack;
int stacksize;
//Since vectors are dynamic and we don't need to declare the size but as per the problem statement I've added it/
cin >> stacksize;
for (int k = 0; k < stacksize; k++)
{
//Taking the input of all the vectors one by one and then adding all the vectors into one vector
int armyheight;
cin>>armyheight; //* Reading a single integer is trivial, so no need for another function here.
stack.push_back(armyheight);
}
return stack;
}
Presto! Problem solved. All you had to do was be more organized.
Addendum: The reason this solves the problem is that extra step of moving the declaration of stack. In the original code, this variable was declared outside the outer loop, and it was never cleared. The result was that it accumulated values from each line that was read. In this version, the variable is re-initialized before reading each line, so values do not accumulate. You could get the same result by moving the line in the original code, without splitting off a new function. However, splitting off a new function is a good habit to get into, as it almost forces you to declare stack at the right level, avoiding the problem in the first place.
bool CompareVectors(vector<vector<int>> st)
{
bool result = true;
for (int k = 0; k < st.size(); k++)
{
if (k != st.size() - 1)
{
if (result)
{
for (auto i = st[k].begin(); i != st[k].end(); ++i)
{
for (auto j = st[k+1].begin(); j != st[k+1].end(); ++j)
{
if (*i < *j)
{
result = true;
break;
}
else
{
result = false;
}
}
if (result)
{
break;
}
}
}
}
}
return result;
}
I made a vector of constant size to store negative values, and then printing the values all I got was zeroes. I just want to know why it is not storing negative values.
#include <iostream>
#include <vector>
int main() {
std::vector<int> v(5);
v.push_back(-1);
v.push_back(-2);
v.push_back(-3);
v.push_back(-4);
v.push_back(-5);
for (int i=0; i<5; i++)
std::cout << v[i] << " "; // All I got was zeroes
}
That's because push_back puts new elements onto the end of the vector.
You can see the effect by running i to 9: the negative numbers will occupy v[5] to v[9].
Writing
std::vector<int> v{-1, -2, -3, -4, -5};
instead is a particularly elegant fix.
The constructor that you invoke fills the first 5 elements with zeros, see here (#3 in the list of overloads):
Constructs the container with count default-inserted instances of T
(where the "default-inserted instance" of an int is 0). What you might have wanted is
std::vector<int> v;
v.reserve(5); /* Prevent unnecessary allocations as you know the desired size. */
v.push_back(-1);
/* ... */
An alternative using the original constructor call is
#include <numeric>
std::vector<int> v(5);
std::iota(v.rbegin(), v.rend(), -v.size());
though this does more work than necessary as every element is first default constructed and then assigned to a new value again.
This is a case where the DRY principle would help you understand your mistake.
vector<int> v(5);
...
for(int i=0;i<5;i++)
Here you are creating an array, for which you think you reserve space for 5 elements. Then you insert those 5 elements. After that you wanted to print contents of the whole array, but instead of just writing v.size(), you repeated the 5, so that your code now reads like "Print first five elements of v", instead of "Print all elements of v".
If you instead wrote what you mean, you'd see that the array actually has 10 elements, not 5.
BTW, since C++11 you can loop over all elements in a more straightforward way:
for(int x : v)
or, if the elements were some more copy-expensive type, you could use references to the elements, even auto-type references:
for(auto& x : v)
This new for-loop syntax is called the range-based for loop.
You can consider the vector a flexible version of the primitive array in C/C++. When you initialize a vector with a size n, the constructed vector has size of n (or maybe larger in the memory, but you don't know since it's implicitly handled by compiler). Note that here n represents the number of entries, but not the actual memory usage (i.e. bytes). If you do not initialize it with a size parameter, the vector is empty with size 0, but in the memory it would have some implicit default memory size.
Let's say your current vector has size 5. And you want to push_back() in another element, then the vector internally will reallocate the entire array into a new memory location which could hold all its old entries plus the new one. So you don't need to reallocate the memory manually by yourself, like what you have to do in C.
Here, in your example, to fill in those 5 negative integers in your vector, there are a couple of ways.
1) You can initialize a vector without specifying the size. And then push in each element you want.
vector<int> v;
for (int i = -1; i >= -5; --i) {
v.push_back(i);
}
2) You can initialize the vector in your way with that size parameter. And then assign them with new values.
vector<int> v(5);
for (int i = 0; i < v.size(); ++i) {
v[i] = -i;
}
3) You can also initialize the vector with those entries when it is constructed.
vector<int> v{-1, -2, -3, -4, -5};
or vector<int> v = {-1, -2, -3, -4, -5};
When you declared the vector with
std::vector<int> v(5);
You made v store five 4-byte spaces in memory (assuming an int = 4 bytes on your system), and by default all of these 4-byte spaces store the bits representing 0's. Then, you pushed 5 more ints (-1, -2, -3, -4, -5) onto the end of the vector with:
v.push_back(-1);
v.push_back(-2);
v.push_back(-3);
v.push_back(-4);
v.push_back(-5);
At this point the vector has 10 elements, the first five being the unknown ints that happen to store 0's on the instance you ran the program. Since your for loop prints the first five elements in the vector, this is why it printed all 0's.
When you declared this line vector<int>v(5) it created a vector of size 5 with default value of all elements 0, at this stage your vector looks like this -
{0, 0, 0, 0, 0}
Now when you called v.push_back(-1) what it did is it pushed -1 to the back of vector increasing its size, now your vector looks like this -
{0, 0, 0, 0, 0, -1}
After every push back is performed your final vector will be - {0, 0, 0, 0, 0, -1, -2, -3, -4, -5}
The size of your vector is now increased to 10 from 5 and when you looped from index 0 to 4 it only printed 0. I hope I explained it well and fix for this is already provided in previous answers.
I am going to start the new question. I posed the question yesterday and wanted to know what's the problem in my program. The program is given below and you people pointed out that this following program does only one pass of the sorting and needs an outer loop as well. At that time I was good like OK. But again when I looked the program I got confused and need to ask Why we need Outer loop as well for the sort since only a single loop can do the sorting(In my opinion). First see program below then I present my logic at the end of the program.
#include <iostream.h>
#include <conio.h>
using namespace std;
main()
{
int number[10];
int temp = 0;
int i = 0;
cout << "Please enter any ten numbers to sort one by one: "
<< "\n";
for (i = 0; i < 10; i++)
{
cin >> number[i];
}
i = 0;
for (i = 0; i < 9; i++)
{
if (number[i] > number[i + 1])
{
temp = number[i + 1];
number[i + 1] = number[i];
number[i] = temp;
}
}
i = 0;
cout << "The sorted numbers are given below:"
<< "\n";
for (i = 0; i < 10; i++)
{
cout << number[i] << "\n";
}
getch();
}
I think the ONLY loop with the bubble condition should do the sorting. Look at the following loop of the program:
for (i=0;i<9;i++)
if(number[i]>number[i+1])
{
temp=number[i+1];
number[i+1]=number[i];
number[i]=temp;
}
Now I explain what I am thinking what this loop "should" do. It will first compare number[0] with number[1]. If the condition is satisfied it will do what is in IF statement's body. Then i will be incremented by 1(i++). Then on next iteration the values compared will be number[1] with number[2]. Then why it does not happen and the loop exits after only pass? In other words may be I'm trying to ask IF statement does not repeat itself in for loop? In my opinion it does. I'm very thankful for help and views, my question might be of small level but that is how I will progress.
Let me give you an example let's only take 3 numbers. So you input
13, 3 ,1
Now you start sorting how you did it. so it compares 13 and 3
13 > 3 so switch both of them.
now we have.
3, 13, 1
Now it'll compare as you said the next pair = 13 and 1
13 > 1 so the new order would be
3, 1, 13
now your loop is finished and you missed to compare 3 and 1
Actually the first loop only sorts the greatest number!
since only a single loop can do the sorting(In my opinion)
This is not correct. Without getting to details, a constant number of loops is not enough to sort, since sorting is Omega(nlogn) problem. Meaning, an O(1) (constant, including 1) number of loops is not enough for it - for any algorithm1,2.
Consider the input
5, 4, 3, 2, 1
a single loop of bubble sort will do:
4, 5, 3, 2, 1
4, 3, 5, 2, 1
4, 3, 2, 5, 1
4, 3, 2, 1, 5
So the algorithm will end up with the array: [ 4, 3, 2, 1, 5], which is NOT sorted.
After one loop of bubble sort, you are only guaranteed to have the last element in place (which indeed happens in the example). The second iteration will make sure the 2 last elements are in place, and the nth iteration will make sure the array is indeed sorted, resulting in n loops, which is achieved via a nested loop.
(1) The outer loop is sometimes hidden as a recursive call (quick sort is an example where it happens) - but there is still a loop.
(2) Comparisons based algorithms, to be exact.
For bubble sorting a pass simply moves the largest element to the end of array. So you need n-1 passes to get a sorted array, thats why you need other loop. Now for your code 1 pass means
if(number[0]>number[0+1])
{
temp=number[0+1];
number[0+1]=number[0];
number[0]=temp;
}
if(number[1]>number[1+1])
{
temp=number[1+1];
number[1+1]=number[1];
number[1]=temp;
}
.....6 more times
if(number[8]>number[8+1])
{
temp=number[8+1];
number[8+1]=number[8];
number[8]=temp;
}
so as you can see IF statement repeats itself, its just that after all 9 IFs the largets element moves to the end of array
This is not correct because
The algorithm gets its name from the way smaller elements "bubble" to the top of the list. (Bubble sort)
So, at the end of the first loop, we get the smallest element. So, for complete sorting, we have to keep total n loops. (where n = total size of the numbers)
I'm very new to C++ and I'm trying to learn the vector in C++..
I wrote the small program as below. I like to foreach(var sal in salaries) like C# but it doesn't allow me to do that so I googled it and found that I have to use iterator.. Im able to compile and run this program but I dont get the expected output.. I'm getting "0 0 0 0 0 0 1 2 3 4 5 6 7 8 9" instead of "0 1 2 3 4 5 6 7 8 9"..
Could anyone please explain me why? Thanks.
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;
void show(int i)
{
cout << i << " ";
}
int main(){
vector<int> salaries(5);
for(int i=0; i < 10; i++){
salaries.push_back(i);
}
for_each(salaries.begin(), salaries.end(), show);
}
You created a vector with 5 elements, then you push 10 more onto the end. That gives you a total of 15 elements, and the results you're seeing. Try changing your definition of the vector (in particular the constructor call), and you'll be set. How about:
vector<int> salaries;
This code creates a vector with a size of 5, and with each of those 5 elements initialized to their default value (0):
vector<int> salaries(5);
push_back inserts a new element, so here, you insert 10 new elements, ending up with a vector with 15 elements:
for(int i=0; i < 10; i++){
salaries.push_back(i);
}
You can create your vector like this instead:
vector<int> salaries;
and you'll get a vector with size 0.
Alternatively, you could initialize it with size 10, and then overwrite each element, instead of inserting new ones:
vector<int> salaries(10);
for(int i=0; i < 10; i++){
salaries[i] = i;
}
In some cases, it may be more efficient to write something like this:
vector<int> salaries; // create a vector with size 0
// allocate space for 10 entries, but while keeping a size of 0
salaries.reserve(10);
for(int i=0; i < 10; i++){
// because we reserved space earlier, these new insertions happen without
// having to copy the vector contents to a larger array.
salaries.push_back(i);
}
When you declare salaries(5), it's adding 5 entries into the vector with values of 0. Then your loop adds 0..9. Therefore you have 15 elements in your vector instead of just 10. Try declaring the vector without the 5.
vector<int> salaries;
vector<int> salaries(5); means, that you are creating the vector which contains 5 int objects from the start, and each int object is initialized with default constructor, and in the case of int contructor sets zero value. That's why you have 5 zero integers at the beginning of the vector container.
#Michael: Which book is that? I'd say
it's wrong. Using resize() is a good
practice if you know in advance how
big you need the vector to be, but
don't set the size at creation unless
you need the vector to contain
default-initialized values.
You can also reserve some capacity in the array in advance which is subtely different than re-size. Reserving simply reserves "at least" that much capacity for the vector (but does not change the size of the vector), while resize adds\removes elements to\from the vector to make it the requested size.
vector<int> salaries(5);
This creates a vector of 5 zeros for its elements. [0, 0, 0, 0, 0]
for(int i=0; i < 10; i++){
salaries.push_back(i);
}
This adds 10 more elements at the end ranging from 0 to 9 [0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Finally, I don't recommend using foreach so much. Functional programming has the downside of decentralizing code. It is extremely useful in some cases, but for these cases, and especially considering how you're starting out, I'd recommend:
for (vector<int>::const_iterator it = salaries.begin(), end = salaries.end(); it != end; ++it){
salaries.push_back(i);
}
Using this technique, you'll be able to iterate through any collection in the standard library without having to write separate functions or function objects for the loop body.
With C++0x you'll get a lot of goodies to make this easier:
for (int salary: salaries)
cout << salary << endl;
There's also BOOST_FOR_EACH already which is almost as easy if you can use boost.