Moving values in Array to the left/right to k positions? - c++

Question

I have an array of n values. I have a value p which can be 1 or -1. I have to move the positions in the Array k positions to the left if p = 1 or to the right if p = -1.
1 ≤ N ≤ 1.000.000
0 ≤ K ≤ 1.000.000
I tried:
#include <iostream>
using namespace std;
int main() {
int n, p, k;
int v[1000002];
cin >> n >> k >> p;
for ( int i = 1; i <= n; i++ ) {
cin >> v[i];
}
if( p == 1) {
for ( int i = k + 1; i <= n ; i++ ) {
cout << v[i] << " ";
}
for ( int i = 1; i <= k; i++ ) {
cout << v[i] << " ";
}
}
else {
for ( int i = n - k + 1; i <= n; i++ ) {
cout << v[i] << " ";
}
for ( int i = 1; i <= n - k; i++ ) {
cout << v[i] << " ";
}
}
return 0;
}
The problem is I use an online tester to do tests on it to see if it applies to all possible cases, but this solution passes all but 3 tests. The results are correct, but it says that I have surpassed the time limit for the respective test. I cannot comprehend how because I only use essential code like reading and printing the array.

Answer 1

I'm not sure if it will help resolve your timeouts, but I can see the following to reduce your calls a tiny bit:
After you read in n, p and k, calculate your offset.
Create a vector that is offset in size.
Read into vector until it's full.
Pipe all input directly to stdout.
Write out of the vector to stdout.
This process will reduce the readin->writeout to a single step for (n - offset elements, where offset = k or n - k depending on p).
If this still fails, you can build a stringstream and write to stringstream, and flush to cout once in a while (i.e. every 8 k or so) to reduce calls to your io channel (cout).
Sample code:
int n, p, k;
cin >> n >> k >> p;
int offset = p == 1 ? k : n - k;
std::vector tmp(offset);
std::stringstream sout;
for (size_t i = 0; i < offset; i++)
{
cin >> tmp[i];
}
for (size_t i = offset; i < n; i++)
{
cin >> sout << ' ';
if (sout.size() > _BUFFER_MAX)
{
cout << sout.str();
std::stringstream().swap(sout);
}
}
cout << sout.str();
std::stringstream().swap(sout);
for (size_t i = 0; i < tmp.size(); i++)
{
sout << tmp[i] << ' ';
if (sout.size() > _BUFFER_MAX)
{
cout << sout.str();
std::stringstream().swap(sout);
}
}
cout << sout.str();

Answer 2

I appreciate the way you're trying to "cheat". There will be many times in your programming career when the best plan is to provide what the customer asks for, completely ignoring the way they think you should provide it.
That enormous array at the beginning is awful, though -- you should use a std::vector with n elements, not 10000002.
I'm surprised that you are running out of time, since all the time is in I/O, and you are making the "obvious" I/O calls that will be similar to what everyone else is doing... but there are a lot of details you don't know about the test environment.
Try putting this at he start of main:
std::ios_base::sync_with_stdio(false);
See: https://en.cppreference.com/w/cpp/io/ios_base/sync_with_stdio
By default, every time you write to cout, the data is pushed into the stdout C stream. It's possible that the test environment is set up in a way the makes this slow.
If you turn off the syncing, you can use cout's internal bufferring, which will make all those little writes a lot faster in those cases.

Answer 3

I have managed to find the problem in my code.
Apparently the problem was that the number of positions to move could be bigger than the actual number of values. This led to some very long for's.
I feel pretty stupid , but I have learned many new things through your answers. Thank you very much!

Answer 4

if k is very large you can simply take it to the mod n (k%n)
you can perform placements in O(n) by calculating the final index of each element after rotation using the formulas:
(i + k) % n for right rotation and (i - k) % n for left rotation.
note: you should add an extra 'n' to handle negative modulo.
here's the code:
int v[1000001];
int main() {
int n, p, k;
cin >> n >> k >> p;
k %= n;
for (int i = 0; i < n; i++) {
if (p == -1) //right
cin >> v[(i + k) % n];
else if (p == 1) //left
cin >> v[(i - k + n) % n];
}
for (int i = 0; i < n; i++) {
cout << v[i] << " ";
}
return 0;
}