Rohan Shahane Rohan Shahane - 3 months ago 44
C++ Question

Why am I getting 'Terminated due to timeout' error here?

I'm solving this problem on Hackerrank.

John Watson performs an operation called a right circular rotation on an array of integers, [a[0],a[ 1]......a[n]].After performing one right circular rotation operation, the array is transformed from [a[0],a[ 1]......a[n]] to [a[n- 1],a[0]......a[n-2]].

Watson performs this operation 'k' times. To test Sherlock's ability to identify the current element at a particular position in the rotated array, Watson asks 'q' queries, where each query consists of a single integer, 'm', for which you must print the element at index in the rotated array (i.e., the value of a[m] ).

Input Format

The first line contains 3 space-separated integers,'n' ,'k' , and , 'q' respectively.
The second line contains 'n' space-separated integers, where each integer 'i' describes array element a[n] (where 0<=i

Output Format

For each query, print the value of the element at index 'm' of the rotated array on a new line.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int n,k,q,temp;
cin>>n>>k>>q; //Input of n,k,q;
int arr[n],qur[q];
for(int i=0;i<n;i++) //Input of array
for(int i=0;i<q;i++) //Input of query numbers

for(int z=0;z<k;z++)
temp = arr[n-1];
for(int i=n-1;i>0;i--)

for(int i=0;i<q;i++)

return 0;

The code's logic seems to be flawless. I'm a newbie.Thanks in advance.


The reason is that you don't really need to perform the rotation, you can use The Ancient Power of Mathematics.
(Performing the rotation can require 10 billion arr[i]=arr[i-1]s, and doing that 500 times would take a while.)

If you start with the sequence

Element: | 12 | 34 | ... | 56 | 78 |           (1)
Index:      0    1   ...   k-1   k

and right-rotate by one, you get

Element: | 78 | 12 | 34 | ... | 56 |           (2)
Index:      0    1    2   ...    k

or, changing the viewpoint just a little bit:

Element: | 12 | 34 | ... | 56 | 78 |           (3)
Index:      1    2   ...    k    0

There's a fairly simple relationship between the indices of the elements in (1) and (3), and you only need some arithmetic to get from one to the other.

Discovering the relationship and performing the arithmetic left as an exercise.