arrayRotations.cpp

/*
A left rotation operation on an array of size shifts each of the array's elements unit to the left. For example, if left rotations are performed on array , then the array would become .

Given an array of integers and a number, , perform left rotations on the array. Then print the updated array as a single line of space-separated integers.

Input Format

The first line contains two space-separated integers denoting the respective values of (the number of integers) and (the number of left rotations you must perform).
The second line contains space-separated integers describing the respective elements of the array's initial state.

Constraints

Output Format

Print a single line of space-separated integers denoting the final state of the array after performing left rotations.

Sample Input

5 4
1 2 3 4 5

Sample Output

5 1 2 3 4

Explanation

When we perform left rotations, the array undergoes the following sequence of changes:

Thus, we print the array's final state as a single line of space-separated values, which is 5 1 2 3 4. 
 */

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

vector<int> array_left_rotation(vector<int> a, int n, int k) {
    vector<int> o(n);
    int o_i = ((n - k) == 0)? 0 : (n-k);
    for (int a_i = 0; a_i < n; a_i++) {
        o[o_i] = a[a_i];
        o_i++;
        if (o_i == n)
            o_i = 0;
    }
    return o;
}

int main(){
    int n;
    int k;
    cin >> n >> k;
    if (n < 1 || n > 100000)
        return -1;
    if (k > n)
        return -1;
    vector<int> a(n);
    for(int a_i = 0;a_i < n;a_i++){
        cin >> a[a_i];
    }
    vector<int> output = array_left_rotation(a, n, k);
    for(int i = 0; i < n;i++)
        cout << output[i] << " ";
    cout << endl;
    return 0;
}