Arrays: Left Rotation | Editorial by coderboy

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Simply you can use the concept of vector (STL).

push_back the first element and then remove first element from the vector and do this for 'd' times.

Explanation :

let suppose an array of size 5 having elements [1, 2, 3, 4, 5] and number of rotations is 4.

Pseudo code :

```
 for (int i = 0; i < d; i++)
    {
        v.push_back(v[0]);
        v.erase(v.begin());
    }
```

When we perform d = 4 left rotations, the array undergoes the following sequence of changes:

[2, 3, 4, 5, 1], [3, 4, 5, 1, 2], [4, 5, 1, 2, 3], [5, 1, 2, 3, 4]

In first iteration when i = 0 : After push_back operation vector 'v' will be [1, 2, 3, 4, 5, 1] and then 'erase' function which delete the element, 
after erase operation vector 'v' will be [2, 3, 4, 5, 1]

In second iteration when i = 1; firstly, vector will be [2, 3, 4, 5, 1, 2] and then [3, 4, 5, 1, 2]

In third iteration when i = 2; firstly, vector will be [3, 4, 5, 1, 2, 3] and then [4, 5, 1, 2, 3]

In third iteration when i = 3; firstly, vector will be [4, 5, 1, 2, 3, 4] and then [5, 1, 2, 3, 4]

```
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n, d;
    cin >> n >> d;

vector <int> v;
    int input;

for (int i = 0; i < n; i++)
    {
        cin >> input;
        v.push_back(input);
    }
    for (int i = 0; i < d; i++)
    {
        v.push_back(v[0]);
        v.erase(v.begin());
    }
    for (int i = 0; i < v.size(); i++)
    {
        cout << v[i] << " ";
    }
    cout << endl;
}

```

Arrays: Left Rotation | Editorial by coderboy

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