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# How to left shift k elements of an array Taimur Hassan

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### Problem statement

This problem requires us to shift/rotate a fixed-size array to the left by k positions, where k is a user-determined variable. For example, given an array [1,2,3,4,5], a shift of 3 positions should result in the final state of the array being [4,5,1,2,3] . There are several different ways to achieve this.

### A simple solution

One simple solution is to write an algorithm to shift/rotate the array once and then repeat this a total of k times.

Let's look at the code below:

#include <iostream>
using namespace std;

void shiftArr(int* arr , int n)
{
// Initially: [1 2 3 4 5]

// swapping the first element with the last element
int temp = arr;
arr = arr[n-1];
arr[n-1] = temp;

// Now: [5 2 3 4 1]

// swapping the first element with the (N-i)th element for the entire length of the array (where i starts from 1)
for(int i=n-2 ; i>=0 ; i--)
{
int temp2 = arr;
arr = arr[i];
arr[i] = temp2;

// Array in each iteration:
// [4 2 3 5 1]
// [3 2 4 5 1]
// [2 3 4 5 1]
}
}

void shiftArrByK(int* arr , int n , int k)
{
for(int i=0 ; i<k%n ; i++) // we take a modulus in case the value of k is >= N
{
shiftArr(arr , n) ;
}
}

int main() {
int arr[] = {1,2,3,4,5};
int size_of_array = 5 ;
int k = 2 ;
cout << "Before shifting: " ;
for(int i=0 ; i<size_of_array ; i++)
{
cout << arr[i] << " " ;
}
cout << endl ;
shiftArrByK(arr , size_of_array , k);

cout << "After shifting: " ;
for(int i=0 ; i<size_of_array ; i++)
{
cout << arr[i] << " " ;
}
cout << endl ;
}

However, this is a very inefficient way to perform this task. In the worst case, it will take $O(N2)$ time, where $N$ is the size of the array. This is because this approach involves moving all the array elements to left-shift the array just once (moving a single element takes O(1) time, and there are a total of $N$ elements), and we may need to left-shift the elements up to $N$ times.

### An efficient solution

A better approach would be one in which the number of times the elements need to be shifted is minimized. The diagram below depicts the algorithm implementing this approach with an example:

An algorithm to shift an array to the left k times (k=2 in this example)

Note: In this method, the maximum number of times that a single element is shifted is $2$, whereas in the previous method, it was $N$.

Let's implement this algorithm below:

#include <iostream>
using namespace std;

void reverseArray(int* arr , int low , int high)
{
while(low<high)
{
swap(arr[low] , arr[high]);
low++;
high--;
}
}

void shiftLeftByN(int* arr , int n , int k)
{
k = k%n ; // we take a modulus in case the value of k is >= N
reverseArray(arr , 0 , k-1);
reverseArray(arr , k , n-1);
reverseArray(arr , 0 , n-1);
}

int main() {
int arr = {1,2,3,4,5} ;
int size_of_array=5 ;
int k=2 ;
cout << "Before rotation: " ;
for(int i=0 ; i<size_of_array ; i++)
{
cout << arr[i] << " " ;
}
cout << endl ;

shiftLeftByN(arr , size_of_array , k) ;

cout << "After rotation: " ;
for(int i=0 ; i<size_of_array ; i++)
{
cout << arr[i] << " " ;
}
cout << endl ;
}

This method takes $O(N)$ time in the worst case. This is because it involves reversing the whole array just once and reversing 2 sub-arrays once each. That makes a total of 3 reversals, and each reversal takes $O(N)$ time. Therefore, this method provides the most efficient way to left shift k elements of an array.

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CONTRIBUTOR Taimur Hassan 