Styjun

I have a question where I need to rotate an array left k times.

i.e. if k = 2, [1, 2, 3, 4, 5] . -> [3, 4, 5, 1, 2]

So, my code is:

def array_left_rotation(a, n, k):
 for i in range(n):
 t = a[i]
 a[i] = a[(i+n-1+k)%n]
 a[(i+n-1+k)%n] = t

 return a

where n = length of the array.

I think the problem is a mapping problem, a[0] -> a[n-1] if k = 1.

What is wrong with my solution?

Another way to do this with the help of indexing is shown below..

def rotate(l, n):
 return l[n:] + l[:n]

print(rotate([1, 2, 3, 4, 5], 2))

#output : [3, 4, 5, 1, 2]

This will only return the original list if n is outside the range [-len(l), len(l)]. To make it work for all values of n, use:

def rotate(l, n):
 return l[-n % len(l):] + l[:-n % len(l)]

One way you can do is by using collections.deque

from collections import deque
k = 2
l = [1, 2, 3, 4, 5]
dl = deque(l)
dl.rotate(k+1)
list(dl)
#[3, 4, 5, 1, 2]

This solution requires constant extra memory, and runs in O(nk).

If the array has zero length, or there are zero rotations, skip the loop and return arr.

For every rotation we store the first element, then shift every other element to the left, finally placing the first element at the back of the list. We save some work for large values of k by recognizing that every rotation after the nth is a repeat solution -> k % n.

def array_left_rotation(arr, n, k):
 for _ in range(0 if 0 in [n, k] else k % n):
 temp = arr[0]
 for idx in range(n - 1):
 arr[idx] = arr[idx + 1]
 arr[-1] = temp
 return arr

N, d = map(int, input().split()) #taking length N and no. of rotations d
a = list(input().split()) #taking input array as list
r = a[d % N : N] + a[0 : d % N] #rotating the array(list)
print(r) 

5 2 #giving N,D as input
1 2 3 4 5 #giving input array
['3', '4', '5', '1', '2'] #output

def leftRotation(a, d, n):
 i=0
 while i < n:
 print (a[(i+d)%n], end = ' ')
 i+=1
 return i

if __name__ == '__main__':
 a =[1, 2, 3, 4, 5, 6, 7]
 d = 4
 n = 7 or len(a)
 leftRotation(a, d, n)

This is how I solved my hackerrank left rotation challenge

def rotate_left3(nums):
temp=[]
for i in range(len(nums)-1):
 temp=nums[i]
 nums[i]=nums[i+1]
 nums[i+1]=temp

return nums

This is how I did it:

def rotate_left3(nums):
 return [ nums[1] , nums[2] , nums[0] ]

It worked perfect 100%, but if you add more sums, you'd have to add its just set for the 3 that the problem specified.