hackerrank Circular Array Rotation Python

Question

I tried to solve the Circular Array Rotation problem on HackerRank. https://www.hackerrank.com/challenges/circular-array-rotation/problem

The following code passes all test cases except case #4, which gets a runtime error. Could someone point out the problem?

def circularArrayRotation(a, k, queries):
 if k < len(a):
 k = k
 elif k == len(a):
 k = 0
 else:
 k = k%a
 newList = []
 for val in queries:
 newInd = -k+val
 if abs(newInd) > len(a):
 newInd = newInd - (len(a)-1)
 newList += [a[newInd]]
 else:
 newList += [a[newInd]] 
 return newList

• sahasrara62: you need to make it fast as it is exceeding the required time limit for that case
• P Song: I don't know how to make it faster
• sahasrara62: Try to get the rotated array at once and in new list store the value of that query index from rotated array

Answer 1

Your solution is correct. But is not running within that time limit for that case 4 only.

You are calculating the value each time for new queries, which is taking time.

What you can do is take the rotated array at once . and then run the queries on the rotated array. Save the result in the list and return it back.

def circularArrayRotation(a, k, queries):
 new_arr = a[-k%len(a):] + a[:-k%len(a)]
 # list slicing is done here. it will get the right rotated array 
 result = []
 for i in queries:
 result.append(new_arr[i])
 # running queries on rotated array
 return result

using above method , list slicing is done in o(n) time . and then running the queries is done is o(1) time.

Answer 2

def circularArrayRotation(a, k, queries):
 j= len(a)
 for x in range(k):
 n=a[j-1]
 a.insert(0,n)
 a.pop(j)
 newList= []
 for m in queries:
 newList.append(a[m])
 return newList

Answer 3

def circularArrayRotation(a, k, queries):
 #rotation
 for _ in range(k):
 a.insert(0, a.pop())
 #putting answer according to query
 ans = []
 for i in queries:
 ans.append(a[i]) 
 return ans

• Thavas Antonio: While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value.

Answer 4

Well you can use deque A deque is a double-ended queue. It can be used to add or remove elements from both ends.

from collections import deque
 def circularArrayRotation(a, k, queries):
 result=[]
 a = deque(a) 
 a.rotate(k) 
 a = list(a) 
 for v in queries:
 result.append(a[v])
 return(result)

Answer 5

I know this is very old, but I was just solving this, so I figured might as well drop by, your else case is wrong! It should be k = k % len(a), you wrote k % a, now I write c++, so don't know what that even does in python, but I'm pretty sure that doesn't automatically put in len(a) Besides, why are you bothering putting that in if else block just modulo every time, if it's smaller, no value change, for => it fixes it to the right value.

Answer 6

I believe a deque is the approach to go, as AnkushRasgon's and shyamantha's answers. Here I post my version in Java. All tests passed.

static int[] circularArrayRotation(int[] a, int k, int[] queries) {
 LinkedList<Integer> list = Arrays.stream(a).boxed()
 .collect(Collectors.toCollection(LinkedList::new));
 
 for (int i = 0; i < k; i++) {
 list.push(list.pollLast());
 }
 
 for (int i = 0; i < queries.length; i++) {
 if (queries[i] < list.size()) {
 queries[i] = list.get(queries[i]);
 }
 }
 
 return queries;
}

Answer 7

def circularArrayRotation(a, k, queries):
 reverse_a = list(reversed(a))
 reverse_a_copy = reverse_a.copy()
 for x in range(k):
 item = reverse_a[x]
 reverse_a_copy.remove(item)
 reverse_a_copy.append(item)
 line = []
 for x in queries:
 line.append(list(reversed(reverse_a_copy))[x])
 return line

• Brad Figueroa: You should always try to provide some explanation when you try to help someone solving a problem, it's much better and appreciated

Answer 8

a=[1,2,3,4,5]
s=2
def rotateList(arr,d,n):
 arr[:]=arr[d-1:n]+arr[0:d-1]
 return arr
print(rotateList(a,5,len(a)))

Answer 9

 for i in range(k):
 random_a = len(a)
 random_b = random_a - 1
 random_c = a.pop(random_b)
 a = [random_c] + a
 oz = a
 random_list = []
 for i in queries:
 z = oz[i]
 y = str(z)
 random_list.append(y)
 return random_list```
is this the correct answer
 

• Frank: Better to give an explanation of your answer.