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AlgorithmFriday_Week6
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def shiftLeft(arr, numm): | |
for _ in range(numm): # O(k) | |
temp = arr[0] | |
for i in range(len(arr)-1): # O(n) | |
arr[i] = arr[i+1] | |
arr[-1] = temp | |
return arr | |
def shiftRight(arr, numm): | |
for _ in range(numm): # O(k) | |
temp = arr[-1] | |
for i in range(len(arr)-1,0,-1): # O(n) | |
arr[i] = arr[i-1] | |
arr[0] = temp | |
return arr | |
def shuffleClass(array, num): | |
try: | |
# Check if array None or empty | |
if array == None or len(array)==0: | |
return [] | |
# Check if num==0 or array contains only one element | |
elif num ==None or num%len(array)==0 or len(array)<=1: | |
return array | |
else: | |
# Assume we're given an array of 5 elements, | |
# Shifting right 7 times is the same as shifting right 2 times i.e. 7%5 of which the later is more optimised | |
# Likewise, shifting right 4 times is the same as shifting left 1 time of which the later is more optimised | |
if num<0: | |
# convert to positive if negative | |
num = len(array)+num | |
num %= len(array) | |
else: | |
num %= len(array) | |
# Given an array of size n, we'll only shift n/2 times at most | |
if num <= len(array)//2: | |
# print(f"I shift Right {num} times") | |
return shiftRight(array, num) | |
else: | |
# print(f"I shift Left {len(array)-num} times") | |
return shiftLeft(array, len(array)-num) | |
except: | |
return [] |
Thank you @meekg33k.
I have spotted the error and made adjustments.
My mistake was not testing my solution on all test cases when I revised the code, I only assumed its correct.
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Hello @Julisam, thank you participating in Week 6 of #AlgorithmFridays.
You came really close to coming up with the most optimal solution for this problem with the optimization you had on line 26 - 41. That was really smart.
However, your solution didn't pass some of the test cases, more specifically the cases where
num
had a negative value. I think the error is traceable to line 31:num %= len(array)
. What do you think?That said, you are still in the running for the $50 award if you're able to correct in on/before 3.00 pm WAT today.
Thanks once again for participating and see you later today for Week 7 of #AlgorithmFridays.