Created
July 11, 2017 06:12
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Larry's Array - Check if an array can be sort by rotating 3 consecituve elements (all elements are unique)
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''' | |
Given an array, all elements are unique, you can only perform rotation on 3 consecutive elements: | |
A B C | |
Example | |
A B C -> B C A -> C A B -> A B C | |
Given an array, find out if it is possible to sort it by only performing rotation. | |
Solution: | |
https://www.hackerrank.com/challenges/larrys-array/editorial | |
Basically, count the inversions in the array. | |
If it is even, then the array can be sorted | |
Otheswise, it cant be sorted. | |
Using the old code for counting inversions using merge sort | |
''' | |
def merge_sort(arr, inverse_count): | |
if len(arr) <= 1: | |
return arr | |
# divide into two parts | |
mid = (len(arr)+1)//2 | |
left = arr[:mid] | |
right = arr[mid:] | |
# recursive call on both parts | |
left = merge_sort(left, inverse_count) | |
right = merge_sort(right, inverse_count) | |
# merge and return the result | |
return merge(left, right, mid, inverse_count) | |
def merge(left, right, mid, inverse_count): | |
result = [None]*(len(left) + len(right)) | |
i, j, k = 0, 0, 0 | |
count = 0 | |
# one by one pick smallest elements from left and right | |
#print(left, right, end = ' ') | |
while i < len(left) and j < len(right): | |
if left[i] <= right[j]: | |
result[k] = left[i] | |
k += 1 | |
i += 1 | |
else: | |
result[k] = right[j] | |
k += 1 | |
j += 1 | |
count += mid - i | |
# get the rest from left (if any) | |
while i < len(left): | |
result[k] = left[i] | |
k += 1 | |
i += 1 | |
# get the rest from right (if any) | |
while j < len(right): | |
result[k] = right[j] | |
k += 1 | |
j += 1 | |
#print(count) | |
inverse_count[0] += count | |
return result | |
if __name__ == '__main__': | |
from sys import stdin, stdout | |
t = int(stdin.readline()) | |
for _ in range(t): | |
n = int(stdin.readline()) | |
arr = [int(x) for x in stdin.readline().split()] | |
inversions = [0] | |
merge_sort(arr, inversions) | |
inv_count = inversions[0] | |
print('YES' if inv_count % 2 == 0 else 'NO') |
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