Pancake Sorting
Created: September 15, 2019 by [lek-tin]
Last updated: September 15, 2019
Given an array A, we can perform a pancake flip: We choose some positive integer k <= A.length, then reverse the order of the first k elements of A. We want to perform zero or more pancake flips (doing them one after another in succession) to sort the array A.
Return the k-values corresponding to a sequence of pancake flips that sort A. Any valid answer that sorts the array within 10 * A.length flips will be judged as correct.
Example 1
Input: [3,2,4,1]
Output: [4,2,4,3]
Explanation:
We perform 4 pancake flips, with k values 4, 2, 4, and 3.
Starting state: A = [3, 2, 4, 1]
After 1st flip (k=4): A = [1, 4, 2, 3]
After 2nd flip (k=2): A = [4, 1, 2, 3]
After 3rd flip (k=4): A = [3, 2, 1, 4]
After 4th flip (k=3): A = [1, 2, 3, 4], which is sorted.
Example 2
Input: [1,2,3]
Output: []
Explanation: The input is already sorted, so there is no need to flip anything.
Note that other answers, such as [3, 3], would also be accepted.
Note
1 <= A.length <= 100A[i]is a permutation of[1, 2, ..., A.length]
Solution
class Solution:
def pancakeSort(self, nums: List[int]) -> List[int]:
n = len(nums)
ks = []
# We can confidently skip the the last element, which is at index 0, because it is guaranteed to be 1 after the last flip
for i in range(n-1, 0, -1):
# suppose the max value in at index 0
MAX = 0
for j in range(1, i+1):
if nums[j] > nums[MAX]:
MAX = j
# MAX != 0 means MAX is not at index 0
# MAX != i means the number at index i is indeed the WRONG number
if MAX != 0 and MAX != i:
ks.append(MAX+1)
nums[:MAX+1] = reversed(nums[:MAX+1])
ks.append(i+1)
nums[:i+1] = reversed(nums[:i+1])
elif MAX == 0:
ks.append(i+1)
nums[:i+1] = reversed(nums[:i+1])
return ks