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Given an array of integers arr, sort the array by performing a series of pancake flips.

In one pancake flip we do the following steps:

Choose an integer k where 1 <= k <= arr.length. Reverse the sub-array arr[0…k-1] (0-indexed). For example, if arr = [3,2,1,4] and we performed a pancake flip choosing k = 3, we reverse the sub-array [3,2,1], so arr = [1,2,3,4] after the pancake flip at k = 3.

Return an array of the k-values corresponding to a sequence of pancake flips that sort arr. Any valid answer that sorts the array within 10 * arr.length flips will be judged as correct.

Example 1:

Input: arr = [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: arr = [3, 2, 4, 1]
After 1st flip (k = 4): arr = [1, 4, 2, 3]
After 2nd flip (k = 2): arr = [4, 1, 2, 3]
After 3rd flip (k = 4): arr = [3, 2, 1, 4]
After 4th flip (k = 3): arr = [1, 2, 3, 4], which is sorted.

Example 2:

Input: arr = [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. 

Solution

Idea: find the biggest item, flip it to top, then flip the entire stack to make the biggest to the bottom. Move to the next biggest one.

def pancakeSort(arr):
    res = []
    n = len(arr)

    for x in range(n, 1, -1):
        # general case: idx = find_max(arr, x)
        # this problem is special because
        # the values range from 0 to k
        idx = arr.index(x)
        res.extend([idx + 1, x])
        arr = arr[:idx:-1] + arr[:idx]
    return res