Pancake Sorting

Problem Description

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.

Constraints:

• 1 <= arr.length <= 100
• 1 <= arr[i] <= arr.length
• All integers in arr are unique (i.e. arr is a permutation of the integers from 1 to arr.length).

Solution (Go)

func pancakeSort(arr []int) []int {
	flips := make([]int, 0)
	for i := len(arr) - 1; i > 0; i-- {
		maxIndex := maxIdx(arr[:i+1])
		if maxIndex != i {
			flips = append(flips, maxIndex+1)
			arr = flip(arr, maxIndex)

			flips = append(flips, i+1)
			arr = flip(arr, i)
		}
	}
	return flips
}

func maxIdx(array []int) int {
	i := 0
	max := array[0]
	for index, value := range array {
		if max < value {
			i = index
			max = value
		}
	}
	return i
}

func flip(array []int, index int) []int {
	for i := 0; i < (index+1)/2; i++ {
		array[i], array[index-i] = array[index-i], array[i]
	}
	return array
}