## 969. Pancake Sorting

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).

## Rust Solution

struct Solution;

impl Solution {
fn pancake_sort(mut a: Vec<i32>) -> Vec<i32> {
let n = a.len();
let mut res = vec![];
for i in 0..n {
let (_, j) = (0..n - i).fold(
(a[0], 0),
|acc, j| {
if a[j] > acc.0 {
(a[j], j)
} else {
acc
}
},
);
a[0..=j].reverse();
a[0..n - i].reverse();
res.push(j + 1);
res.push(n - i);
}
res.iter().map(|&x| x as i32).collect()
}
}

#[test]
fn test() {
let a = vec![3, 2, 4, 1];
let res = vec![3, 4, 2, 3, 1, 2, 1, 1];
assert_eq!(Solution::pancake_sort(a), res);
}

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