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Formatted question description: https://leetcode.ca/all/969.html

# 969. Pancake Sorting (Medium)

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. 1 <= A.length <= 100
2. A[i] is a permutation of [1, 2, ..., A.length]

Companies:
Square

Related Topics:
Array, Sort

## Solution 1.

• class Solution {
public List<Integer> pancakeSort(int[] A) {
List<Integer> list = new ArrayList<Integer>();
int length = A.length;
for (int i = length; i >= 1; i--) {
if (A[i - 1] != i) {
int flipIndex = 0;
for (int j = length - 1; j >= 0; j--) {
if (A[j] == i) {
flipIndex = j;
break;
}
}
if (flipIndex > 0) {
pancakeFlip(A, flipIndex);
}
pancakeFlip(A, i - 1);
}
}
return list;
}

public void pancakeFlip(int[] array, int index) {
int left = 0, right = index;
while (left < right) {
int temp = array[left];
array[left] = array[right];
array[right] = temp;
left++;
right--;
}
}
}

############

class Solution {
public List<Integer> pancakeSort(int[] arr) {
int n = arr.length;
List<Integer> ans = new ArrayList<>();
for (int i = n - 1; i > 0; --i) {
int j = i;
for (; j > 0 && arr[j] != i + 1; --j)
;
if (j < i) {
if (j > 0) {
reverse(arr, j);
}
reverse(arr, i);
}
}
return ans;
}

private void reverse(int[] arr, int j) {
for (int i = 0; i < j; ++i, --j) {
int t = arr[i];
arr[i] = arr[j];
arr[j] = t;
}
}
}

• // OJ: https://leetcode.com/problems/pancake-sorting
// Time: O(N^2)
// Space: O(1)
class Solution {
public:
vector<int> pancakeSort(vector<int>& A) {
vector<int> ans;
for (int i = A.size(); i > 0; --i) {
int j = i - 1;
for (; j >= 0 && A[j] != i; --j);
reverse(A.begin(), A.begin() + j + 1);
ans.push_back(j + 1);
reverse(A.begin(), A.begin() + i);
ans.push_back(i);
}
return ans;
}
};

• class Solution:
def pancakeSort(self, arr: List[int]) -> List[int]:
def reverse(arr, j):
i = 0
while i < j:
arr[i], arr[j] = arr[j], arr[i]
i, j = i + 1, j - 1

n = len(arr)
ans = []
for i in range(n - 1, 0, -1):
j = i
while j > 0 and arr[j] != i + 1:
j -= 1
if j < i:
if j > 0:
ans.append(j + 1)
reverse(arr, j)
ans.append(i + 1)
reverse(arr, i)
return ans

############

class Solution(object):
def pancakeSort(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
N = len(A)
res = []
for x in range(N, 0, -1):
i = A.index(x)
res.extend([i + 1, x])
A = A[:i:-1] + A[:i]
return res

• func pancakeSort(arr []int) []int {
var ans []int
n := len(arr)
reverse := func(j int) {
for i := 0; i < j; i, j = i+1, j-1 {
arr[i], arr[j] = arr[j], arr[i]
}
}
for i := n - 1; i > 0; i-- {
j := i
for ; j > 0 && arr[j] != i+1; j-- {
}
if j < i {
if j > 0 {
ans = append(ans, j+1)
reverse(j)
}
ans = append(ans, i+1)
reverse(i)
}
}
return ans
}

• function pancakeSort(arr: number[]): number[] {
let ans = [];
for (let n = arr.length; n > 1; n--) {
let index = 0;
for (let i = 1; i < n; i++) {
if (arr[i] >= arr[index]) {
index = i;
}
}
if (index == n - 1) continue;
reverse(arr, index);
reverse(arr, n - 1);
ans.push(index + 1);
ans.push(n);
}
return ans;
}

function reverse(nums: Array<number>, end: number): void {
for (let i = 0, j = end; i < j; i++, j--) {
[nums[i], nums[j]] = [nums[j], nums[i]];
}
}


• impl Solution {
pub fn pancake_sort(mut arr: Vec<i32>) -> Vec<i32> {
let mut res = vec![];
for n in (1..arr.len()).rev() {
let mut max_idx = 0;
for idx in 0..=n {
if arr[max_idx] < arr[idx] {
max_idx = idx;
}
}
if max_idx != n {
if max_idx != 0 {
arr[..=max_idx].reverse();
res.push(max_idx as i32 + 1);
}
arr[..=n].reverse();
res.push(n as i32 + 1);
}
}
res
}
}