Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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:
k where 1 <= k <= arr.length.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 <= 1001 <= arr[i] <= arr.lengtharr are unique (i.e. arr is a permutation of the integers from 1 to arr.length).Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Two Pointers · Greedy
[3,2,4,1]
[1,2,3]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #969: Pancake Sorting
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) {
ans.add(j + 1);
reverse(arr, j);
}
ans.add(i + 1);
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;
}
}
}
// Accepted solution for LeetCode #969: Pancake Sorting
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
}
# Accepted solution for LeetCode #969: Pancake Sorting
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
// Accepted solution for LeetCode #969: Pancake Sorting
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
}
}
// Accepted solution for LeetCode #969: Pancake Sorting
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]];
}
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.