0845MediumArrayTwo PointersGreedySorting

Spatula Flips to Sort the Stack

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A cook has a stack of pancakes described by arr, where arr[0] is the pancake on top and arr[len(arr) - 1] the one on the bottom. The pancakes carry the sizes 1 through len(arr), each exactly once.

A flip of depth d slides a spatula under the d-th pancake from the top and turns that whole top portion over, reversing the order of its first d pancakes.

The cook sorts the stack so sizes increase from top to bottom, using this routine. Take the largest size not yet settled. If it is not already in its final place, flip at its depth to bring it to the top, then flip at the depth of its final place to drop it there. A flip is never performed when it would change nothing. Repeat with the next largest size.

Return the depths of the flips the routine performs, in the order it performs them.

Examples

Example 1

Input: arr = [5, 1, 4, 2, 3]
Output: [5, 3, 4, 2, 3, 2]

The size 5 is already on top, so the routine flips at depth 5 to send it to the bottom, then goes on to settle the size 4, and so on down.

Example 2

Input: arr = [1, 2]
Output: []

Both pancakes already sit in their final places, so no flip is performed.

Example 3

Input: arr = [2, 1]
Output: [2]

The size 2 is already on top, so no flip is needed to bring it up, and a flip of depth 2 sends it to the bottom.

Constraints

1 <= arr.length <= 100
1 <= arr[i] <= 100
arr holds each of the sizes 1 through arr.length exactly once

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def spatula_flips(arr: list[int]) -> list[int]:

Java

public List<Integer> spatulaFlips(int[] arr)

The intended approach

Time O(n^2) for n pancakes: each round searches for one size and reverses a prefix.Space O(n) for the working stack, plus the returned list of depths.

Settle the largest unsettled size each round, because once a pancake is in its final place the rounds that follow never reach that deep. Find where that size currently sits: a flip at its depth brings it to the top, and a flip at the depth of its final place sends it down to where it belongs. Skipping a flip that would change nothing is what keeps the answer well defined, and it covers both a pancake already on top and one already in its final place.

Pythonentry point: spatula_flips

def spatula_flips(arr):
    stack = list(arr)
    flips = []
    for size in range(len(stack), 1, -1):
        target = size - 1
        if stack[target] == size:
            continue                      # already in its final place
        at = stack.index(size)
        if at != 0:
            flips.append(at + 1)
            stack[:at + 1] = stack[:at + 1][::-1]
        flips.append(size)
        stack[:size] = stack[:size][::-1]
    return flips

The obvious version

Follow the same routine but locate each size by scanning the stack position by position and carry out each flip by building a fresh list rather than reversing in place, so the bookkeeping is done the long way round.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_spatula_flips(arr):
    stack = list(arr)
    flips = []
    n = len(stack)
    remaining = n
    while remaining > 1:
        where = -1
        for i in range(n):
            if stack[i] == remaining:
                where = i
                break
        if where == remaining - 1:
            remaining -= 1
            continue
        if where != 0:
            depth = where + 1
            flips.append(depth)
            top = stack[:depth]
            top = top[::-1]
            stack = top + stack[depth:]
        depth = remaining
        flips.append(depth)
        top = stack[:depth]
        top = top[::-1]
        stack = top + stack[depth:]
        remaining -= 1
    return flips

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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