Pancake Sorting

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ArraysGreedy Algorithms

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).

Try coding it in LeetCode

Pancake Sorting

Medium

Block

1 view

Topics:

ArraysGreedy Algorithms

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).

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array `arr`?
Is the input array `arr` guaranteed to contain a permutation of integers from 1 to n, or could there be missing or repeated numbers?
If there are multiple possible sequences of flips that sort the array, is any valid sequence acceptable?
What should be returned if the input array is already sorted?
Are the integers in the array guaranteed to be positive?

Brute Force Solution

Approach

Pancake sorting is like arranging pancakes in a stack from largest to smallest, using only flips of the spatula. The brute force way tries out all possible flip combinations until it finds one that sorts the stack correctly.

Here's how the algorithm would work step-by-step:

Consider all possible sequences of flips you can make.
For each sequence of flips, actually perform those flips on the pancake stack to see what the resulting order of pancakes is.
Check if the resulting pancake stack is sorted from largest at the bottom to smallest at the top.
If a sequence of flips results in a sorted stack, remember that sequence.
Repeat this process for all possible sequences of flips until you've checked them all.
Once you've found all the sequences that sort the stack, choose the shortest sequence of flips.

Code Implementation

def pancake_sort_brute_force(pancakes):
    # Generate all possible flip sequences.
    def generate_flip_sequences(num_pancakes):
        if num_pancakes == 0:
            return [[]]
        
        smaller_sequences = generate_flip_sequences(num_pancakes - 1)
        new_sequences = []
        for sequence in smaller_sequences:
            for i in range(1, num_pancakes + 1):
                new_sequences.append(sequence + [i])
        return new_sequences

    # Function to flip a pancake stack
    def flip(pancake_stack, flip_index):
        return pancake_stack[:flip_index][::-1] + pancake_stack[flip_index:]

    # Function to check if a pancake stack is sorted
    def is_sorted(pancake_stack):
        for i in range(len(pancake_stack) - 1):
            if pancake_stack[i] > pancake_stack[i + 1]:
                return False
        return True

    # This section finds the best solution
    shortest_solution = None
    all_flip_sequences = generate_flip_sequences(len(pancakes))

    # Iterate through all possible flip sequences
    for flip_sequence in all_flip_sequences:
        current_stack = pancakes[:]
        for flip_index in flip_sequence:
            current_stack = flip(current_stack, flip_index)

        # Check if this flip sequence resulted in a sorted stack
        if is_sorted(current_stack):

            # Update if it's the shortest solution found so far
            if shortest_solution is None or len(flip_sequence) < len(shortest_solution):
                shortest_solution = flip_sequence

    # Return the shortest sequence of flips
    if shortest_solution is not None:
        return shortest_solution
    else:
        return []

Big(O) Analysis

Time Complexity

O((n!)*n^2) – The brute force approach considers all possible flip sequences. There are n! possible permutations of the initial pancake stack that can be reached by flips. For each of these permutations, we simulate the flips, which takes O(n) time to flip a sub-array. We also need to check if the resulting stack is sorted, taking O(n) time. Therefore, for each of the n! permutations, it takes O(n) + O(n) = O(n) to flip and verify. Because we are evaluating the flips for all permutations, we multiply the number of permutations by the time spent per permutation: n! * n. We need to find the shortest sequence and in worst case need to evaluate all possible sequences for sorting, thus we get (n! * n * n) which is O((n!)*n^2). The dominant factor is n!.

Space Complexity

O(N!) – The algorithm explores all possible sequences of flips. In the worst-case scenario, it needs to store all sequences of flips that lead to a sorted stack. The number of possible flip sequences grows factorially with the number of pancakes, N. Therefore, the space required to store these sequences is proportional to N!, where N is the number of pancakes. No other significant auxiliary space is used beyond storing these sequences of flips. This is because the in-place flip operations don't contribute extra space. Finally, storing possible sequence means the space complexity is O(N!).

Optimal Solution

Approach

The goal is to sort a stack of pancakes using only flips. The most efficient approach involves strategically placing the largest unsorted pancake in its correct position with carefully chosen flips, repeating this process until all pancakes are sorted.

Here's how the algorithm would work step-by-step:

Find the largest unsorted pancake.
Flip the stack so the largest unsorted pancake is on top.
Flip the stack again so the largest unsorted pancake is at the bottom of the unsorted part of the stack, putting it in its correct position.
Repeat this process for the next largest unsorted pancake, working your way down to the smallest pancake.
Continue until all pancakes are sorted from largest to smallest, ensuring that at each stage, the largest unsorted item is placed correctly.

Code Implementation

def pancake_sort(arr):
    current_array_size = len(arr)
    while current_array_size > 1:
        # Find the index of the largest unsorted pancake
        maximum_index = 0
        for i in range(1, current_array_size):
            if arr[i] > arr[maximum_index]:
                maximum_index = i

        # Flip to put the largest pancake at the beginning
        if maximum_index != current_array_size - 1:
            if maximum_index != 0:

                flip(arr, maximum_index + 1)

            # Flip to put the largest pancake in its correct position
            flip(arr, current_array_size)

        current_array_size -= 1

def flip(arr, end_index):
    start_index = 0
    while start_index < end_index:
        arr[start_index], arr[end_index - 1] = arr[end_index - 1], arr[start_index]
        start_index += 1
        end_index -= 1

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the array from right to left, effectively processing n elements. For each element, the algorithm searches for the maximum value within the unsorted portion of the array, requiring a traversal of at most n elements in the worst case. Then, it performs at most two flips, each taking O(n) time, but since they are performed a constant number of times within the loop, they do not change the overall complexity. Therefore, the nested operation of finding the maximum and potentially flipping within the outer loop results in a time complexity proportional to approximately n * n, which simplifies to O(n²).

Space Complexity

O(1) – The algorithm operates in-place, modifying the original pancake stack directly. It doesn't create any auxiliary data structures like temporary arrays, hash maps, or trees to store additional information. Only a few integer variables are used to keep track of indices during the flipping process. Therefore, the space complexity remains constant regardless of the input size N (the number of pancakes).

Edge Cases

Case	How to Handle
Null or empty input array	Return an empty list, as there are no flips needed for an empty array.
Input array with a single element	Return an empty list, as a single-element array is already sorted.
Already sorted array	Return an empty list, as no flips are needed to sort the array.
Array with elements in reverse sorted order (worst case)	The algorithm should still correctly sort the array, potentially requiring the maximum number of flips.
Maximum sized input array (close to constraints)	Ensure the algorithm's time and space complexity are efficient enough to handle large arrays without exceeding time or memory limits.
Array contains numbers outside the range [1, n] where n is array length	The problem states it's a permutation of 1 to n; however we should validate input array for number range and throw an error if the range assumption is broken.
Integer overflow in calculation of indices during flip operations	While unlikely given problem constraints, use appropriate data types (e.g., `int`) and ensure no intermediate calculations could cause overflow.
Multiple possible solutions exist (different flip sequences)	The problem asks for any valid solution, so the algorithm only needs to find one correct sequence of flips.