Sort Array by Moving Items to Empty Space

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ArraysGreedy Algorithms

You are given an integer array nums. You are allowed to reorder nums however you want.

Initially, there is an empty array called empty. You are also given the following operation:

Choose any index i from the array nums and move the element at the i^th index to the end of empty.

Return the minimum number of operations to make nums sorted in non-decreasing order.

Example 1:

Input: nums = [4,7,2,3,9]
Output: 3
Explanation: We can perform the following operations:
- Move 4 to the end of empty, nums is now equal to [7,2,3,9] and empty is equal to [4].
- Move 7 to the end of empty, nums is now equal to [2,3,9] and empty is equal to [4,7].
- Move 9 to the end of empty, nums is now equal to [2,3] and empty is equal to [4,7,9].
nums is now sorted, and the number of operations performed is 3. 
It can be proven that 3 is the minimum number of operations we need to perform.

Example 2:

Input: nums = [1,2,3,4,5]
Output: 0
Explanation: nums is already sorted, so we don't need to perform any operations.

Example 3:

Input: nums = [3,2,1]
Output: 3
Explanation: We need to move all the elements to empty, so that nums is sorted.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

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 range of values within the array, and are there any constraints on the values themselves (e.g., only positive integers)?
What should I return if the input array is empty or null?
Does the array contain only integers, or are other data types (like floats) possible?
Is it always possible to sort the array by moving items to the empty space, or are there cases where the array is fundamentally unsortable with this method?
Is there any initial configuration of the array that would result in infinite loops or require an excessively large number of moves, and if so, how should I handle it?

Brute Force Solution

Approach

The brute force approach aims to solve the array sorting problem by trying out every possible sequence of moves. We explore all ways to shift elements into the empty space, one after another, until the array is sorted. This is like trying every possible puzzle combination until the picture is right.

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

Start with the initial arrangement of items in the array.
Find the location of the empty space.
Consider each item next to the empty space as a candidate to move into that space.
For each candidate, imagine moving it into the empty space, creating a new arrangement of items.
Check if the new arrangement is the sorted array.
If the new arrangement is sorted, you're done!
If not, repeat the process of finding the empty space and considering possible moves for the new arrangement.
Keep track of every possible arrangement you create and how you got there.
If you've created an arrangement you already saw before, stop exploring that possibility, as it will lead to an infinite loop.
Eventually, either you find a sorted arrangement, or you've tried every possible arrangement. If you've tried every possible arrangement and none are sorted, it means that it is impossible to sort the array using this method.

Code Implementation

def sort_array_by_moving_items_to_empty_space(array):
    empty_space = array.index(0)
    target_array = sorted(array)
    visited_arrangements = {tuple(array)}
    queue = [(array, [])]

    while queue:
        current_array, moves = queue.pop(0)
        empty_space = current_array.index(0)

        if current_array == target_array:
            return moves

        # Explore possible moves: left, right, up, down
        possible_moves = []

        if empty_space > 0:
            possible_moves.append(empty_space - 1)
        if empty_space < len(current_array) - 1:
            possible_moves.append(empty_space + 1)

        for move_to in possible_moves:
            new_array = current_array[:]

            # Simulate moving the item to the empty space
            new_array[empty_space], new_array[move_to] = new_array[move_to], new_array[empty_space]
            new_array_tuple = tuple(new_array)

            # Avoid cycles by checking if already visited
            if new_array_tuple not in visited_arrangements:

                visited_arrangements.add(new_array_tuple)
                queue.append((new_array, moves + [(move_to, empty_space)]))

    return None

Big(O) Analysis

Time Complexity

O(n!) – The described brute-force approach explores all possible arrangements of the array elements by moving items into the empty space. In the worst-case scenario, we might need to generate and check every possible permutation of the array to find the sorted arrangement. The number of possible permutations for an array of size n is n! (n factorial). Therefore, the time complexity is O(n!), as we potentially have to examine all n! arrangements.

Space Complexity

O(N!) – The brute force approach explores all possible arrangements of the array. To avoid cycles (step 9), it needs to keep track of visited arrangements. In the worst case, where no arrangement is repeated until the very end, the algorithm could potentially visit every possible permutation of the array. Since there are N! possible permutations of an array of size N, the space complexity to store these visited arrangements becomes O(N!). The 'keeping track of every possible arrangement' (step 8) is what makes the space complexity O(N!).

Optimal Solution

Approach

The goal is to sort numbers in a list using only one empty spot to move things around. The best way to do this is to think about putting each number in its correct place, one at a time, using the empty space as our helper.

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

Find the number that belongs in the very first position of the list.
If that number isn't already in the first spot, bring it there. Do this by moving the empty spot around until it's next to the number you want to move, then swap them.
Now, find the number that belongs in the second position.
If it's not there already, use the empty spot to bring it to the second position, similar to what you did for the first number.
Keep repeating this process for each position in the list, going from left to right.
Each time, you're putting the correct number into the right spot by cleverly using the empty space to shuffle things around.
When you get to the last position, the correct number should already be there, meaning the list is now sorted.

Code Implementation

def sort_array_by_moving_items(array_to_sort):
    empty_position = array_to_sort.index(0)
    array_length = len(array_to_sort)

    for i in range(array_length):
        if array_to_sort[i] != i + 1 and array_to_sort[i] != 0:
            # Find the position where the current number should be.
            correct_number_position = array_to_sort.index(i + 1)

            # Move the correct number to its proper index.
            while correct_number_position != i:
                neighbor_position = -1
                if correct_number_position > 0:
                    neighbor_position = correct_number_position - 1
                else:
                    neighbor_position = correct_number_position + 1

                if correct_number_position - empty_position == 1 or correct_number_position - empty_position == -1:
                    array_to_sort[correct_number_position], array_to_sort[empty_position] = array_to_sort[empty_position], array_to_sort[correct_number_position]
                    empty_position = correct_number_position
                else:

                    #Move empty space to be next to correct number.
                    if empty_position > correct_number_position:
                        array_to_sort[empty_position], array_to_sort[empty_position - 1] = array_to_sort[empty_position - 1], array_to_sort[empty_position]
                        empty_position -= 1

                    else:
                        array_to_sort[empty_position], array_to_sort[empty_position + 1] = array_to_sort[empty_position + 1], array_to_sort[empty_position]
                        empty_position += 1

    return array_to_sort

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through the array of size n, placing each element in its correct sorted position. For each element, finding the correct value and moving it to its desired index can take up to n moves of the empty space in the worst case. Therefore, sorting each of the n elements requires at most n operations, leading to a time complexity of approximately n * n, which simplifies to O(n²).

Space Complexity

O(1) – The algorithm sorts the array in-place, meaning it modifies the original array directly without creating a copy. It only requires a few constant space variables such as the index of the empty space, and the current position being considered. Therefore, the auxiliary space used does not depend on the input size N, the number of elements in the array. Thus, the space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty array	Return an empty list or null if the input array is null or empty, indicating no moves are possible.
Array with only one element	Return an empty list since no move is required to sort a single-element array.
Array already sorted	Return an empty list as no moves are needed; the array is already in the desired state.
Array with all elements the same	Return an empty list as moving any element to the empty space will not change the order.
Array with duplicates affecting optimal move sequence	The chosen algorithm must consider all possible move sequences when duplicates are present and select the shortest one.
Maximum-sized array exceeding memory limits	Handle potential out-of-memory errors by choosing algorithms with lower memory footprints or utilizing techniques like divide-and-conquer if appropriate.
Array where the 'empty space' index is not within the bounds of the array	Throw an exception or return an error code to indicate that the input is invalid if the empty space index is out of bounds.
Integer overflow when calculating distances or indices	Use appropriate data types (e.g., long) to prevent integer overflows when calculating distances between elements or accessing array indices.

Null or empty array

How to Handle:

Return an empty list or null if the input array is null or empty, indicating no moves are possible.

Array with only one element

How to Handle:

Return an empty list since no move is required to sort a single-element array.

Array already sorted

How to Handle:

Return an empty list as no moves are needed; the array is already in the desired state.

Array with all elements the same

How to Handle:

Return an empty list as moving any element to the empty space will not change the order.

Array with duplicates affecting optimal move sequence

How to Handle:

The chosen algorithm must consider all possible move sequences when duplicates are present and select the shortest one.

Maximum-sized array exceeding memory limits

How to Handle:

Handle potential out-of-memory errors by choosing algorithms with lower memory footprints or utilizing techniques like divide-and-conquer if appropriate.

Array where the 'empty space' index is not within the bounds of the array

How to Handle:

Throw an exception or return an error code to indicate that the input is invalid if the empty space index is out of bounds.

Integer overflow when calculating distances or indices

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflows when calculating distances between elements or accessing array indices.