Moving Stones Until Consecutive II

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ArraysGreedy AlgorithmsSliding Windows

There are some stones in different positions on the X-axis. You are given an integer array stones, the positions of the stones.

Call a stone an endpoint stone if it has the smallest or largest position. In one move, you pick up an endpoint stone and move it to an unoccupied position so that it is no longer an endpoint stone.

In particular, if the stones are at say, stones = [1,2,5], you cannot move the endpoint stone at position 5, since moving it to any position (such as 0, or 3) will still keep that stone as an endpoint stone.

The game ends when you cannot make any more moves (i.e., the stones are in three consecutive positions).

Return an integer array answer of length 2 where:

answer[0] is the minimum number of moves you can play, and
answer[1] is the maximum number of moves you can play.

Example 1:

Input: stones = [7,4,9]
Output: [1,2]
Explanation: We can move 4 -> 8 for one move to finish the game.
Or, we can move 9 -> 5, 4 -> 6 for two moves to finish the game.

Example 2:

Input: stones = [6,5,4,3,10]
Output: [2,3]
Explanation: We can move 3 -> 8 then 10 -> 7 to finish the game.
Or, we can move 3 -> 7, 4 -> 8, 5 -> 9 to finish the game.
Notice we cannot move 10 -> 2 to finish the game, because that would be an illegal move.

Constraints:

3 <= stones.length <= 10⁴
1 <= stones[i] <= 10⁹
All the values of stones are unique.

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 for the stone positions in the input array?
Can the input array `stones` be empty or contain duplicate values?
If the stones are already consecutive, what should the function return?
Are the stone positions integers? What data type should I use for them?
Could you provide a specific example input and the expected output to ensure I understand the problem correctly?

Brute Force Solution

Approach

The goal is to arrange stones consecutively using the fewest moves. The brute force strategy involves trying every possible arrangement of the stones, calculating the moves needed for each arrangement, and then choosing the best one.

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

Consider all possible final positions of the stones. Imagine sliding all the stones to different spots on the number line until they are right next to each other.
For each of these possible final arrangements, calculate how many moves it would take to get the stones into that specific consecutive order.
To calculate the moves for a specific arrangement, figure out how many stones are already in their correct consecutive spots.
The number of moves needed is then the total number of stones minus the number of stones that were already in the right place.
Repeat steps 2-4 for every possible final arrangement.
After examining all the possibilities, find the arrangement that required the fewest moves.
This arrangement represents the minimum number of moves to make the stones consecutive, and that's your answer.

Code Implementation

def moving_stones_until_consecutive_ii_brute_force(stones):
    number_of_stones = len(stones)
    stones.sort()

    minimum_moves = float('inf')

    # Iterate through all possible starting positions
    for starting_position_index in range(number_of_stones):
        # Calculate the end position based on the starting position
        end_position = stones[starting_position_index] + number_of_stones - 1

        # Iterate through all possible arrangements
        potential_arrangement = list(range(stones[starting_position_index], end_position + 1))
        current_moves = 0
        stones_in_place = 0

        # Count stones that are already in the correct consecutive spot
        for stone_index in range(number_of_stones):
            if stones[stone_index] in potential_arrangement:
                stones_in_place += 1

        current_moves = number_of_stones - stones_in_place

        # Update the minimum_moves if a better arrangement is found
        minimum_moves = min(minimum_moves, current_moves)

    # Iterate through all possible ending positions
    for ending_position_index in range(number_of_stones):
        starting_position = stones[ending_position_index] - number_of_stones + 1

        potential_arrangement = list(range(starting_position, stones[ending_position_index] + 1))
        current_moves = 0
        stones_in_place = 0

        # Count stones that are already in the correct consecutive spot
        for stone_index in range(number_of_stones):
            if stones[stone_index] in potential_arrangement:
                stones_in_place += 1

        current_moves = number_of_stones - stones_in_place

        minimum_moves = min(minimum_moves, current_moves)

    return minimum_moves

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible consecutive arrangements of the n stones. For each arrangement, it iterates through all n stones to determine how many are already in the correct position. The outer loop considers n possible starting positions for the consecutive sequence. The inner loop, used to calculate correctly placed stones for each starting position, also iterates up to n times. Therefore, the total operations approximate n * n, resulting in a time complexity of O(n²).

Space Complexity

O(1) – The brute force strategy described focuses on iterating through possible arrangements and calculating moves. The steps involve checking existing positions and counting. No auxiliary data structures like lists, hash maps, or stacks are created to store intermediate arrangements or visited states. Therefore, the extra space used is limited to a few variables for indexing or counting, which remains constant regardless of the number of stones (N). This results in a constant space complexity.

Optimal Solution

Approach

The goal is to arrange stones in consecutive positions with minimal moves. The key is to realize that the fewest moves occur when we pack the stones as tightly as possible, leaving the fewest gaps. We efficiently find this optimal packing by focusing on the densest arrangements within the sorted stones.

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

First, sort the stone positions from smallest to largest.
Consider a sliding window across the sorted stone positions. This window represents a potential consecutive arrangement.
For each window, calculate how many stones are currently *not* inside the window. These are the stones we need to move *into* the window to make it consecutive.
The minimum number of moves will be the lowest number of stones we need to move *into* any of the windows that we considered.
However, there's a special case: if a window already contains all but one of the stones, and the empty spot is at either end of the entire possible range of stone positions, then you can always fill that spot with one move to win. So we compare that case's moves against all the previous window moves we checked.
Return the smallest number of required moves found.

Code Implementation

def movingStonesII(stones): 
    stones.sort()
    number_of_stones = len(stones)
    max_position = stones[-1] - stones[0] + 1

    minimum_moves = number_of_stones

    for i in range(number_of_stones):
        for j in range(i, number_of_stones):
            window_size = stones[j] - stones[i] + 1

            if window_size > max_position:
                continue

            stones_in_window = j - i + 1

            stones_outside_window = number_of_stones - stones_in_window
            minimum_moves = min(minimum_moves, stones_outside_window)

    # Special case: almost full window with empty space at the end.
    if stones[-1] - stones[0] + 1 == number_of_stones - 1 and \
       (stones[-1] - stones[1] == number_of_stones - 2 or stones[-2] - stones[0] == number_of_stones - 2):

        minimum_moves = min(minimum_moves, 2)

    maximum_moves = max(stones[-1] - stones[1] - (number_of_stones - 2),
                        stones[-2] - stones[0] - (number_of_stones - 2))

    return [minimum_moves, maximum_moves]

def movingStonesUntilConsecutiveII(stones):
    stones.sort()
    number_of_stones = len(stones)

    maximum_moves = max(stones[-1] - stones[1] - (number_of_stones - 2),
                        stones[-2] - stones[0] - (number_of_stones - 2))

    minimum_moves = number_of_stones
    i = 0
    for j in range(number_of_stones):
        while stones[j] - stones[i] + 1 > number_of_stones:
            i += 1

        stones_in_window = j - i + 1
        stones_outside_window = number_of_stones - stones_in_window
        minimum_moves = min(minimum_moves, stones_outside_window)

        # Handle almost complete window with missing edge case
        if stones_in_window == number_of_stones - 1 and stones[j] - stones[i] + 1 == number_of_stones - 1:
            minimum_moves = 2

    return [minimum_moves, maximum_moves]

Big(O) Analysis

Time Complexity

O(n log n) – The algorithm's time complexity is dominated by the sorting step, which typically takes O(n log n) time using efficient sorting algorithms like merge sort or quicksort. The subsequent sliding window iteration takes O(n) time, as it involves iterating through the sorted stone positions once. Since O(n log n) grows faster than O(n), the overall time complexity is O(n log n). The steps within the sliding window are constant time operations. Therefore, the sort is the main driver of the time complexity.

Space Complexity

O(1) – The provided solution primarily utilizes in-place operations or constant space variables. The sorting step might use O(log N) space depending on the specific sorting algorithm implementation. However, the sliding window approach itself does not create any auxiliary data structures that scale with the input size N (number of stones). Therefore, considering only the described steps, the space complexity is dominated by constant variables used for indexing and calculations.

Edge Cases

Case	How to Handle
Null or empty stones array	Return [0, 0] immediately, as there are no stones to move and thus no moves possible.
Stones array with only one element	Return [0, 0] since only one stone is not enough to create gaps that need filling.
Stones already consecutive	Return [0, 0] since no moves are required to make them consecutive.
Stones array with many duplicates	The solution sorts the stones and only considers unique positions, making duplicates irrelevant.
Stones array with a large range and few stones	The sliding window approach efficiently handles sparse distributions as it calculates the gaps between stones.
Integer overflow when calculating gaps between stones	Use long data type to store intermediate calculations of gaps to prevent overflows during subtraction.
Maximum array size causing performance issues with sorting	Ensure the sorting algorithm used is efficient (e.g., mergesort or quicksort with O(n log n) complexity).
All stones clustered at one end, with one stone far away on the other end	The min calculation in the sliding window dynamically adapts to this uneven stone distribution to find the minimum moves.

Null or empty stones array

How to Handle:

Return [0, 0] immediately, as there are no stones to move and thus no moves possible.

Stones array with only one element

How to Handle:

Return [0, 0] since only one stone is not enough to create gaps that need filling.

Stones already consecutive

How to Handle:

Return [0, 0] since no moves are required to make them consecutive.

Stones array with many duplicates

How to Handle:

The solution sorts the stones and only considers unique positions, making duplicates irrelevant.

Stones array with a large range and few stones

How to Handle:

The sliding window approach efficiently handles sparse distributions as it calculates the gaps between stones.

Integer overflow when calculating gaps between stones

How to Handle:

Use long data type to store intermediate calculations of gaps to prevent overflows during subtraction.

Maximum array size causing performance issues with sorting

How to Handle:

Ensure the sorting algorithm used is efficient (e.g., mergesort or quicksort with O(n log n) complexity).

All stones clustered at one end, with one stone far away on the other end

How to Handle:

The min calculation in the sliding window dynamically adapts to this uneven stone distribution to find the minimum moves.