Moving Stones Until Consecutive

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Arrays

There are three stones in different positions on the X-axis. You are given three integers a, b, and c, the positions of the stones.

In one move, you pick up a stone at an endpoint (i.e., either the lowest or highest position stone), and move it to an unoccupied position between those endpoints. Formally, let's say the stones are currently at positions x, y, and z with x < y < z. You pick up the stone at either position x or position z, and move that stone to an integer position k, with x < k < z and k != y.

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: a = 1, b = 2, c = 5
Output: [1,2]
Explanation: Move the stone from 5 to 3, or move the stone from 5 to 4 to 3.

Example 2:

Input: a = 4, b = 3, c = 2
Output: [0,0]
Explanation: We cannot make any moves.

Example 3:

Input: a = 3, b = 5, c = 1
Output: [1,2]
Explanation: Move the stone from 1 to 4; or move the stone from 1 to 2 to 4.

Constraints:

1 <= a, b, c <= 100
a, b, and c have different values.

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 integers in the `stones` array? Are negative values possible?
Are there any constraints on the length of the `stones` array? Can it be empty or contain duplicate stone positions?
If the stones are already consecutive, what should the function return? Specifically, what are the minimum and maximum moves in this scenario?
Can you provide a specific example where the initial stone positions are not sorted to illustrate the expected behavior?
Should I return the result as an array of two integers, where the first integer is the minimum number of moves and the second is the maximum?

Brute Force Solution

Approach

The brute force approach for moving stones is like trying every possible stone arrangement to see which one gets us closest to having them in consecutive positions. We'll consider all combinations of moves, no matter how inefficient. We then pick the one that requires the least amount of movements.

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

First, we need to consider all possible final positions for the stones to be consecutive. Imagine sliding the stones along a number line and consider all possible arrangements.
For each possible final position, we determine how many moves it would take to rearrange the stones into that consecutive arrangement.
To find the number of moves for a given final position, we need to calculate how far each stone needs to be moved to reach its place in the consecutive arrangement.
Add up all these individual move distances to find the total number of moves for that particular arrangement.
Repeat this process for every possible final position that the stones could be placed in consecutively.
Once we've considered all possible final positions and calculated the number of moves needed for each, we pick the arrangement that requires the fewest moves.
The smallest number of moves among all possibilities is our answer.

Code Implementation

def moving_stones_until_consecutive_brute_force(stones):
    stones.sort()
    minimum_moves = float('inf')

    # Iterate through all possible starting positions for consecutive stones
    for starting_position in range(stones[2] - stones[0] - 1):

        total_moves_for_position = 0
        # Calculate the moves required to reach the current starting position
        for i in range(3):
            total_moves_for_position += abs(stones[i] - (starting_position + i + stones[0]))

        minimum_moves = min(minimum_moves, total_moves_for_position)

    # Iterate through all possible starting positions for consecutive stones
    for starting_position in range(min(stones) , max(stones) + 1):

        total_moves_for_position = 0
        # Calculate the moves required to reach the current starting position
        for i in range(3):
            total_moves_for_position += abs(stones[i] - (starting_position + i))

        minimum_moves = min(minimum_moves, total_moves_for_position)

    # Since min_moves is the smallest number of moves among all possibilities, return it
    return minimum_moves

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach iterates through all possible permutations of stone positions to find the optimal arrangement. Generating all permutations of n stones requires calculating n factorial (n!) arrangements. For each permutation, we calculate the moves needed to make the stones consecutive, which takes O(n) time. Since we need to consider all n! permutations, the overall time complexity is O(n! * n). The n! term dominates, so the time complexity simplifies to O(n!).

Space Complexity

O(1) – The brute force approach, as described, doesn't explicitly use any auxiliary data structures like arrays, hash maps, or lists. It calculates move distances and keeps track of the minimum moves found so far, which likely involves a few integer variables. The amount of extra memory used is independent of the number of stones (N), remaining constant regardless of the input size. Therefore, the space complexity is O(1).

Optimal Solution

Approach

The goal is to minimize moves to arrange stones consecutively. The key idea is to sort the stone positions and focus on the largest gap between them. By strategically placing the remaining stone into this largest gap, we minimize the required moves.

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

First, find out where all the stones are currently located and put those numbers in order from smallest to largest.
Next, find the biggest space between any two stones that are next to each other in the sorted list.
Figure out how many moves it would take to put the remaining stone(s) into that largest gap to make all stones touch each other in a row.
There are some tricky edge cases depending on the size of gaps. Special cases need to be considered where the stones are already close together.
The minimum number of moves is the smaller of either one move, or the calculations that involve biggest gap or the tricky edge cases described above.
The maximum number of moves is found by considering the largest combined amount of moves needed to fill each gap, less one move to account for the edges.
Return the minimum and maximum number of moves.

Code Implementation

def moving_stones_until_consecutive(stone_a, stone_b, stone_c):
    stones = sorted([stone_a, stone_b, stone_c])
    stone_a = stones[0]
    stone_b = stones[1]
    stone_c = stones[2]

    maximum_moves = (stone_c - stone_b - 1) + (stone_b - stone_a - 1)

    # Check for edge cases where stones are already close.
    if stone_c - stone_a == 2:
        minimum_moves = 0

    elif stone_b - stone_a <= 2 or stone_c - stone_b <= 2:
        minimum_moves = 1
    # Otherwise, calculate the minimum moves based on the gaps.
    else:
        minimum_moves = 2

    return [minimum_moves, maximum_moves]

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation in this approach is sorting the stone positions initially, which typically takes O(n log n) time using efficient sorting algorithms like merge sort or quicksort, where n is the number of stones. Finding the largest gap, calculating minimum and maximum moves then involves iterating through the sorted stone positions, which takes O(n) time. Since O(n log n) dominates O(n), the overall time complexity is O(n log n).

Space Complexity

O(1) – The algorithm sorts the stone positions which might involve a temporary array during the sorting process. However, the problem states we are given the stone positions as input, and it doesn't explicitly mention creating an auxiliary data structure for the sorted stones; it is assumed to modify the input array in place. The auxiliary space is therefore dominated by a few integer variables to store the stone positions, biggest gap, number of moves and potentially swap variables during sorting. Thus, the space used remains constant regardless of the input size, N.

Edge Cases

Case	How to Handle
Empty input array	Return [0, 0] immediately as no moves are possible with no stones.
Input array with size 1	Return [0, 0] immediately as no moves are possible with only one stone.
Input array with size 2	Calculate the difference and return [difference - 1, difference - 1] because at most one move can be made to place stones consecutively.
Stones are already consecutive (e.g., [1, 2, 3])	Return [0, 0] since no moves are necessary.
Stones are nearly consecutive (e.g., [1, 2, 5])	The minimum moves would be 1, and the maximum would be number of gaps - 1 (here 2).
Stones are far apart (e.g., [1, 10, 100])	Calculate min moves by reducing largest gap to 2 and max moves by moving each stone towards the middle.
Input array with duplicate values (e.g., [1, 1, 5])	Duplicates will be treated as separate stones since the number of stones determine the consecutive sequence length.
Integer overflow potential when calculating differences	Ensure the input integers are within a range that prevents overflow during subtraction, or use a larger data type.

Empty input array

How to Handle:

Return [0, 0] immediately as no moves are possible with no stones.

Input array with size 1

How to Handle:

Return [0, 0] immediately as no moves are possible with only one stone.

Input array with size 2

How to Handle:

Calculate the difference and return [difference - 1, difference - 1] because at most one move can be made to place stones consecutively.

Stones are already consecutive (e.g., [1, 2, 3])

How to Handle:

Return [0, 0] since no moves are necessary.

Stones are nearly consecutive (e.g., [1, 2, 5])

How to Handle:

The minimum moves would be 1, and the maximum would be number of gaps - 1 (here 2).

Stones are far apart (e.g., [1, 10, 100])

How to Handle:

Calculate min moves by reducing largest gap to 2 and max moves by moving each stone towards the middle.

Input array with duplicate values (e.g., [1, 1, 5])

How to Handle:

Duplicates will be treated as separate stones since the number of stones determine the consecutive sequence length.

Integer overflow potential when calculating differences

How to Handle:

Ensure the input integers are within a range that prevents overflow during subtraction, or use a larger data type.