2766. Relocate Marbles

Problem Description

You have an array of integers, nums, which represents the initial positions of some marbles, indexed from 0. Additionally, you have two more integer arrays of the same length called moveFrom and moveTo, also 0-indexed. These arrays represent a sequence of steps you'll take to move the marbles from one position to another, respectively. For every i in the range of moveFrom, you will move all marbles located at the position given by moveFrom[i] to the position moveTo[i]. Your goal is to determine which positions are occupied by at least one marble after completing all the move steps. The final output should be a sorted list of these occupied positions.

It's important to note a couple of aspects of the problem:

• An occupied position is any position where there's at least one marble.
• Positions can be occupied by more than one marble, which means they don't necessarily decrease or increase in count after a move, as moves can be to the same position from different starting points.

Intuition

To solve this problem, the intuition is to track the positions of marbles as moves are applied. The challenge is to do this efficiently – adjusting the positions step-by-step could become costly if we had to move each marble individually or maintain a list of marble counts at each position.

The key insight is realizing that to determine the occupied positions, we don't need to track the number of marbles at each position, just whether a position is occupied or not. A set data structure is perfect for this task because it can hold unique values and match the behavior we want: when marbles move from a position, we simply remove the initial position from the set, and we add the new position to the set.

A set is also useful because it automatically handles cases where multiple moves involve the same positions, as duplicate positions in a set are not possible. This means if we add an already existing moveTo position to the set, the set remains unchanged, correctly reflecting the nature of the problem.

The final step is to return a sorted list of the set elements, as we want the occupied positions in ascending order.

Here's the intuition behind each step of the provided solution:

1. Initialize a set with the starting positions from nums.
2. Iterate through each moveFrom and moveTo pair.
3. For each pair, remove the moveFrom position and add the moveTo position to the set.
4. Since the set only contains unique elements, it'll accurately represent all distinct occupied positions after all moves.
5. Convert the set to a sorted list to get the final positions in the required order.

Learn more about Sorting patterns.

Solution Approach

The solution approach involves using a set data structure and simple for-loop iteration over the arrays that command marble moves. The choice of a set is due to its inherent properties where it stores unique elements and allows for efficient insertion and removal operations.

Let's break down the key parts of the implementation:

• Initialization of Position Set: The initial positions of the marbles (nums) are inserted into a set named pos. This is to ensure we have unique positions noted.

  1pos = set(nums)

• Processing Moves: A for-loop iterates over the zip(moveFrom, moveTo) to get pairs of start and end positions for each move. For each pair:

  • We remove the starting position f from the set pos, as that position is no longer occupied by any marble after the move:

    1pos.remove(f)

  • Then, we add the new position t to pos to indicate that position is now occupied by at least one marble. If it's already occupied (i.e., already in the set), the set doesn't change, which is in line with the problem's constraints.

    1pos.add(t)

This continues for all moves, iteratively updating the positions of the marbles.

• Returning Sorted Positions: Finally, after all moves have been applied, the set pos holds all the unique positions currently occupied by the marbles. We convert the set to a list and sort it to get the final answer in the required ascending order:

  1return sorted(pos)

No explicit hash table is used here, but the set is internally implemented as a hash table which allows the remove and add operations to be performed in constant average time complexity, O(1). The sorting at the end is O( n log n ), where n is the number of occupied positions.

Hence, the solution combines the efficiency of hash tables for updating marble positions and the necessity of returning a sorted list of the final positions. This makes the approach both intuitive and optimal for the problem at hand.

Example Walkthrough

Let's use a small example to illustrate the solution approach:

Suppose we have:

• nums array as [1, 3, 5] representing the initial positions of the marbles.
• moveFrom array as [5, 1] representing positions from where marbles are moved.
• moveTo array as [3, 4] representing positions to where marbles are moved.

Following the solution steps:

• We first initialize a set with the initial marble positions, pos, which will look like this: {1, 3, 5}.
• Next, we need to process the moves:
  • For the first move, we get moveFrom[0] as 5 and moveTo[0] as 3. We remove 5 from pos and try to add 3. After this step, pos looks like this: {1, 3} because 3 is already present.
  • For the second move, we get moveFrom[1] as 1 and moveTo[1] as 4. We remove 1 from pos and add 4 to pos. Now, pos looks like this: {3, 4}.
• Finally, we convert pos to a sorted list to get the final answer. Sorting {3, 4} will simply give us [3, 4] as the sorted list of occupied positions after all moves are completed.

So, the output for this example will be [3, 4] indicating these final positions are occupied by at least one marble.

Solution Implementation

Time and Space Complexity

The time complexity of the provided code can be broken down as follows:

• Creating the pos set from nums takes O(n) time, where n is the length of the nums.
• The zip operation does not add to the complexity since it's only creating an iterator.
• The for loop runs for the length of moveFrom and moveTo lists. In the worst case, this will be O(n) – assuming that each movement is unique.
• Within the loop, the remove operation on a set takes O(1) time on average, and the add operation also takes O(1) time.
• Finally, sorting the resulting set pos will take O(n * log n) time since it can have at most n elements from nums.

So, the overall time complexity is dominated by the sorting operation, which results in O(n * log n).

The space complexity of the code is a result of the following:

• The set pos, which in the worst-case will hold all unique elements of nums, hence has a space complexity of O(n).
• The sorted list returned is also O(n) space complexity since it holds the same number of elements as in the set pos.

Therefore, the overall space complexity is O(n) as it is dominated by the space required to store the unique elements from nums.

Learn more about how to find time and space complexity quickly using problem constraints.