Relocate Marbles

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Arrays

You are given a 0-indexed integer array nums representing the initial positions of some marbles. You are also given two 0-indexed integer arrays moveFrom and moveTo of equal length.

Throughout moveFrom.length steps, you will change the positions of the marbles. On the i^th step, you will move all marbles at position moveFrom[i] to position moveTo[i].

After completing all the steps, return the sorted list of occupied positions.

Notes:

We call a position occupied if there is at least one marble in that position.
There may be multiple marbles in a single position.

Example 1:

Input: nums = [1,6,7,8], moveFrom = [1,7,2], moveTo = [2,9,5]
Output: [5,6,8,9]
Explanation: Initially, the marbles are at positions 1,6,7,8.
At the i = 0th step, we move the marbles at position 1 to position 2. Then, positions 2,6,7,8 are occupied.
At the i = 1st step, we move the marbles at position 7 to position 9. Then, positions 2,6,8,9 are occupied.
At the i = 2nd step, we move the marbles at position 2 to position 5. Then, positions 5,6,8,9 are occupied.
At the end, the final positions containing at least one marbles are [5,6,8,9].

Example 2:

Input: nums = [1,1,3,3], moveFrom = [1,3], moveTo = [2,2]
Output: [2]
Explanation: Initially, the marbles are at positions [1,1,3,3].
At the i = 0th step, we move all the marbles at position 1 to position 2. Then, the marbles are at positions [2,2,3,3].
At the i = 1st step, we move all the marbles at position 3 to position 2. Then, the marbles are at positions [2,2,2,2].
Since 2 is the only occupied position, we return [2].

Constraints:

1 <= nums.length <= 10⁵
1 <= moveFrom.length <= 10⁵
moveFrom.length == moveTo.length
1 <= nums[i], moveFrom[i], moveTo[i] <= 10⁹
The test cases are generated such that there is at least a marble in moveFrom[i] at the moment we want to apply the i^th move.

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 initial configuration of marbles and holes represented as, and what are the data types and range of values for the marbles' positions and move instructions?
Are there any constraints on the number of move operations (e.g., upper bound)?
If a move instruction attempts to place a marble into a hole that is already occupied, what should happen? Should the move be skipped, or is there another rule?
What should the output be? Should it be the final positions of the marbles after all moves, or some other representation?
Is the order of marbles relevant after relocation or are we just concerned with the final positions?

Brute Force Solution

Approach

The brute force way to relocate marbles is like physically trying out every possible move. We consider each possible swap of marbles between boxes until we find a configuration that meets the desired condition. It’s thorough but not efficient.

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

Start with the marbles in their original boxes.
Consider moving one marble from its current box to every other box.
For each of these moves, see if the marbles are now in the correct boxes.
If the marbles are not yet in the correct boxes, try moving a marble from one of the boxes to every other box again, until it matches the correct order.
Keep repeating this process of trying all possible moves until the marbles are arranged correctly.
If we can't find the solution with a certain number of moves, we try increasing the number of moves to explore all combinations.

Code Implementation

def relocate_marbles_brute_force(initial_state, target_state, max_moves=5):
    def states_equal(state1, state2):
        return state1 == state2

    def get_possible_next_states(current_state):
        possible_states = []
        number_of_boxes = len(current_state)
        for from_box_index in range(number_of_boxes):
            for to_box_index in range(number_of_boxes):
                if from_box_index != to_box_index:
                    new_state = current_state[:]
                    new_state[from_box_index], new_state[to_box_index] = \
                        new_state[to_box_index], new_state[from_box_index]
                    possible_states.append(new_state)
        return possible_states

    def search(current_state, moves_remaining):
        if states_equal(current_state, target_state):
            return True

        if moves_remaining == 0:
            return False

        # Iterate through all possible next states and recursively search
        for next_state in get_possible_next_states(current_state):
            if search(next_state, moves_remaining - 1):
                return True

        return False

    # Try increasing the number of moves until max_moves is reached
    for moves in range(1, max_moves + 1):
        if search(initial_state, moves):
            return True

    # No solution found within the maximum number of moves
    return False

Big(O) Analysis

Time Complexity

O((n^n)!) – The algorithm considers all possible swaps of marbles between n boxes to find the correct arrangement. In the worst case, it explores all possible permutations of marbles across boxes. Moving one marble to other locations generates n possibilities. Repeating this multiple times for each such arrangement explodes combinatorially. Therefore, the number of operations grows factorially with the number of possible arrangements, which themselves grow exponentially with n. This massive exploration results in a time complexity approximated by O((n^n)!).

Space Complexity

O(1) – The described brute force approach primarily involves iterative swapping and checking. It does not explicitly mention storing intermediate states or configurations in auxiliary data structures like lists, arrays, or hash maps. Only a fixed number of variables are needed to track current marble positions and move counts, irrespective of the number of marbles or boxes (N). Therefore, the auxiliary space complexity remains constant, resulting in O(1).

Optimal Solution

Approach

To efficiently move marbles, we use a counting-based approach. Instead of directly swapping marbles, we keep track of how many marbles are supposed to be in each location after all the moves are complete. This allows us to quickly determine the final configuration without actually moving anything.

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

First, create a way to keep track of how many marbles are initially in each location.
Next, go through the list of moves. For each move, decrease the count of marbles at the original location and increase the count at the destination.
After processing all the moves, use the updated counts to determine the final arrangement of marbles.
Since we know exactly how many marbles should be in each location, we can directly place them without actually moving any physical marble.

Code Implementation

def relocate_marbles(locations, moves):
    marble_counts = {}
    for location in locations:
        marble_counts[location] = marble_counts.get(location, 0) + 1

    # Update marble counts based on moves
    for move_from, move_to in moves:
        if move_from in marble_counts:

            # Reduce count from original location
            marble_counts[move_from] -= 1
            if marble_counts[move_from] == 0:
                del marble_counts[move_from]

            # Increase count to new location
            marble_counts[move_to] = marble_counts.get(move_to, 0) + 1

    final_locations = []

    # Reconstruct the final marble arrangement
    for location in sorted(marble_counts.keys()):
        for _ in range(marble_counts[location]):
            final_locations.append(location)

    return final_locations

Big(O) Analysis

Time Complexity

O(m + n) – The algorithm first initializes a count of marbles at each location, which takes O(n) time, where n is the number of locations. Then, it iterates through the list of moves, updating the counts, which takes O(m) time, where m is the number of moves. Finally, it constructs the final arrangement based on the updated counts, which takes O(n) time. Therefore, the overall time complexity is O(n + m + n), which simplifies to O(m + n).

Space Complexity

O(N) – The algorithm uses a counting-based approach, which means it needs to keep track of the number of marbles at each location. This is achieved by creating a data structure (e.g., an array or hash map) to store the counts for each location. The size of this data structure is directly proportional to the number of possible locations, which can be considered N. Therefore, the auxiliary space required is O(N).

Edge Cases

Case	How to Handle
Null input array	Throw IllegalArgumentException or return an empty array to signal invalid input.
Empty 'move' list or 'moveFrom' and 'moveTo' lists of zero length	If no moves are specified, return the original marbles array unchanged.
'moveFrom' and 'moveTo' lists of different lengths	Throw IllegalArgumentException because the moves cannot be applied if the lists have different sizes.
Invalid 'moveFrom' index (out of bounds for the marbles array)	Throw IndexOutOfBoundsException when the 'moveFrom' index is less than 0 or greater than or equal to the marbles array's length.
Applying moves results in an index conflict (moving two marbles to the same position)	Ensure that the 'moveTo' indices are unique to prevent overwriting marbles in the final arrangement.
Large input array and many moves, potentially impacting performance.	Use an efficient data structure like an array to directly update the marbles positions based on the moves.
The 'moveFrom' and 'moveTo' contain the same index	Ignore the moves where the `moveFrom` index is the same as the `moveTo` index as it doesn't change anything.
Integer overflow if indices or array lengths are very large	Use long type for indexing or lengths if the size of input arrays or number of moves are sufficiently large to potentially cause integer overflow

Null input array

How to Handle:

Throw IllegalArgumentException or return an empty array to signal invalid input.

Empty 'move' list or 'moveFrom' and 'moveTo' lists of zero length

How to Handle:

If no moves are specified, return the original marbles array unchanged.

'moveFrom' and 'moveTo' lists of different lengths

How to Handle:

Throw IllegalArgumentException because the moves cannot be applied if the lists have different sizes.

Invalid 'moveFrom' index (out of bounds for the marbles array)

How to Handle:

Throw IndexOutOfBoundsException when the 'moveFrom' index is less than 0 or greater than or equal to the marbles array's length.

Applying moves results in an index conflict (moving two marbles to the same position)

How to Handle:

Ensure that the 'moveTo' indices are unique to prevent overwriting marbles in the final arrangement.

Large input array and many moves, potentially impacting performance.

How to Handle:

Use an efficient data structure like an array to directly update the marbles positions based on the moves.

The 'moveFrom' and 'moveTo' contain the same index

How to Handle:

Ignore the moves where the `moveFrom` index is the same as the `moveTo` index as it doesn't change anything.

Integer overflow if indices or array lengths are very large

How to Handle:

Use long type for indexing or lengths if the size of input arrays or number of moves are sufficiently large to potentially cause integer overflow