Minimum Number of Operations to Move All Balls to Each Box

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ArraysStringsDynamic Programming

You have n boxes. You are given a binary string boxes of length n, where boxes[i] is '0' if the i^th box is empty, and '1' if it contains one ball.

In one operation, you can move one ball from a box to an adjacent box. Box i is adjacent to box j if abs(i - j) == 1. Note that after doing so, there may be more than one ball in some boxes.

Return an array answer of size n, where answer[i] is the minimum number of operations needed to move all the balls to the i^th box.

Each answer[i] is calculated considering the initial state of the boxes.

Example 1:

Input: boxes = "110"
Output: [1,1,3]
Explanation: The answer for each box is as follows:
1) First box: you will have to move one ball from the second box to the first box in one operation.
2) Second box: you will have to move one ball from the first box to the second box in one operation.
3) Third box: you will have to move one ball from the first box to the third box in two operations, and move one ball from the second box to the third box in one operation.

Example 2:

Input: boxes = "001011"
Output: [11,8,5,4,3,4]

Constraints:

n == boxes.length
1 <= n <= 2000
boxes[i] is either '0' or '1'.

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 maximum possible length of the `boxes` string?
Can the input string `boxes` be empty or null?
If the input string `boxes` contains only zeros (no balls), what should the output array be?
Is the input string `boxes` guaranteed to only contain '0' and '1' characters, or do I need to handle other characters?
Are we optimizing for space complexity as well as time complexity?

Brute Force Solution

Approach

The brute force approach for this problem involves checking every single box individually. For each box, we consider it as the destination and calculate the total number of moves required to bring all the balls to that specific box.

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

Imagine you pick the first box.
Go through all the other boxes, one by one, and check if there's a ball in that box.
If there's a ball in a different box, calculate how many moves it would take to move that single ball to the first box.
Add up the number of moves from all the other boxes with balls to the first box. This gives you the total moves if you make the first box the destination.
Now, do the exact same thing for the second box. Figure out how many moves it would take to move all balls to the second box.
Repeat this process for every single box in the row.
After calculating the total moves for each box, you'll have a list showing the total moves needed for each possible destination box.

Code Implementation

def min_operations(boxes):
    number_of_boxes = len(boxes)
    result = []

    for destination_box in range(number_of_boxes):
        total_moves = 0

        # Iterate through each box to calculate moves to destination_box
        for source_box in range(number_of_boxes):

            # Only calculate moves if there's a ball in the source box
            if boxes[source_box] == '1':
                total_moves += abs(source_box - destination_box)

        result.append(total_moves)

    return result

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n boxes in the input string. For each box, it then iterates through all n boxes again to calculate the distance to each box containing a ball. Therefore, the dominant operation is a nested loop structure where the outer loop runs n times and the inner loop also runs n times for each outer loop iteration. The total number of operations approximates n * n, resulting in a time complexity of O(n²).

Space Complexity

O(N) – The brute force approach calculates the number of moves for each box and stores it in a result array. This result array has a size equal to the number of boxes, which is the input size N. Therefore, the auxiliary space required is directly proportional to the input size N, leading to a space complexity of O(N).

Optimal Solution

Approach

The key to solving this efficiently is to avoid recalculating the distances for each box individually. We'll precompute some information to make the process much faster. Think of it like setting up tools before starting a construction project.

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

First, calculate the total number of steps needed to move all balls to the very first box. This is our initial reference point.
Then, as we move from one box to the next, figure out how the total steps change. Notice that balls to the left of the current box get one step further away, while balls to the right get one step closer.
Count how many balls are to the left and right of the current box. This allows us to easily update the total steps: add the number of balls on the left and subtract the number of balls on the right.
Repeat this process for each box, using the previous box's step count as a starting point. This avoids recomputing the distances from scratch each time, saving a lot of effort.
The step count at each box represents the minimum number of operations to move all balls to that box. By tracking this number at each position, you have the final answer.

Code Implementation

def min_operations(boxes):
    number_of_boxes = len(boxes)
    result = [0] * number_of_boxes

    #Calculate initial steps needed to move to the first box.
    initial_steps = 0
    for i in range(number_of_boxes):
        if boxes[i] == '1':
            initial_steps += abs(i - 0)

    result[0] = initial_steps

    left_count = 0
    right_count = 0

    #Count initial balls to the right of the first box
    for i in range(1, number_of_boxes):
        if boxes[i] == '1':
            right_count += 1

    #Iterate through the boxes and update the step count.
    for i in range(1, number_of_boxes):

        #Update steps, balls on left get further, balls on right closer.
        result[i] = result[i-1] + left_count - right_count

        #Update left and right counts.
        if boxes[i] == '1':
            right_count -= 1
            left_count += 1

    return result

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string of length n once to calculate the initial number of steps to move all balls to the first box. Subsequently, it iterates through the remaining boxes (n-1), updating the step count based on the number of balls to the left and right. Both operations are performed in linear time with respect to n. Therefore, the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm uses an output array (result) to store the minimum number of operations for each box. The size of this array is the same as the length of the input string 'boxes', which we can denote as N. While the steps mentioned involve calculations using counts of balls on either side of the current box, these counts can be maintained in constant space. The dominant space usage is from the output array of size N.

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty array since there are no boxes to operate on.
String with only one box (n=1)	Return an array containing only zero, as no moves are required.
String with all boxes empty ('0000')	Return an array of all zeros, as no moves are needed from empty boxes.
String with all boxes containing balls ('1111')	The solution correctly calculates the increasing cost from each box to every other box.
String with a single ball in the first box ('1000')	The solution accurately calculates the distances from the first box to all other boxes.
String with a single ball in the last box ('0001')	The solution computes the correct distances moving all balls to the location of the last box.
Maximum input string length (n=200)	The solution needs to have O(n^2) time complexity or less to pass all test cases; using prefix sums is recommended.
Integer overflow in calculating operations	Use 64-bit integers (long in Java/C++) to store the number of operations, especially when the string length is large.

Null or empty input string

How to Handle:

Return an empty array since there are no boxes to operate on.

String with only one box (n=1)

How to Handle:

Return an array containing only zero, as no moves are required.

String with all boxes empty ('0000')

How to Handle:

Return an array of all zeros, as no moves are needed from empty boxes.

String with all boxes containing balls ('1111')

How to Handle:

The solution correctly calculates the increasing cost from each box to every other box.

String with a single ball in the first box ('1000')

How to Handle:

The solution accurately calculates the distances from the first box to all other boxes.

String with a single ball in the last box ('0001')

How to Handle:

The solution computes the correct distances moving all balls to the location of the last box.

Maximum input string length (n=200)

How to Handle:

The solution needs to have O(n^2) time complexity or less to pass all test cases; using prefix sums is recommended.

Integer overflow in calculating operations

How to Handle:

Use 64-bit integers (long in Java/C++) to store the number of operations, especially when the string length is large.