Problem Description

In this problem, we're working with a simulated array of n boxes, each represented by 0 if the box is empty, or by 1 if the box contains one ball. We are tasked with calculating the minimum number of operations required to move all balls into each individual box one by one. In each operation, we are allowed to move one ball to an adjacent box (to the left or right). The goal is to figure out for every box i the number of moves needed if all balls were to be moved to box i. The answer should be returned as an array where each index i contains the minimum operations for box i.

The problem tests one's ability to efficiently compute cumulative operations while taking into account the both directions—left to right and right to left.

Intuition

To approach this problem, we need to come up with a way to count the operations for all positions without simulating each individual move, as that would be too slow. We do so in two sweeps, first from left to right, and then from right to left.

In the first sweep, we start from the leftmost box and move right, counting how many balls we've seen so far and adding that count to our operations because for each ball we pass, it would take one more operation to move it to the current box. We keep track of the total operations we'd need if we were moving all the balls to the right.

In the second pass, we do the opposite. We start from the rightmost box and move left, also counting balls and accumulating the operations needed to move all balls to the current position from the right. We then add this value to the corresponding position in our result from the first pass.

The accumulation of these movements gives us the minimum operation count for each box because it folds in the moves needed from both sides.

This approach allows us to calculate the minimum number of operations for each box in a single iteration from each direction, optimizing the solution to run with linear time complexity.

Solution Approach

The given solution takes a two-pass approach, which can be thought of as dynamic programming to some extent where we are building up the answer based on previous computations.

First Pass: Left to Right

• We initialize a counter cnt to keep track of the number of balls seen so far and an array ans of zeros with the same length as the boxes string to store the operations required for each box.
• Starting with the second element (since the first box would always require zero moves to get to itself), we iterate through the boxes string from left to right.
  • If the previous box (at index i-1) contains a ball ('1'), we increment cnt as we have encountered another ball.
  • The operations to move all balls encountered so far to the current box i is the sum of operations needed for the previous box ans[i-1] plus the number of balls encountered so far cnt. We set this value in the ans array at index i.

Second Pass: Right to Left

• We then initialize another counter s that will serve a similar purpose as cnt but for the right-to-left pass. We also reset cnt back to zero.
• Now, we iterate through the string from right to left starting from the second-to-last box.
  • If the next box (at index i+1) contains a ball, we increment cnt by one because we've encountered another ball.
  • We then add cnt to s. The reasoning behind this is the same as the left to right pass but in reverse; s keeps a running total of the number of operations needed to bring the balls from the right to the current box.
  • Finally, we add s to the existing value in ans[i] to account for the operations needed from the right side.

Final Answer

• After completing both passes, the ans array now contains the sum of the operations needed from both sides (left and right) for each box and thus gives us the correct minimum number of operations for each box.
• The function returns the array ans as the final result.

This algorithm has a time complexity of O(n) because it passes through the boxes string only twice, irrespective of the number of balls. It uses additional space for the array ans, giving it a space complexity of O(n) as well.

Example Walkthrough

Let's consider a small example to illustrate the solution approach. Suppose we have an array of boxes: boxes = "001011". Let's walk through the algorithm.

First Pass: Left to Right

1. We start with cnt = 0 and ans = [0, 0, 0, 0, 0, 0] representing the initial minimum operations for each box.

2. At i = 1 (boxes[1] = "0"), the previous box is empty, so cnt remains 0 and ans[1] = ans[0] + cnt = 0.

3. At i = 2 (boxes[2] = "1"), we find the first ball. We increment cnt to 1 since we've encountered a ball, and ans[2] = ans[1] + cnt = 0 + 1 = 1.

4. At i = 3 (boxes[3] = "0"), we've seen one ball so far, so ans[3] = ans[2] + cnt = 1 + 1 = 2.

5. At i = 4 (boxes[4] = "1"), we encounter another ball, increment cnt to 2, and ans[4] = ans[3] + cnt = 2 + 2 = 4.

6. At i = 5 (boxes[5] = "1"), we once again increment cnt to 3 as we've found another ball, and ans[5] = ans[4] + cnt = 4 + 3 = 7.

Now, ans represents the operations to move the balls from left to right, and it looks like this: [0, 0, 1, 2, 4, 7].

Second Pass: Right to Left

1. We reset cnt to 0 and introduce s = 0. We start from the rightmost box and move leftward.

2. At i = 4 (boxes[4] = "1"), since the next box contains a ball, we increment cnt to 1. We add cnt to s, which makes s = 1, and add s to ans[4]: ans[4] = ans[4] + s = 4 + 1 = 5.

3. At i = 3 (boxes[3] = "0"), we have seen one ball on the right, so s becomes 1 + 1 = 2, and ans[3] = ans[3] + s = 2 + 2 = 4.

4. At i = 2 (boxes[2] = "1"), we encounter another ball, and hence increment cnt to 2. Now, s = s + cnt = 2 + 2 = 4, and ans[2] = ans[2] + s = 1 + 4 = 5.

5. At i = 1 (boxes[1] = "0"), s = s + cnt = 4 + 2 = 6, and ans[1] = ans[1] + s = 0 + 6 = 6.

6. At i = 0 (boxes[0] = "0"), we continue and s = s + cnt = 6 + 2 = 8, but since it's the leftmost box, we don't need to add it to ans[0], which remains 0.

The final ans array after accumulating operations from both sides is: [0, 6, 5, 4, 5, 7].

This array represents the minimum number of moves to get all balls in front of each respective box. For example, ans[1] = 6 means if we want to move all balls to box 1, it will take us a total of 6 moves.

Thus, our function would return [0, 6, 5, 4, 5, 7] as the final output for the input string "001011". This example demonstrates the efficiency of the algorithm by avoiding the brute force approach of simulating each move, which would be impractical for large strings.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code consists of two separate for loops that each iterate through the array boxes. Each loop runs in linear time relative to the number of elements n in the input string boxes.

The first loop runs from index 1 to n, with constant-time operations within the loop, leading to an O(n) complexity.

The second loop similarly runs with constant-time operations but in reverse order, from n-2 down to 0. This loop also results in an O(n) complexity.

Since these loops do not nest and run in sequence, the overall time complexity of the function is O(n) + O(n), which simplifies to O(n).

Space Complexity

The space complexity is determined by the additional memory used by the algorithm besides the input. In this case, the algorithm only allocates space for one additional array, ans, which stores the result and has the same length as the input string boxes.

Thus, the space complexity is directly proportional to the length of the input, resulting in a space complexity of O(n).

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