How to Find the Minimum Number of Operations to Move All Balls to Each Box
When working on problems involving arrays and movement costs, a common question in tech interviews is: "Given a series of boxes each holding some balls, what is the minimum number of operations required to move all balls to every individual box?" This question tests your understanding of array manipulation, prefix sums, and optimization techniques. Let’s explore how to solve this efficiently with clear explanations and examples.
Understanding the Problem
Suppose you have an array representing boxes, where each element is either 0 (no ball) or 1 (a ball present). For example:
Html
Your goal is to compute an array where each element indicates the minimum total operations needed to move all balls to that position. Each operation is defined as moving a single ball from its current box to the target box, costing 1 per move.
For the example array:
- Moving all balls to the first box (index 0): move balls from boxes 2 and 4.
- Moving all balls to the third box (index 2): move balls from boxes 0 and 4.
And so on.
Brute Force Approach
A naive solution involves two nested loops, where for each box, you sum the absolute distance to all other boxes with balls. The implementation looks like this:
Python
This approach works correctly but is inefficient for large datasets, as it has a time complexity of approximately O(n^2).
Optimized Approach Using Prefix Sums
To improve efficiency, we can use prefix sums. The idea involves calculating the total moves for each box based on previously computed results, reducing the number of calculations.
Here's how it works:
- Count the total number of balls on the left and right sides of a current box.
- Calculate the total moves to gather all balls at the current position based on previous computations.
- Update the counts as you move from left to right and right to left.
Implementation
The following code demonstrates this approach:
Python
However, this code snippet can be simplified for clarity by two passes:
First pass: accumulate counts from the left.
Second pass: from the right, accumulate costs based on the counts.
Complete Solution
Python
In this version, the code calculates the minimal operations efficiently in O(n) time.
Finding the minimum number of operations to move all balls to each box involves working with array prefix sums to avoid redundant calculations. By doing two passes—one from the left and one from the right
