Formatted question description: https://leetcode.ca/all/1769.html

# 1769. Minimum Number of Operations to Move All Balls to Each Box

## Level

Medium

## Description

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'`

.

## Solution

For each index of `boxes`

, calculate the number of balls to the left of the index and the number of balls to the right of the index. Then for each index, calculate the number of operations to move all balls to the left to the current index, and the number of operations to move all balls to the right to the current index.

```
class Solution {
public int[] minOperations(String boxes) {
int length = boxes.length();
int[] leftBalls = new int[length];
int[] rightBalls = new int[length];
for (int i = 1; i < length; i++)
leftBalls[i] = leftBalls[i - 1] + boxes.charAt(i - 1) - '0';
for (int i = length - 2; i >= 0; i--)
rightBalls[i] = rightBalls[i + 1] + boxes.charAt(i + 1) - '0';
int[] leftOperations = new int[length];
int[] rightOperations = new int[length];
for (int i = 1; i < length; i++)
leftOperations[i] = leftOperations[i - 1] + leftBalls[i];
for (int i = length - 2; i >= 0; i--)
rightOperations[i] = rightOperations[i + 1] + rightBalls[i];
int[] operations = new int[length];
for (int i = 0; i < length; i++)
operations[i] = leftOperations[i] + rightOperations[i];
return operations;
}
}
```