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Minimum Number of Operations to Move All Balls to Each Box

You have n boxes. You are given a binary string boxes of length n, where boxes[i] is '0' if the ith 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 ith 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'.

Minimum Number of Operations to Move All Balls to Each Box

This problem asks to find the minimum number of operations needed to move all balls in a row of boxes to each box individually. The input is a binary string boxes where '1' represents a ball and '0' an empty box. An operation involves moving one ball to an adjacent box.

Approach

The most efficient approach uses prefix sums to avoid redundant calculations. We can calculate the minimum operations for each box in two passes:

  1. Left Pass: Calculate the total operations needed to move balls from the left side to the current box. For each box, the cost of moving a ball from a box to the left is the distance between boxes.
  2. Right Pass: Similarly, calculate the total operations needed to move balls from the right side.

Finally, sum the left and right operation counts for each box to get the minimum operations.

Time and Space Complexity Analysis

  • Time Complexity: O(n), where n is the length of the boxes string. We iterate through the string twice.
  • Space Complexity: O(n) to store the left, right, and ans arrays.

Code Implementation (Python)

class Solution:
    def minOperations(self, boxes: str) -> List[int]:
        n = len(boxes)
        left = [0] * n
        right = [0] * n
        cnt = 0
        for i in range(1, n):
            if boxes[i - 1] == '1':
                cnt += 1
            left[i] = left[i - 1] + cnt  #Cumulative sum of balls to the left
        cnt = 0
        for i in range(n - 2, -1, -1):
            if boxes[i + 1] == '1':
                cnt += 1
            right[i] = right[i + 1] + cnt #Cumulative sum of balls to the right
        return [a + b for a, b in zip(left, right)] #Sum left and right operations

Code Implementation (Java)

class Solution {
    public int[] minOperations(String boxes) {
        int n = boxes.length();
        int[] left = new int[n];
        int[] right = new int[n];
        int cnt = 0;
        for (int i = 1; i < n; ++i) {
            if (boxes.charAt(i - 1) == '1') {
                cnt++;
            }
            left[i] = left[i - 1] + cnt;
        }
        cnt = 0;
        for (int i = n - 2; i >= 0; --i) {
            if (boxes.charAt(i + 1) == '1') {
                cnt++;
            }
            right[i] = right[i + 1] + cnt;
        }
        int[] ans = new int[n];
        for (int i = 0; i < n; ++i) {
            ans[i] = left[i] + right[i];
        }
        return ans;
    }
}

This efficient solution provides a clear and concise way to solve the problem with optimal time and space complexity. Other solutions exist (e.g., brute force), but they're less efficient. The provided code is well-commented and easy to understand. The use of prefix sums is a key optimization.