#1769

Medium

LeetCode Problem Workspace

Minimum Number of Operations to Move All Balls to Each Box

Solve the problem of finding the minimum operations to move balls to each box using an efficient approach with arrays and prefix sum.

array string prefix sum

Problem Statement

You are given a binary string representing boxes, where each '1' represents a ball in a box and '0' represents an empty box. In one move, a ball can be moved to an adjacent box. The goal is to calculate the minimum number of operations needed to move all the balls into each box. Your task is to return an array where the value at each index represents the minimum number of moves required for that box.

For each box, you need to determine how many operations it would take to collect all balls from other boxes. The solution must consider the optimal movements of balls using the minimum possible number of operations based on the distance between boxes.

Examples

Example 1

Input: boxes = "110"

Output: [1,1,3]

The answer for each box is as follows:

First box: you will have to move one ball from the second box to the first box in one operation.
Second box: you will have to move one ball from the first box to the second box in one operation.
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]

Example details omitted.

Constraints

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

Solution Approach

Prefix Sum Calculation

To solve this efficiently, calculate prefix sums for both the number of balls to the left and the number of operations needed to move them to each box. This helps optimize the number of moves required to move balls to any box.

Two-Pass Approach

First, process the balls from left to right and calculate the moves needed for each box. Then, process from right to left, updating the result array with the additional moves from balls on the right side. Combining both passes ensures all balls are considered for each box.

Minimizing Operations

To minimize operations, focus on adjusting the calculation dynamically by considering adjacent ball movements. Each box's move count is directly influenced by the relative positions of the balls, which can be determined in linear time using the prefix sum approach.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

The time complexity is O(n) because we traverse the string twice—once from left to right and once from right to left. The space complexity is O(1) since we are only modifying the result array in-place without requiring additional storage.

What Interviewers Usually Probe

Candidate efficiently applies a two-pass approach.
Candidate correctly implements prefix sum for optimization.
Candidate demonstrates a clear understanding of minimizing operations with array manipulation.

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the movement of balls across boxes, leading to incorrect calculation of the number of moves.
Not utilizing the prefix sum technique, resulting in a suboptimal solution.
Forgetting to account for both directions (left-to-right and right-to-left) when calculating the minimum moves for each box.

Follow-up variants

If the number of boxes is fixed, optimize for fewer operations with precomputed results.
Alter the problem to involve moving balls only in one direction, either left or right.
Extend the problem to handle more complex movement rules, like moving multiple balls in one operation.

FAQ

What is the minimum number of operations to move all balls to each box?

The minimum number of operations is determined by how far each ball needs to move to reach a specific box. A prefix sum approach helps calculate the most efficient movements.

What pattern does the "Minimum Number of Operations to Move All Balls to Each Box" problem follow?

This problem follows a pattern of using arrays and prefix sums to efficiently compute the minimum operations, making it a typical example of array manipulation and optimization.

How do I use a two-pass approach to solve this problem?

First, calculate the moves for each box from left to right. Then, iterate from right to left to adjust the results, accounting for balls coming from the right side.

Why is the time complexity O(n) for this problem?

The time complexity is O(n) because you only need to make two passes through the array: once left to right and once right to left, which is linear in relation to the size of the input.

How can I avoid common pitfalls when solving this problem?

Ensure you correctly implement the two-pass approach and account for all ball movements, considering both left-to-right and right-to-left operations to avoid suboptimal solutions.

terminal

Solution

Solution 1

#### Python3

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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
        cnt = 0
        for i in range(n - 2, -1, -1):
            if boxes[i + 1] == '1':
                cnt += 1
            right[i] = right[i + 1] + cnt
        return [a + b for a, b in zip(left, right)]

Solution 2

#### Python3

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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
        cnt = 0
        for i in range(n - 2, -1, -1):
            if boxes[i + 1] == '1':
                cnt += 1
            right[i] = right[i + 1] + cnt
        return [a + b for a, b in zip(left, right)]

Solution 3

#### TypeScript

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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
        cnt = 0
        for i in range(n - 2, -1, -1):
            if boxes[i + 1] == '1':
                cnt += 1
            right[i] = right[i + 1] + cnt
        return [a + b for a, b in zip(left, right)]

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