Introduction

In this blog post, we will explore how to solve the LeetCode problem "1769" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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'.

Explanation

Here's the breakdown of the solution:

• Calculate Prefix/Suffix Costs: Efficiently precompute the cost of moving all balls to the left and right of each box. This avoids redundant calculations in the main loop.
• Iterate and Sum: Iterate through each box, summing the precomputed left and right costs to determine the total cost of moving all balls to that box.

• Dynamic Programming Principles: Leverage dynamic programming principles to update the prefix/suffix costs iteratively, building upon previous calculations.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def minOperations(boxes: str) -> list[int]:
    n = len(boxes)
    answer = [0] * n

    # Calculate cost to move all balls to the left of each box
    left_cost = [0] * n
    left_count = 0
    left_sum = 0
    for i in range(1, n):
        if boxes[i - 1] == '1':
            left_count += 1
        left_sum += left_count
        left_cost[i] = left_sum

    # Calculate cost to move all balls to the right of each box
    right_cost = [0] * n
    right_count = 0
    right_sum = 0
    for i in range(n - 2, -1, -1):
        if boxes[i + 1] == '1':
            right_count += 1
        right_sum += right_count
        right_cost[i] = right_sum

    # Calculate the total cost for each box
    for i in range(n):
        answer[i] = left_cost[i] + right_cost[i]

    return answer