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

Description

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'.

Solution

Brute Force

Iterate once to get all the indexes of balls, then iterate again to calculate: ∑ ∣ ball_index − i ∣ \sum |\text{ball\_index} - i| ∑∣ball_index−i∣

Time complexity: o ( n 2 ) o(n^2) o(n2)

Space complexity: o ( n ) o(n) o(n)

Prefix sum

Like 238. Product of Array Except Self, we can have left_cnt to denote the number of balls at the left of the current index, when moving one step forward, we have an extra left_cnt cost to move all the balls. We could do the same for right_cnt, and the final result would be left_cost + right_cost.

Time complexity: o ( n ) o(n) o(n)

Space complexity: o ( 1 ) o(1) o(1)

Code

Brute Force

python3 复制代码

class Solution:
    def minOperations(self, boxes: str) -> List[int]:
        ball_indexes = []
        for i in range(len(boxes)):
            if boxes[i] == '1':
                ball_indexes.append(i)
        res = []
        for i in range(len(boxes)):
            cur_res = 0
            for each_ball_index in ball_indexes:
                cur_res += abs(each_ball_index - i)
            res.append(cur_res)
        return res

Prefix sum

python3 复制代码

class Solution:
    def minOperations(self, boxes: str) -> List[int]:
        res = [0] * len(boxes)
        left_cnt = 0
        left_cost = 0
        for i in range(1, len(boxes)):
            if boxes[i - 1] == '1':
                left_cnt += 1
            left_cost += left_cnt
            res[i] += left_cost
        right_cnt = 0
        right_cost = 0
        for i in range(len(boxes) - 2, -1, -1):
            if boxes[i + 1] == '1':
                right_cnt += 1
            right_cost += right_cnt
            res[i] += right_cost
        return res