C++ ] leetCode 1769 - 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'.

 

 

 

 

class Solution {
public:
    vector<int> minOperations(string boxes) {

        int len = boxes.size();
        vector<int> answer(len);
        vector<int> idxBallBox;

        for(int i=0; i< len; i++)
        {
            if(boxes[i] == '1') idxBallBox.push_back(i);
        }

        for(int i = 0; i< len; i++)
        {
            int sum = 0;
            for(int elem : idxBallBox)
            {                
                sum += abs(i-elem);
            }
            answer[i]=sum;
        }

        return answer;

    }
};

 

먼저 공이 들어있는 인덱스를 찾아 벡터에 넣어둔다.

그리고 박스의 시작부터 끝까지 반복하는데, 해당 인덱스와 공이 들어있는 박스 인덱스들의 절대값 차이를 누적하여 answer 벡터에 넣는다.