// Source : https://leetcode.com/problems/minimum-number-of-operations-to-move-all-balls-to-each-box/
// Author : Hao Chen
// Date   : 2021-03-20

/***************************************************************************************************** 
 *
 * 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) {
        vector<int> result(boxes.size());
        //minOperations01(boxes, result); //128ms
        minOperations02(boxes, result); //4s
        return result;
    }
    
    void minOperations01(string& boxes, vector<int>& result ) {
        vector<int> balls;
        for(int i=0; i<boxes.size(); i++) {
            if(boxes[i] == '1') balls.push_back(i);
        }
        
        for (int i=0; i<boxes.size(); i++) {
            int steps = 0;
            for (int j=0; j<balls.size(); j++) {
                steps += abs(balls[j] - i);
            }
            result[i] = steps;
        }
    }
    void minOperations02(string& boxes, vector<int>& result ) {
        //from left to right
        for(int i=0, ops=0, balls=0; i< boxes.size(); i++) {
            result[i] += ops;
            balls += (boxes[i] == '1' ? 1 : 0);
            ops += balls;
        }
        //from right to left
        for(int i=boxes.size()-1, ops=0, balls=0; i>=0; i--) {
            result[i] += ops;
            balls += (boxes[i] == '1' ? 1 : 0);
            ops += balls;
        }
    }
};