// 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 minOperations(string boxes) { vector result(boxes.size()); //minOperations01(boxes, result); //128ms minOperations02(boxes, result); //4s return result; } void minOperations01(string& boxes, vector& result ) { vector balls; for(int i=0; i& 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; } } };