New Problem Solution -"Minimum Number of Operations to Move All Balls to Each Box"

Author: haoel

README.md

@@ -22,6 +22,7 @@ LeetCode
 |1779|[Find Nearest Point That Has the Same X or Y Coordinate](https://leetcode.com/problems/find-nearest-point-that-has-the-same-x-or-y-coordinate/) | [C++](./algorithms/cpp/findNearestPointThatHasTheSameXOrYCoordinate/FindNearestPointThatHasTheSameXOrYCoordinate.cpp)|Easy|
 |1774|[Closest Dessert Cost](https://leetcode.com/problems/closest-dessert-cost/) | [C++](./algorithms/cpp/closestDessertCost/ClosestDessertCost.cpp)|Medium|
 |1773|[Count Items Matching a Rule](https://leetcode.com/problems/count-items-matching-a-rule/) | [C++](./algorithms/cpp/countItemsMatchingARule/CountItemsMatchingARule.cpp)|Easy|
+|1769|[Minimum Number of Operations to Move All Balls to Each Box](https://leetcode.com/problems/minimum-number-of-operations-to-move-all-balls-to-each-box/) | [C++](./algorithms/cpp/minimumNumberOfOperationsToMoveAllBallsToEachBox/MinimumNumberOfOperationsToMoveAllBallsToEachBox.cpp)|Medium|
 |1768|[Merge Strings Alternately](https://leetcode.com/problems/merge-strings-alternately/) | [C++](./algorithms/cpp/mergeStringsAlternately/MergeStringsAlternately.cpp)|Easy|
 |1763|[Longest Nice Substring](https://leetcode.com/problems/longest-nice-substring/) | [C++](./algorithms/cpp/longestNiceSubstring/LongestNiceSubstring.cpp)|Easy|
 |1761|[Minimum Degree of a Connected Trio in a Graph](https://leetcode.com/problems/minimum-degree-of-a-connected-trio-in-a-graph/) | [C++](./algorithms/cpp/minimumDegreeOfAConnectedTrioInAGraph/MinimumDegreeOfAConnectedTrioInAGraph.cpp)|Hard|

algorithms/cpp/minimumNumberOfOperationsToMoveAllBallsToEachBox/MinimumNumberOfOperationsToMoveAllBallsToEachBox.cpp

@@ -0,0 +1,78 @@
+// 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;
+        }
+    }
+};