New Problem Solution - "1739. Building Boxes"

README.md

@@ -66,6 +66,7 @@ LeetCode
 |1748|[Sum of Unique Elements](https://leetcode.com/problems/sum-of-unique-elements/) | [C++](./algorithms/cpp/sumOfUniqueElements/SumOfUniqueElements.cpp)|Easy|
 |1743|[Restore the Array From Adjacent Pairs](https://leetcode.com/problems/restore-the-array-from-adjacent-pairs/) | [C++](./algorithms/cpp/restoreTheArrayFromAdjacentPairs/RestoreTheArrayFromAdjacentPairs.cpp)|Medium|
 |1742|[Maximum Number of Balls in a Box](https://leetcode.com/problems/maximum-number-of-balls-in-a-box/) | [C++](./algorithms/cpp/maximumNumberOfBallsInABox/MaximumNumberOfBallsInABox.cpp)|Easy|
+|1739|[Building Boxes](https://leetcode.com/problems/building-boxes/) | [C++](./algorithms/cpp/buildingBoxes/BuildingBoxes.cpp)|Hard|
 |1738|[Find Kth Largest XOR Coordinate Value](https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/) | [C++](./algorithms/cpp/findKthLargestXorCoordinateValue/FindKthLargestXorCoordinateValue.cpp)|Medium|
 |1736|[Latest Time by Replacing Hidden Digits](https://leetcode.com/problems/latest-time-by-replacing-hidden-digits/) | [C++](./algorithms/cpp/latestTimeByReplacingHiddenDigits/LatestTimeByReplacingHiddenDigits.cpp)|Easy|
 |1734|[Decode XORed Permutation](https://leetcode.com/problems/decode-xored-permutation/) | [C++](./algorithms/cpp/decodeXORedPermutation/DecodeXoredPermutation.cpp)|Medium|

algorithms/cpp/buildingBoxes/BuildingBoxes.cpp

@@ -0,0 +1,121 @@
+// Source : https://leetcode.com/problems/building-boxes/
+// Author : Hao Chen
+// Date   : 2021-04-09
+
+/***************************************************************************************************** 
+ *
+ * You have a cubic storeroom where the width, length, and height of the room are all equal to n 
+ * units. You are asked to place n boxes in this room where each box is a cube of unit side length. 
+ * There are however some rules to placing the boxes:
+ * 
+ * 	You can place the boxes anywhere on the floor.
+ * 	If box x is placed on top of the box y, then each side of the four vertical sides of the 
+ * box y must either be adjacent to another box or to a wall.
+ * 
+ * Given an integer n, return the minimum possible number of boxes touching the floor.
+ * 
+ * Example 1:
+ * 
+ * Input: n = 3
+ * Output: 3
+ * Explanation: The figure above is for the placement of the three boxes.
+ * These boxes are placed in the corner of the room, where the corner is on the left side.
+ * 
+ * Example 2:
+ * 
+ * Input: n = 4
+ * Output: 3
+ * Explanation: The figure above is for the placement of the four boxes.
+ * These boxes are placed in the corner of the room, where the corner is on the left side.
+ * 
+ * Example 3:
+ * 
+ * Input: n = 10
+ * Output: 6
+ * Explanation: The figure above is for the placement of the ten boxes.
+ * These boxes are placed in the corner of the room, where the corner is on the back side.
+ * 
+ * Constraints:
+ * 
+ * 	1 <= n <= 10^9
+ ******************************************************************************************************/
+
+/*
+
+    At first, let's  build the perfect pyramid at the corner.
+
+    we can find the following sequence:
+
+        height    cubes
+        1         1
+        2         1 + 2 = 3
+        3         1 + 2 + 3 = 6
+        4         1 + 2 + 3 + 4 = 10
+        5         1 + 2 + 3 + 4 + 5 = 15
+
+    total(height) = total(height - 1) + sum( from 1 to height )
+    
+    sum ( from 1 to height) = (height * (height+1)) / 2
+                            = height^2/2 + height/2
+    So,
+    
+    total(height) =  (1+2+...+height)/2 + ( 1^2 + 2^2 +...+ height^2 ) / 2
+    
+    we know,  Σn^2 = [n(n+1)(2n+1)]/6  (ref: https://brilliant.org/wiki/sum-of-n-n2-or-n3/)
+    
+    So, 
+    
+    total(height) = (height * (height+1)) / 4 + (height(height+1)(2height+1))/12
+                  = height * (height + 1) * (height + 2) / 6
+    
+
+    for the rest cubes, we can place them like this
+
+             (10) 
+            (6) (9) 
+          (3) (5) (8) 
+        (1) (2) (4) (7) 
+        
+    sum ( for 1 to n ) = n(n+1)/2
+    
+
+*/
+
+
+class Solution {
+private:
+    int total(long h){
+        return h * (h+1) * (h+2) / 6;
+    }
+public:
+    int minimumBoxes(int n) {
+        //find the maxiumn height which total(height) <= n
+        //binary search
+        int left = 1, right = pow(6l*n, 1.0/3) ;
+        while(left <= right){
+            int mid = left + (right - left) / 2;
+            int t = total(mid);
+            if ( t == n ) return mid*(mid+1l)/2;
+            if ( t < n) left = mid + 1;
+            else right = mid - 1;
+        }
+        int height = right;
+        int remind = n - total(height);
+        int bottom = height * (height+1l)/2 ;
+        //cout << "n=" << n << ", height=" << height << 
+        //    ", bottom = " << bottom << ", remind=" << remind << endl;
+
+        //find teh maxium h which sum(1..h) <= remind
+        //binary search
+        left = 1; right = sqrt(2*remind);
+        while ( left <= right) {
+            int mid = left + (right - left)/2;
+            int h = mid*(mid+1)/2;
+            if ( h == remind) return bottom + mid;
+            if ( h < remind) left = mid + 1;
+            else right = mid -1;
+        }
+        //cout << "left=" << left << ", right=" << right << endl;
+        return bottom + left;
+    }
+};