You are given a set of n types of rectangular 3-D boxes, where the i^th box has height h(i), width w(i) and depth d(i) (all real numbers). You want to create a stack of boxes which is as tall as possible, but you can only stack a box on top of another box if the dimensions of the 2-D base of the lower box are each strictly larger than those of the 2-D base of the higher box. Of course, you can rotate a box so that any side functions as its base. It is also allowable to use multiple instances of the same type of box.
Source: http://people.csail.mit.edu/bdean/6.046/dp/. The link also has a video for an explanation of the solution.
The Box Stacking problem is a variation of LIS problem. We need to build a maximum height stack.
Following are the key points to note in the problem statement:
1) A box can be placed on top of another box only if both width and depth of the upper placed box are smaller than width and depth of the lower box respectively.
2) We can rotate boxes such that width is smaller than depth. For example, if there is a box with dimensions {1x2x3} where 1 is height, 2×3 is base, then there can be three possibilities, {1x2x3}, {2x1x3} and {3x1x2}
3) We can use multiple instances of boxes. What it means is, we can have two different rotations of a box as part of our maximum height stack.
Following is the solution based on DP solution of LIS problem.
Method 1 : dynamic programming using tabulation
1) Generate all 3 rotations of all boxes. The size of rotation array becomes 3 times the size of the original array. For simplicity, we consider width as always smaller than or equal to depth.
2) Sort the above generated 3n boxes in decreasing order of base area.
3) After sorting the boxes, the problem is same as LIS with following optimal substructure property.
MSH(i) = Maximum possible Stack Height with box i at top of stack
MSH(i) = { Max ( MSH(j) ) + height(i) } where j < i and width(j) > width(i) and depth(j) > depth(i).
If there is no such j then MSH(i) = height(i)
4) To get overall maximum height, we return max(MSH(i)) where 0 < i < n
Following is the implementation of the above solution.
C++
/* Dynamic Programming implementation of Box Stacking problem */#include<stdio.h>#include<stdlib.h>/* Representation of a box */struct Box{ // h --> height, w --> width, d --> depth int h, w, d; // for simplicity of solution, always keep w <= d};// A utility function to get minimum of two integersint min (int x, int y){ return (x < y)? x : y; }// A utility function to get maximum of two integersint max (int x, int y){ return (x > y)? x : y; }/* Following function is needed for library function qsort(). We use qsort() to sort boxes in decreasing order of base area. Refer following link for help of qsort() and compare()int compare (const void *a, const void * b){ return ( (*(Box *)b).d * (*(Box *)b).w ) - ( (*(Box *)a).d * (*(Box *)a).w );}/* Returns the height of the tallest stack that can be formed with give type of boxes */int maxStackHeight( Box arr[], int n ){ /* Create an array of all rotations of given boxes For example, for a box {1, 2, 3}, we consider three instances{{1, 2, 3}, {2, 1, 3}, {3, 1, 2}} */ Box rot[3*n]; int index = 0; for (int i = 0; i < n; i++) { // Copy the original box rot[index].h = arr[i].h; rot[index].d = max(arr[i].d, arr[i].w); rot[index].w = min(arr[i].d, arr[i].w); index++; // First rotation of box rot[index].h = arr[i].w; rot[index].d = max(arr[i].h, arr[i].d); rot[index].w = min(arr[i].h, arr[i].d); index++; // Second rotation of box rot[index].h = arr[i].d; rot[index].d = max(arr[i].h, arr[i].w); rot[index].w = min(arr[i].h, arr[i].w); index++; } // Now the number of boxes is 3n n = 3*n; /* Sort the array 'rot[]' in non-increasing order of base area */ qsort (rot, n, sizeof(rot[0]), compare); // Uncomment following two lines to print all rotations // for (int i = 0; i < n; i++ ) // printf("%d x %d x %d\n", rot[i].h, rot[i].w, rot[i].d); /* Initialize msh values for all indexes msh[i] --> Maximum possible Stack Height with box i on top */ int msh[n]; for (int i = 0; i < n; i++ ) msh[i] = rot[i].h; /* Compute optimized msh values in bottom up manner */ for (int i = 1; i < n; i++ ) for (int j = 0; j < i; j++ ) if ( rot[i].w < rot[j].w && rot[i].d < rot[j].d && msh[i] < msh[j] + rot[i].h ) { msh[i] = msh[j] + rot[i].h; } /* Pick maximum of all msh values */ int max = -1; for ( int i = 0; i < n; i++ ) if ( max < msh[i] ) max = msh[i]; return max;}/* Driver program to test above function */int main(){ Box arr[] = { {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32} }; int n = sizeof(arr)/sizeof(arr[0]); printf("The maximum possible height of stack is %d\n", maxStackHeight (arr, n) ); return 0;} |
Java
/* Dynamic Programming implementation of Box Stacking problem in Java*/import java.util.*;public class GFG { /* Representation of a box */ static class Box implements Comparable<Box>{ // h --> height, w --> width, // d --> depth int h, w, d, area; // for simplicity of solution, // always keep w <= d /*Constructor to initialise object*/ public Box(int h, int w, int d) { this.h = h; this.w = w; this.d = d; } /*To sort the box array on the basis of area in decreasing order of area */ @Override public int compareTo(Box o) { return o.area-this.area; } } /* Returns the height of the tallest stack that can be formed with give type of boxes */ static int maxStackHeight( Box arr[], int n){ Box[] rot = new Box[n*3]; /* New Array of boxes is created - considering all 3 possible rotations, with width always greater than equal to width */ for(int i = 0;i < n;i++){ Box box = arr[i]; /* Original Box*/ rot[3*i] = new Box(box.h, Math.max(box.w,box.d), Math.min(box.w,box.d)); /* First rotation of box*/ rot[3*i + 1] = new Box(box.w, Math.max(box.h,box.d), Math.min(box.h,box.d)); /* Second rotation of box*/ rot[3*i + 2] = new Box(box.d, Math.max(box.w,box.h), Math.min(box.w,box.h)); } /* Calculating base area of each of the boxes.*/ for(int i = 0; i < rot.length; i++) rot[i].area = rot[i].w * rot[i].d; /* Sorting the Boxes on the basis of Area in non Increasing order.*/ Arrays.sort(rot); int count = 3 * n; /* Initialize msh values for all indexes msh[i] --> Maximum possible Stack Height with box i on top */ int[]msh = new int[count]; for (int i = 0; i < count; i++ ) msh[i] = rot[i].h; /* Computing optimized msh[] values in bottom up manner */ for(int i = 0; i < count; i++){ msh[i] = 0; Box box = rot[i]; int val = 0; for(int j = 0; j < i; j++){ Box prevBox = rot[j]; if(box.w < prevBox.w && box.d < prevBox.d){ val = Math.max(val, msh[j]); } } msh[i] = val + box.h; } int max = -1; /* Pick maximum of all msh values */ for(int i = 0; i < count; i++){ max = Math.max(max, msh[i]); } return max; } /* Driver program to test above function */ public static void main(String[] args) { Box[] arr = new Box[4]; arr[0] = new Box(4, 6, 7); arr[1] = new Box(1, 2, 3); arr[2] = new Box(4, 5, 6); arr[3] = new Box(10, 12, 32); System.out.println("The maximum possible "+ "height of stack is " + maxStackHeight(arr,4)); }}// This code is contributed by Divyam |
Python3
# Dynamic Programming implementation# of Box Stacking problemclass Box: # Representation of a box def __init__(self, h, w, d): self.h = h self.w = w self.d = d def __lt__(self, other): return self.d * self.w < other.d * other.wdef maxStackHeight(arr, n): # Create an array of all rotations of # given boxes. For example, for a box {1, 2, 3}, # we consider three instances{{1, 2, 3}, # {2, 1, 3}, {3, 1, 2}} rot = [Box(0, 0, 0) for _ in range(3 * n)] index = 0 for i in range(n): # Copy the original box rot[index].h = arr[i].h rot[index].d = max(arr[i].d, arr[i].w) rot[index].w = min(arr[i].d, arr[i].w) index += 1 # First rotation of the box rot[index].h = arr[i].w rot[index].d = max(arr[i].h, arr[i].d) rot[index].w = min(arr[i].h, arr[i].d) index += 1 # Second rotation of the box rot[index].h = arr[i].d rot[index].d = max(arr[i].h, arr[i].w) rot[index].w = min(arr[i].h, arr[i].w) index += 1 # Now the number of boxes is 3n n *= 3 # Sort the array 'rot[]' in non-increasing # order of base area rot.sort(reverse = True) # Uncomment following two lines to print # all rotations # for i in range(n): # print(rot[i].h, 'x', rot[i].w, 'x', rot[i].d) # Initialize msh values for all indexes # msh[i] --> Maximum possible Stack Height # with box i on top msh = [0] * n for i in range(n): msh[i] = rot[i].h # Compute optimized msh values # in bottom up manner for i in range(1, n): for j in range(0, i): if (rot[i].w < rot[j].w and rot[i].d < rot[j].d): if msh[i] < msh[j] + rot[i].h: msh[i] = msh[j] + rot[i].h maxm = -1 for i in range(n): maxm = max(maxm, msh[i]) return maxm# Driver Codeif __name__ == "__main__": arr = [Box(4, 6, 7), Box(1, 2, 3), Box(4, 5, 6), Box(10, 12, 32)] n = len(arr) print("The maximum possible height of stack is", maxStackHeight(arr, n))# This code is contributed by vibhu4agarwal |
C#
using System;class Box{ public int h, w, d, area; public Box(int h, int w, int d) { this.h = h; this.w = w; this.d = d; } public bool IsSmallerThan(Box other) { return this.w * this.d < other.w * other.d; }}class GFG{ public static int MaxStackHeight(Box[] arr, int n) {// Create an array of all rotations of// given boxes. For example, for a box {1, 2, 3},// we consider three instances{{1, 2, 3},// {2, 1, 3}, {3, 1, 2}} Box[] rot = new Box[3 * n]; for (int i = 0; i < n; i++) { Box box = arr[i]; // Copy the original box rot[3 * i] = new Box(box.h, Math.Max(box.w, box.d), Math.Min(box.w, box.d)); // First rotation of the box rot[3 * i + 1] = new Box(box.w, Math.Max(box.h, box.d), Math.Min(box.h, box.d)); // Second rotation of the box rot[3 * i + 2] = new Box(box.d, Math.Max(box.w, box.h), Math.Min(box.h, box.w)); } // Calculating base area ofeach of the boxes for (int i = 0; i < 3*n; i++) rot[i].area = rot[i].w * rot[i].d; // Sort the array 'rot[]' in non-increasing // order of base area Array.Sort(rot, (a, b) => b.IsSmallerThan(a) ? -1 : 1); int count = 3 * n; // Initialize msh values for all indexes // msh[i] --> Maximum possible Stack Height // with box i on top int[] msh = new int[count]; for (int i = 0; i < count; i++) msh[i] = rot[i].h; // Compute optimized msh values // in bottom up manner for (int i = 0; i < count; i++) { msh[i] = 0; Box box = rot[i]; int val = 0; for (int j = 0; j < i; j++) { Box prevBox = rot[j]; if (box.w < prevBox.w && box.d < prevBox.d) { val = Math.Max(val, msh[j]); } } msh[i] = val + box.h; } int max = -1; for (int i = 0; i < count; i++) { max = Math.Max(max, msh[i]); } return max; } public static void Main() { Box[] arr = new Box[4]; arr[0] = new Box(4, 6, 7); arr[1] = new Box(1, 2, 3); arr[2] = new Box(4, 5, 6); arr[3] = new Box(10, 12, 32); Console.WriteLine("The maximum possible " + "height of stack is " + MaxStackHeight(arr, 4)); }} |
Javascript
// Definition of a Boxclass Box { constructor(h, w, d) { this.h = h; // height this.w = w; // width this.d = d; // depth }}// Utility function to get the minimum of two integersfunction min(a, b) { return a < b ? a : b;}// Utility function to get the maximum of two integersfunction max(a, b) { return a > b ? a : b;}// Function to compare two boxes based on their base areafunction compare(box1, box2) { return box2.w * box2.d - box1.w * box1.d;}// Function to find the maximum height of a stack of boxesfunction maxStackHeight(boxes) { // Create an array of all rotations of the boxes const rotations = []; for (let i = 0; i < boxes.length; i++) { const box = boxes[i]; // Original orientation rotations.push(new Box(box.h, max(box.w, box.d), min(box.w, box.d))); // First rotation rotations.push(new Box(box.w, max(box.h, box.d), min(box.h, box.d))); // Second rotation rotations.push(new Box(box.d, max(box.h, box.w), min(box.h, box.w))); } // Sort the rotations in non-increasing order of their base area rotations.sort(compare); // Initialize an array to store the maximum stack height for each box const maxStackHeight = new Array(rotations.length).fill(0); for (let i = 0; i < rotations.length; i++) { maxStackHeight[i] = rotations[i].h; } // Compute the maximum stack height for each box in bottom-up manner for (let i = 1; i < rotations.length; i++) { for (let j = 0; j < i; j++) { if (rotations[i].w < rotations[j].w && rotations[i].d < rotations[j].d) { maxStackHeight[i] = max(maxStackHeight[i], maxStackHeight[j] + rotations[i].h); } } } // Find the maximum stack height let maxHeight = 0; for (let i = 0; i < rotations.length; i++) { if (maxHeight < maxStackHeight[i]) { maxHeight = maxStackHeight[i]; } } return maxHeight;}// Example usageconst boxes = [ new Box(4, 6, 7), new Box(1, 2, 3), new Box(4, 5, 6), new Box(10, 12, 32),];const ans = maxStackHeight(boxes);console.log(ans); // Output: 60 |
Output
The maximum possible height of stack is 60
In the above program, given input boxes are {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32}. Following are all rotations of the boxes in decreasing order of base area.
10 x 12 x 32 12 x 10 x 32 32 x 10 x 12 4 x 6 x 7 4 x 5 x 6 6 x 4 x 7 5 x 4 x 6 7 x 4 x 6 6 x 4 x 5 1 x 2 x 3 2 x 1 x 3 3 x 1 x 2
The height 60 is obtained by boxes { {3, 1, 2}, {1, 2, 3}, {6, 4, 5}, {4, 5, 6}, {4, 6, 7}, {32, 10, 12}, {10, 12, 32}}
Time Complexity: O(n^2)
Auxiliary Space: O(n), since n extra space has been taken.
Method 2 : dynamic programming using memoization (top-down)
C++
/* Dynamic Programming top-down implementation of Box * Stacking problem */#include <bits/stdc++.h>using namespace std;/* Representation of a box */class Box {public: int length; int width; int height;};// dp arrayint dp[303];/* boxes -> vector of Box bottom_box_index -> index of the bottom box index -> index of current box*//* NOTE: we can use only one variable in place of bottom_box_index and index but it has been avoided to make it simple */int findMaxHeight(vector<Box>& boxes, int bottom_box_index, int index){ // base case if (index < 0) return 0; if (dp[index] != -1) return dp[index]; int maximumHeight = 0; // recurse for (int i = index; i >= 0; i--) { // if there is no bottom box if (bottom_box_index == -1 // or if length & width of new box is < that of // bottom box || (boxes[i].length < boxes[bottom_box_index].length && boxes[i].width < boxes[bottom_box_index].width)) maximumHeight = max(maximumHeight, findMaxHeight(boxes, i, i - 1) + boxes[i].height); } return dp[index] = maximumHeight;}/* wrapper function for recursive calls whichReturns the height of the tallest stack that can beformed with give type of boxes */int maxStackHeight(int height[], int width[], int length[], int types){ // creating a vector of type Box class vector<Box> boxes; // Initialize dp array with -1 memset(dp, -1, sizeof(dp)); Box box; /* Create an array of all rotations of given boxes For example, for a box {1, 2, 3}, we consider three instances{{1, 2, 3}, {2, 1, 3}, {3, 1, 2}} */ for (int i = 0; i < types; i++) { // copy original box box.height = height[i]; box.length = max(length[i], width[i]); box.width = min(length[i], width[i]); boxes.push_back(box); // First rotation of box box.height = width[i]; box.length = max(length[i], height[i]); box.width = min(length[i], height[i]); boxes.push_back(box); // Second rotation of box box.height = length[i]; box.length = max(width[i], height[i]); box.width = min(width[i], height[i]); boxes.push_back(box); } // sort by area in ascending order .. because we will be dealing with this vector in reverse sort(boxes.begin(), boxes.end(), [](Box b1, Box b2) { // if area of box1 < area of box2 return (b1.length * b1.width) < (b2.length * b2.width); }); // Uncomment following two lines to print all rotations //for (int i = boxes.size() - 1; i >= 0; i-- ) // printf("%d x %d x %d\n", boxes[i].length, boxes[i].width, boxes[i].height); return findMaxHeight(boxes, -1, boxes.size() - 1);}int main(){ // where length, width and height of a particular box // are at ith index of the following arrays int length[] = { 4, 1, 4, 10 }; int width[] = { 6, 2, 5, 12 }; int height[] = { 7, 3, 6, 32 }; int n = sizeof(length) / sizeof(length[0]); printf("The maximum possible height of stack is %d\n", maxStackHeight(height, length, width, n)); return 0;} |
Java
import java.util.Arrays;public class BoxStacking { static class Box implements Comparable<Box> { int length; int width; int height; public Box(int length, int width, int height) { this.length = length; this.width = width; this.height = height; } public int compareTo(Box b) { return Integer.compare(this.length * this.width, b.length * b.width); } } public static int maxStackHeight(int[] length, int[] width, int[] height) { int n = length.length; Box[] boxes = new Box[3 * n]; int k = 0; for (int i = 0; i < n; i++) { boxes[k++] = new Box(length[i], width[i], height[i]); boxes[k++] = new Box(width[i], height[i], length[i]); boxes[k++] = new Box(height[i], length[i], width[i]); } Arrays.sort(boxes); int[] dp = new int[3 * n]; int maxHeight = 0; for (int i = k - 1; i >= 0; i--) { dp[i] = boxes[i].height; for (int j = i + 1; j < k; j++) { if (boxes[i].length < boxes[j].length && boxes[i].width < boxes[j].width) { dp[i] = Math.max(dp[i], boxes[i].height + dp[j]); } } maxHeight = Math.max(maxHeight, dp[i]); } return maxHeight; } public static void main(String[] args) { int[] length = { 4, 1, 4, 10 }; int[] width = { 6, 2, 5, 12 }; int[] height = { 7, 3, 6, 32 }; System.out.println("The maximum possible height of stack is " + maxStackHeight(length, width, height)); }}// This code contributed SRJ |
Python3
# Dynamic Programming top-down implementation of Box# Stacking problemclass Box: def __init__(self, length, width, height): self.length = length self.width = width self.height = height# dp arraydp = [-1]*303def findMaxHeight(boxes, bottom_box_index, index): # base case if index < 0: return 0 if dp[index] != -1: return dp[index] maximumHeight = 0 # recurse for i in range(index, -1, -1): # if there is no bottom box if bottom_box_index == -1 or (boxes[i].length < boxes[bottom_box_index].length and boxes[i].width < boxes[bottom_box_index].width): maximumHeight = max(maximumHeight, findMaxHeight(boxes, i, i-1) + boxes[i].height) dp[index] = maximumHeight return maximumHeight# Returns the height of the tallest stack that can be# formed with give type of boxes def maxStackHeight(height, width, length, types): # creating a vector of type Box class boxes = [] # Initialize dp array with -1 dp[:] = [-1]*303 # Create an array of all rotations of given boxes # For example, for a box {1, 2, 3}, we consider three # instances{{1, 2, 3}, {2, 1, 3}, {3, 1, 2}} for i in range(types): # copy original box box = Box(length[i], max(length[i], width[i]), min(length[i], width[i])) boxes.append(box) # First rotation of box box = Box(width[i], max(length[i], height[i]), min(length[i], height[i])) boxes.append(box) # Second rotation of box box = Box(length[i], max(width[i], height[i]), min(width[i], height[i])) boxes.append(box) # sort by area in ascending order .. because we will be dealing with this vector in reverse boxes.sort(key=lambda box : (box.length * box.width)) # Uncomment following two lines to print all rotations #for (int i = boxes.size() - 1; i >= 0; i-- ) # printf("%d x %d x %d\n", boxes[i].length, boxes[i].width, boxes[i].height); return findMaxHeight(boxes, -1, len(boxes)-1)# where length, width and height of a particular box# are at ith index of the following arrayslength = [ 4, 1, 4, 10 ]width = [ 6, 2, 5, 12 ]height = [ 7, 3, 6, 32 ]types = len(length)print("The maximum possible height of stack is", maxStackHeight(height, length, width, types))# This code is contributed b factworx412 |
Javascript
class Box { constructor(length, width, height) { this.length = length; this.width = width; this.height = height; } } function maxStackHeight(ength, width, height) { let n = length.length; let boxes = new Array(); let k = 0; for (let i = 0; i < n; i++) { boxes[k++] = new Box(length[i], width[i], height[i]); boxes[k++] = new Box(width[i], height[i], length[i]); boxes[k++] = new Box(height[i], length[i], width[i]); } boxes.sort((b1, b2) => { // if area of box1 < area of box2 return b1.length * b1.width - b2.length * b2.width; }); let dp = new Array(3 * n); let maxHeight = 0; for (let i = k - 1; i >= 0; i--) { dp[i] = boxes[i].height; for (let j = i + 1; j < k; j++) { if ( boxes[i].length < boxes[j].length && boxes[i].width < boxes[j].width ) { dp[i] = Math.max(dp[i], boxes[i].height + dp[j]); } } maxHeight = Math.max(maxHeight, dp[i]); } return maxHeight; } let length = [4, 1, 4, 10]; let width = [6, 2, 5, 12]; let height = [7, 3, 6, 32]; console.log( "The maximum possible height of stack is " + maxStackHeight(length, width, height) ); |
C#
using System;using System.Linq;public class BoxStacking { public class Box : IComparable<Box> { public int length; public int width; public int height; public Box(int length, int width, int height) { this.length = length; this.width = width; this.height = height; } public int CompareTo(Box b) { return (this.length * this.width).CompareTo(b.length * b.width); } } public static int maxStackHeight(int[] length, int[] width, int[] height) { int n = length.Length; Box[] boxes = new Box[3 * n]; int k = 0; for (int i = 0; i < n; i++) { boxes[k++] = new Box(length[i], width[i], height[i]); boxes[k++] = new Box(width[i], height[i], length[i]); boxes[k++] = new Box(height[i], length[i], width[i]); } Array.Sort(boxes); int[] dp = new int[3 * n]; int maxHeight = 0; for (int i = k - 1; i >= 0; i--) { dp[i] = boxes[i].height; for (int j = i + 1; j < k; j++) { if (boxes[i].length < boxes[j].length && boxes[i].width < boxes[j].width) { dp[i] = Math.Max(dp[i], boxes[i].height + dp[j]); } } maxHeight = Math.Max(maxHeight, dp[i]); } return maxHeight; } public static void Main(string[] args) { int[] length = { 4, 1, 4, 10 }; int[] width = { 6, 2, 5, 12 }; int[] height = { 7, 3, 6, 32 }; Console.WriteLine("The maximum possible height of stack is " + maxStackHeight(length, width, height)); }} |
Output
The maximum possible height of stack is 60
Time Complexity: O(n^2)
Auxiliary Space: O(n)
In the above program, for boxes of dimensions of {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32} on giving the input as {4, 1, 4, 10} for length, {6, 2, 5, 12} for width and {7, 3, 6, 32} for height. Following rotations are possible for the boxes in decreasing order of base area.
32 x 12 x 10 <- 32 x 10 x 12 12 x 10 x 32 <- 7 x 6 x 4 <- 6 x 5 x 4 <- 7 x 4 x 6 6 x 4 x 5 6 x 4 x 7 5 x 4 x 6 <- 3 x 2 x 1 <- 3 x 1 x 2 2 x 1 x 3 <- The maximum possible height of stack is 60
The height 60 is obtained by boxes { {2, 1, 3}, {3, 2, 1}, {5, 4, 6}, {6, 5, 4}, {7, 6, 4}, {12, 10, 32}, {32, 12, 10}}
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