# Box Stacking Problem

In this problem a set of different boxes are given, the length, breadth, and width may differ for different boxes. Our task is to find a stack of these boxes, whose height is as much as possible. We can rotate any box as we wish. But there is a rule to maintain.

One can place a box on another box if the area of the top surface for the bottom box is larger than the lower area of the top box.

## Input and Output

Input:
A list of boxes is given. Each box is denoted by (length, bredth, height).
{ (4, 6, 7), (1, 2, 3), (4, 5, 6), (10, 12, 32) }
Output:
The maximum possible height of box stack is: 60

## Algorithm

maxHeight(boxList, n)

Input − The list of different boxes, number of boxes.

Output − maximum height found by making the stacking of boxes.

Begin
define rotation array rot of size 3n.
index := 0

for all boxes i, in the boxList, do
rot[index].len := boxList[i].len
rot[index].hei := maximum of boxList[i].hei and boxList[i].bre
rot[index].bre := minimum of boxList[i].hei and boxList[i].bre
index := index + 1

rot[index].len := boxList[i].bre
rot[index].hei := maximum of boxList[i].len and boxList[i].hei
rot[index].bre := minimum of boxList[i].len and boxList[i].hei
index := index + 1

rot[index].len := boxList[i].hei
rot[index].hei := maximum of boxList[i].len and boxList[i].bre
rot[index].bre := minimum of boxList[i].len and boxList[i].bre
index := index + 1

n := 3n
sort the rot list
define maxHeightTemp array

for i := 1 to n-1, do
for j := 0 to i-1, do
if rot[i].bre < rot[j].bre AND
rot[i].hei < rot[j].hei AND
maxHeightTemp[i] < maxHeightTemp + rot[i].len, then
maxHeightTemp[i] := maxHeightTemp[j] +
rot[i].len
done
done

maxHeight := -1
for i := 0 to n-1, do
if maxHeight < maxHeightTemp[i], then
maxHeight := maxHeightTemp[i]
done
done
return maxHeight
End

## Example

#include<iostream>
#include<algorithm>
using namespace std;

struct Box {
int length, bredth, height;
};

int min (int x, int y) {
return (x < y)? x : y;
}

int max (int x, int y) {
return (x > y)? x : y;
}

bool compare(Box b1, Box b2) {
return b1.height > b2.height;    //to sort the box as descending order of height
}

int maxHeight( Box boxList[], int n ) {
Box rotation[3*n];    //a box can be rotared as 3 type, so there is 3n number of rotations
int index = 0;

for (int i = 0; i < n; i++) {
//store initial position of the box
rotation[index].length = boxList[i].length;
rotation[index].height = max(boxList[i].height, boxList[i].bredth);
rotation[index].bredth = min(boxList[i].height, boxList[i].bredth);
index++;

//dimention after first rotation
rotation[index].length = boxList[i].bredth;
rotation[index].height = max(boxList[i].length, boxList[i].height);
rotation[index].bredth = min(boxList[i].length, boxList[i].height);
index++;

//Dimention after second rotation
rotation[index].length = boxList[i].height;
rotation[index].height = max(boxList[i].length, boxList[i].bredth);
rotation[index].bredth = min(boxList[i].length, boxList[i].bredth);
index++;
}

n = 3*n;    //set n as 3n for 3 rotations of each boxes

sort(rotation, rotation+n, compare);    //sort rotation array as descending order

int maxHTemp[n];    //temporary max height if ith box is stacked

for (int i = 0; i < n; i++ )
maxHTemp[i] = rotation[i].length;

for (int i = 1; i < n; i++ )    //find optimized stack height
for (int j = 0; j < i; j++ )
if ( rotation[i].bredth < rotation[j].bredth && rotation[i].height < rotation[j].height
&& maxHTemp[i] < maxHTemp[j] + rotation[i].length) {
maxHTemp[i] = maxHTemp[j] + rotation[i].length;
}
int maxHeight = -1;
for ( int i = 0; i < n; i++ )    //find the maximum height from all temporary heights

if ( maxHeight < maxHTemp[i] )
maxHeight = maxHTemp[i];

return maxHeight;
}

int main() {
Box arr[] = { {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32} };
int n = 4;
cout<<"The maximum possible height of box stack is: " << maxHeight (arr, n) << endl;
}

## Output

The maximum possible height of box stack is: 60

Updated on: 16-Jun-2020

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