Stacking Boxes is problem 103 in the UVa Online Judge. Even though I include the problem description in this post, I encourage you to visit the UVa Online Judge because there you will be able to submit your solution to get it judged.

I substituted the verb nest with verb fits in the problem statement.

Problem

Consider an n-dimensional “box” given by its dimensions. In two dimensions the box (2,3) might represent a box with length 2 units and width 3 units. In three dimensions the box (4,8,9) can represent a box 4 x 8 x 9 (length, width, and height). In 6 dimensions it is, perhaps, unclear what the box (4,5,6,7,8,9) represents; but we can analyze properties of the box such as the sum of its dimensions.

In this problem you will analyze a property of a group of n-dimensional boxes. You are to determine the longest stack of boxes, that is a sequence of boxes b_1, b_2, ..., b_k such that each box b_i fits in box b_i+1 (1 <= i < k).

A box D = (d_1, d_2, ..., d_n) fits in a box E = (e_1, e_2, ..., e_n) if there is some rearrangement of the d_i such that when rearranged each dimension is less than the corresponding dimension in box E. This loosely corresponds to turning box D to see if it will fit in box E. However, since any rearrangement suffices, box D can be contorted, not just turned (see examples below).

For example, the box D = (2,6) fits in the box E = (7,3) since D can be rearranged as (6,2) so that each dimension is less than the corresponding dimension in E. The box D = (9,5,7,3) does NOT fit in the box E = (2,10,6,8) since no rearrangement of D results in a box that satisfies the fitting property, but F = (9,5,7,1) does fit in box E since F can be rearranged as (1,9,5,7) which fits in E.

Formally, we define that a box fits in another as follows: box D = (d_1, d_2, ..., d_n) fits in box E = (e_1, e_2, ..., e_n) if there is a permutation p of 1 ... n such that (d_p[1], d_p[2], ..., d_p[n]) “fits” in (e_1, e_2, ..., e_n) i.e., if d_p[i] < e_i for all 1 <= i < n.

Input

The input consists of a series of box sequences. Each box sequence begins with a line consisting of the the number of boxes k in the sequence followed by the dimensionality of the boxes, n (on the same line.)

This line is followed by k lines, one line per box with the n measurements of each box on one line separated by one or more spaces. The i-th line in the sequence 1 <= i <= k gives the measurements for the i-th box.

There may be several box sequences in the input file. Your program should process all of them and determine, for each sequence, which of the k boxes determine the longest stack and the length of that stack (the number of boxes in the stack).

In this problem the maximum dimensionality is 10 and the minimum dimensionality is 1. The maximum number of boxes in a sequence is 30.

Output

For each box sequence in the input file, output the length of the longest stack on one line followed on the next line by a list of the boxes that comprise this stack in order. The “smallest” or “innermost” box of the stack should be listed first, the next box (if there is one) should be listed second, etc.

The boxes should be numbered according to the order in which they appeared in the input file (first box is box 1, etc.).

If there is more than one longest stack then any one of them can be output.

Sample Input

5 2
3 7
8 10
5 2
9 11
21 18
8 6
5 2 20 1 30 10
23 15 7 9 11 3
40 50 34 24 14 4
9 10 11 12 13 14
31 4 18 8 27 17
44 32 13 19 41 19
1 2 3 4 5 6
80 37 47 18 21 9

Sample Output

5
3 1 2 4 5
4
7 2 5 6

Approach

We approach the problem as a search problem. We search for an answer in a set of candidates.

An answer is any longest stack. Consider the following sequence of boxes.

(1, 5) (2, 4) (3, 5) (3, 6)

The stacks for the sequence are the following.

  *             *             *             *             *             *
| | | | | |
(1, 5) (2, 4) (1, 5) (2, 4) (3, 5) (3, 6)
| / \
| / \
(3, 6) (3, 5) (3, 6)

Stacks (1, 5) (3, 6), (2, 4) (3, 5), and (2, 4) (3, 6) are the longest stacks. Each of them is an answer for the given sequence of boxes.

The set of candidates consists of one stack for each given box. Each candidate is a longest stack that ends with the corresponding box. For the given sequence, the candidates are the following.

  *             *             *             *
| | | |
(1, 5) (2, 4) (1, 5) (2, 4)
| /
| /
(3, 6) (3, 5)

We choose candidates in the following way. For each given box, we consider the set of stacks that end with the box and choose the first longest stack we find. For example, for box (3, 6) we consider the following stacks.

                              *                 *                       *
| | |
(1, 5) (2, 4) (3, 6)
| \
| \
(3, 6) (3, 6)

Stacks (1, 5) (3, 6) and (2, 4) (3, 6) are longest. We choose the (1, 5) (3, 6) as the candidate for (3, 6).

The set of candidates contains at least one answer. The reason is that each answer is also a longest stack that ends with some box. For example, answers (1, 5) (3, 6) and (2, 4) (3, 6) are the longest stacks that end with (3, 6). By picking any of them as the candidate for box (3, 6), we include an answer in the set of candidates.

We answer with the first longest candidate we find.

Solution

We solve the problem in three steps. We preprocess the input sequence, construct the set of candidates, and search for a longest candidate.

Preprocessing consists in arranging the sequence of boxes so that for each box, the boxes that fit are located to the left of the box. We preprocess by sorting the dimensions of each box and then sorting the boxes lexicographically. Consider the following sequence of boxes.

(5, 3) (1, 5) (3, 6) (4, 2)

After sorting the dimensions, the sequence is the following.

(3, 5) (1, 5) (3, 6) (2, 4)

The preprocessed sequence is the following.

(1, 5) (2, 4) (3, 5) (3, 6)

The preprocessed sequence is arranged. For example, only box (2, 4) fits (3, 5) and box (2, 4) is to the left of (3, 5). Even though box (1, 5) does not fit (3, 5), we do not care that box (1, 5) is to the left of (3, 5).

We sort the dimensions of boxes because one sorted box fits into another sorted box when corresponding dimensions fit. For example, box (2, 4) fits (3, 6) because 2 < 3 and 4 < 6. Another example, box (3, 6) does not fit (2, 4) because 3 does not fit 2. Given that the dimensions are sorted, we already know that the other dimension of the first box does not fit 2. In general, when we consider boxes a := (a1, a2, ..., an) and b := (b1, b2, ..., bn), a fits b if and only if for each 1 <= i <= n, ai < bi. There are two cases, either the condition is satisfied or not. In the case where the condition is satisfied, we satisfy the definition of fits. The following diagram illustrates the other case, the case where for some i, ai is not less than bi.

a1 <= a2 <= ... <= ai-1 <= ai <= ai+1 <= ... <= an

v


b1 <= b2 <= ... <= bi-1 <= bi <= bi+1 <= ... <= bn

In this case there is no way another dimension of a fits bi while the rest of the dimensions of a fit in other dimensions of b. Consider that the only dimensions a1, ..., ai-1 may fit bi. Suppose that aj fits bi as follows.

a1 <= a2 <= ... <= aj-1 <= aj <= aj+1 <= ... <= ai-1 <= ai <= ai+1 <= ... <= an

ʌ v
|__________________________ ‖
\
b1 <= b2 <= ... <= bj-1 <= bj <= bj+1 <= ... <= bi-1 <= bi <= bi+1 <= ... <= bn

Consider that ai, ..., an may only fit bi+1, ..., bn. It is impossible to fit each dimension ai, ..., an into bi+1, ..., bn because number of dimensions is not the same.

Sorting the boxes lexicographically after sorting the dimensions arranges the sequence of boxes. The reason is that for each box, all boxes to the right do not fit the box. Consider box (2, 4) in the following sorted sequence.

(1, 5) (2, 4) (3, 5) (3, 6)

Box (2, 4) comes before (3, 5) because 2 < 3, which is the same reason why (3, 5) does not fit (2, 4). In general, when we consider boxes a := (a1, a2, ..., an) and b := (b1, b2, ..., bn), b does not fit a if a comes before b. If a comes before b, either all dimensions of a and b are the same or there is a dimension ai that is less or equal to dimension bi, as illustrated in the following diagram.

a1 <= a2 <= ... <= ai-1 <= ai <= ai+1 <= ... <= an

ʌ
‖ ‖ ‖ ‖

b1 <= b2 <= ... <= bi-1 <= bi <= bi+1 <= ... <= bn

In either case, the condition that each dimension of b fits the corresponding dimension of a is not satisfied and thus b does not fit a.

We construct the set of candidates by iterating the preprocessed sequence applying the following two rules to each box. If no candidate to the left of the box fits the box, then the candidate for the box is the box itself. Consider the following table that indicates the candidates for (1, 5) and (2, 4).

      Box | (1, 5) | (2, 4) | (3, 5) | (3, 6)
----------|--------|--------|--------|--------
| (1, 5) | (2, 4) | |
Candidate | | | |
| | | |

For box (2, 4), candidate (1, 5) does not fit, thus stack (2, 4) is the candidate for box (2, 4). If at least one candidate to the left fits the box, we stack the box on all candidates that fit and we choose the first longest stack we get. Consider the following table that indicates the candidates for all boxes.

      Box | (1, 5) | (2, 4) | (3, 5) | (3, 6)
----------|--------|--------|--------|--------
| (1, 5) | (2, 4) | (2, 4) | (1, 5)
Candidate | | | (3, 5) | (3, 6)
| | | |

For box (3, 6), candidates (1, 5) and (2, 4) fit the box. Each candidate corresponds to a longest stack that ends with (3, 6), stacks (1, 5), (3, 6) and (2, 4) (3, 6) respectively. We choose (1, 5) (3, 6) as the longest candidate for (3, 6) because we consider candidate (1, 5) before candidate (2, 4).

The construction of the set of candidates belongs to complexity class O(n^2). Our construction corresponds to the O(n^2) algorithm for the Longest Increasing Subsequence Problem.

We return the first longest candidate we find from left to right. Consider the selected candidate from the set of candidates.

      Box | (1, 5) | (2, 4) | (3, 5) | (3, 6)
----------|--------|--------|--------|--------
| (1, 5) | (2, 4) | (2, 4) | (1, 5)
Candidate | | | (3, 5) | (3, 6)
| | | |
^
|
selected candidate

Implementation

We implement our solution in C. We explain our implementation of the preprocessing, construction of the set of candidates, and search for a longest candidate. After that, we show the full code for our implementation.

We apply function merge_sort_ints(int *a, int n) to sorting the dimensions of a given box. Function merge_sort_ints sorts array a of n integers by merge sort.

void merge_sort_ints(int *a, int n) {
int i, j, k;
int m = n / 2;
int tmp[n];
if(n <= 1)
return;
merge_sort_ints(a, m);
merge_sort_ints(a + m, n - m);
for(k = i = 0, j = m; k < n; k++)
if(i < m && j < n)
if(a[i] <= a[j])
tmp[k] = a[i++];
else
tmp[k] = a[j++];
else if(i < m)
tmp[k] = a[i++];
else
tmp[k] = a[j++];
for(i = 0; i < n; i++)
a[i] = tmp[i];
}

Consider the following application of merge_sort_ints in line SORT_DIMS. That application sorts dimensions for each box after reading all the boxes for a sequence.

typedef struct box {
int d[MAX_DIM];
int original_position;
} box;

int main() {
int n, d, i, j;
box box[MAX_BOX];
...
while(Si(n) != EOF) {
/* Read input. */
Si(d);
for(i = 0; i < n; i++) {
box[i].original_position = i + 1;
for(j = 0; j < d; j++)
Si(box[i].d[j]);
}
...

/* For each box, sort dimensions. */
for(i = 0; i < n; i++)
merge_sort_ints(box[i].d, d); /* ..................... SORT_DIMS */

...

We apply function merge_sort_boxes(box *a, int n, int d) to sorting the sequence of boxes. Function merge_sort_boxes sorts array a of n boxes that have d dimensions by merge sort.

void merge_sort_boxes(box *a, int n, int d) {
int i, j, k;
int m = n / 2;
box tmp[n];
if(n <= 1)
return;
merge_sort_boxes(a, m, d);
merge_sort_boxes(a + m, n - m, d);
for(k = i = 0, j = m; k < n; k++)
if(i < m && j < n)
if(le_box(a[i], a[j], d))
copy_box(tmp + k , a + i++, d);
else
copy_box(tmp + k, a + j++, d);
else if(i < m)
copy_box(tmp + k , a + i++, d);
else
copy_box(tmp + k, a + j++, d);
for(i = 0; i < n; i++)
copy_box(a + i, tmp + i, d);
}

Consider the following application of merge_sort_boxes in line SORT_BOXES. That application sorts the sequence of boxes after sorting dimensions.

typedef struct box {
int d[MAX_DIM];
int original_position;
} box;

int main() {
int n, d, i, j;
box box[MAX_BOX];
...
while(Si(n) != EOF) {
/* Read input. */
Si(d);
for(i = 0; i < n; i++) {
box[i].original_position = i + 1;
for(j = 0; j < d; j++)
Si(box[i].d[j]);
}
...

/* For each box, sort dimensions. */
for(i = 0; i < n; i++)
merge_sort_ints(box[i].d, d); /* ..................... SORT_DIMS */

/* Sort boxes lexicographically. */
merge_sort_boxes(box, n, d); /* ........................ SORT_BOXES */

...

We compute the candidate set with the following block of code in main after sorting the sequence.

#define NO_PARENT -1

typedef struct box {
int d[MAX_DIM];
int original_position;
} box;

...

int box_fits(box *this, box *into_this, int d) {
int i;
for(i = 0; i < d; i++)
if(this->d[i] >= into_this->d[i])
return 0;
return 1;
}

...

int main() {
int n, d, i, j;
box box[MAX_BOX];
int longest_here_parent[MAX_BOX];
int longest_here_length[MAX_BOX];
...
while(Si(n) != EOF) {

...

/* For each position, compute one longest increasing subsequence
that ends in the position. */

for(i = 0; i < n; i++) {
longest_here_parent[i] = NO_PARENT;
longest_here_length[i] = 1;
}
for(i = 0; i < n; i++)
for(j = 0; j < i; j++)
if(box_fits(box + j, box + i, d) &&
longest_here_length[i] < longest_here_length[j] + 1) {
longest_here_parent[i] = j;
longest_here_length[i] = longest_here_length[j] + 1;
}

...

We search for a longest candidate and we keep the first we find in the following code.

int main() {
int n, d, i, j;
box box[MAX_BOX];
int longest_here_parent[MAX_BOX];
int longest_here_length[MAX_BOX];
int longest_last;
int longest_length;
int longest[MAX_BOX];
while(Si(n) != EOF) {

...

/* Compute the longest increasing subsequence. */
longest_last = 0;
longest_length = 1;
for(i = 1; i < n; i++)
if(longest_length < longest_here_length[i]) {
longest_last = i;
longest_length = longest_here_length[i];
}
i = longest_last;
j = longest_length - 1;
for( ; longest_here_length[i] != 1; j--) {
longest[j] = i;
i = longest_here_parent[i];
}
longest[j] = i;

The full code of our implementation is the following.

#include <stdio.h>

#define MAX_DIM 10
#define MAX_BOX 30
#define NO_PARENT -1

#define Si(i) scanf("%d", &i)

typedef struct box {
int d[MAX_DIM];
int original_position;
} box;

void merge_sort_ints(int *a, int n) {
int i, j, k;
int m = n / 2;
int tmp[n];
if(n <= 1)
return;
merge_sort_ints(a, m);
merge_sort_ints(a + m, n - m);
for(k = i = 0, j = m; k < n; k++)
if(i < m && j < n)
if(a[i] <= a[j])
tmp[k] = a[i++];
else
tmp[k] = a[j++];
else if(i < m)
tmp[k] = a[i++];
else
tmp[k] = a[j++];
for(i = 0; i < n; i++)
a[i] = tmp[i];
}

int le_box(box a, box b, int d) {
int i;
for(i = 0; i < d; i++) {
if(a.d[i] < b.d[i])
return 1;
else if(a.d[i] > b.d[i])
return 0;
}
return 1;
}

void copy_box(box *to, box *from, int d) {
int i = 0;
to->original_position = from->original_position;
for(i = 0; i < d; i++)
to->d[i] = from->d[i];
}

void merge_sort_boxes(box *a, int n, int d) {
int i, j, k;
int m = n / 2;
box tmp[n];
if(n <= 1)
return;
merge_sort_boxes(a, m, d);
merge_sort_boxes(a + m, n - m, d);
for(k = i = 0, j = m; k < n; k++)
if(i < m && j < n)
if(le_box(a[i], a[j], d))
copy_box(tmp + k , a + i++, d);
else
copy_box(tmp + k, a + j++, d);
else if(i < m)
copy_box(tmp + k , a + i++, d);
else
copy_box(tmp + k, a + j++, d);
for(i = 0; i < n; i++)
copy_box(a + i, tmp + i, d);
}

int box_fits(box *this, box *into_this, int d) {
int i;
for(i = 0; i < d; i++)
if(this->d[i] >= into_this->d[i])
return 0;
return 1;
}

int main() {
int n, d, i, j;
box box[MAX_BOX];
int longest_here_parent[MAX_BOX];
int longest_here_length[MAX_BOX];
int longest_last;
int longest_length;
int longest[MAX_BOX];
while(Si(n) != EOF) {

/* Read input. */
Si(d);
for(i = 0; i < n; i++) {
box[i].original_position = i + 1;
for(j = 0; j < d; j++)
Si(box[i].d[j]);
}

/* For each box, sort dimensions. */
for(i = 0; i < n; i++)
merge_sort_ints(box[i].d, d); /* ..................... SORT_DIMS */

/* Sort boxes lexicographically. */
merge_sort_boxes(box, n, d); /* ........................ SORT_BOXES */

/* For each position, compute one longest increasing subsequence
that ends in the position. */

for(i = 0; i < n; i++) {
longest_here_parent[i] = NO_PARENT;
longest_here_length[i] = 1;
}
for(i = 0; i < n; i++)
for(j = 0; j < i; j++)
if(box_fits(box + j, box + i, d) &&
longest_here_length[i] < longest_here_length[j] + 1) {
longest_here_parent[i] = j;
longest_here_length[i] = longest_here_length[j] + 1;
}

/* Compute the longest increasing subsequence. */
longest_last = 0;
longest_length = 1;
for(i = 1; i < n; i++)
if(longest_length < longest_here_length[i]) {
longest_last = i;
longest_length = longest_here_length[i];
}
i = longest_last;
j = longest_length - 1;
for( ; longest_here_length[i] != 1; j--) {
longest[j] = i;
i = longest_here_parent[i];
}
longest[j] = i;

/* Output longest increasing subsequence. */
printf("%d\n", longest_length);
printf("%d", box[longest[0]].original_position);
for(i = 1; i < longest_length; i++)
printf(" %d", box[longest[i]].original_position);
printf("\n");
}
return 0;
}

Our implementation was accepted by the UVa Online Judge.

Summary

We approach Stacking Boxes as a search problem. We restrict the search space to a set of candidates. The construction of the set of candidates belongs to complexity class O(n^2) and is the most expensive part of our solution. Our solution to the problem consists of three steps.

  1. Sort dimensions for each box so that comparing any two boxes reduces to comparing their corresponding dimensions.
  2. Sort the sequence of boxes lexicographically so that for each box, the boxes to the left are the boxes that may fit.
  3. Construct the set of candidates in a similar way to the O(n^2) algorithm for the Longest Increasing Subsequence Problem.

A faster solution

I have explored the possibility of solving the problem in at most O(n log n) steps.

A quick look at Algorithmist shows that Stacking Boxes is a particular case of the Longest Increasing Subsequence Problem. There I found an O(n log n) algorithm that computes the longest increasing subsequence of an array of integers, which I tried to apply to the problem. The algorithm looked promising because it refines a single solution as it processes the sequence of integers one by one. The problem is that, unlike integers, boxes are not always comparable by means of the relation ‘fits’. For example, neither (1, 5) fits (3, 5) nor vice versa. In mathematicalese that means that the relation ‘fits’ is not a total relation.

Given that not all boxes are comparable, instead of refining only one solution, I could only think of an adaptation of the algorithm that refines a set of candidates. The refinement consists in searching for each box a stack where you can place it. The procedure for searching a stack for a box is similar to patiente sorting. The difference is that finding the stack for each box cannot be done by binary search applying relation ‘fits’. The worst case scenario is that all boxes in the input sequence are incomparable and for that reason my adaptation belongs to class O(n^2).

Do you have any idea how to solve Stacking Boxes faster? Let me know in the comments.

Want to read more?

I love to explain and answer questions on programming problems, the kind you find in coding interviews. I publish a new programming problem and its solution every Sunday. Did I mention that I love to answer questions?

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